{"id":27952,"date":"2026-11-09T15:08:11","date_gmt":"2026-11-09T15:08:11","guid":{"rendered":"https:\/\/tradefoxapp.com\/?page_id=27952"},"modified":"2026-11-09T15:56:39","modified_gmt":"2026-11-09T15:56:39","slug":"electrical","status":"publish","type":"page","link":"https:\/\/tradefoxapp.com\/fr\/calculators\/electrical\/","title":{"rendered":"Electrical calculators"},"content":{"rendered":"\n<style>\n\/* Tradefox calculators \u2014 Apple-style, scoped to .tfc so it can sit inside any WordPress page *\/\n.tfc{--fg:#1d1d1f;--fg2:#6e6e73;--line:#d2d2d7;--soft:#f5f5f7;--brand:#F15927;--brand-ink:#C2410C;--radius:18px;\n font-family:-apple-system,BlinkMacSystemFont,\"SF Pro Text\",\"SF Pro Display\",\"Helvetica Neue\",Helvetica,Arial,sans-serif;\n color:var(--fg);-webkit-font-smoothing:antialiased;font-size:17px;line-height:1.47;letter-spacing:-.022em;text-align:left}\n.tfc *,.tfc *::before,.tfc *::after{box-sizing:border-box}\n.tfc a{color:var(--brand-ink);text-decoration:none}\n.tfc 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v=String(value).trim();if(!\/^[+-]?(?:\\d+(?:\\.\\d*)?|\\.\\d+)(?:e[+-]?\\d+)?$\/i.test(v))return null;const n=Number(v);return Number.isFinite(n)?n:null;}\nfunction isCorrect(value,q){const n=numericAnswer(value);if(n===null)return false;if(q.integer)return n===q.answer;return Math.abs(n-q.answer)<=Math.max(Math.abs(q.answer)*0.005,0.00000001);}\nif(typeof module!=='undefined')module.exports={calculate,numericAnswer,isCorrect};\n\n\n\/* ---- original engine.js (verbatim) ---- *\/\n'use strict';\nconst fs=typeof require==='function'?require('fs'):null;\nfunction expressionValue(expression,values){return calculate(expression.replace(\/\\b[a-z][a-z0-9]*\\b\/gi,t=>Object.prototype.hasOwnProperty.call(values,t)?'('+String(values[t])+')':t));}\nfunction condition(text,v){const m=text.match(\/^(.*?)(<=|>=|<|>)(.*)$\/);if(!m)throw Error('Invalid rule');const a=expressionValue(m[1],v),b=expressionValue(m[3],v);return m[2]==='<='?a<=b:m[2]==='>='?a>=b:m[2]==='<'?a<b:a>b;}\nfunction validateField(f,n){if(!Number.isFinite(n))throw Error(f.label+': enter a finite number.');if(f.rule==='positive'&&n<=0)throw Error(f.label+' must be greater than zero.');if(f.rule==='nonnegative'&#038;&#038;n<0)throw Error(f.label+' cannot be negative.');if(f.rule==='fraction0'&#038;&#038;(n<0||n>1))throw Error(f.label+' must be between 0 and 1 (100%).');if(f.rule==='fraction'&&(n<=0||n>1))throw Error(f.label+' must be above 0 and at most 1 (100%).');if(f.rule==='integer'&&(!Number.isInteger(n)||n<=0))throw Error(f.label+' must be a positive whole number.');if(f.rule==='poles'&#038;&#038;(!Number.isInteger(n)||n<2||n%2))throw Error('Total poles must be an even whole number of at least 2.');if(f.rule==='angle180'&#038;&#038;(n<0||n>180))throw Error(f.label+' must be between 0 and 180 degrees.');if(f.kind==='temp'&&n< -273.15)throw Error('Temperature cannot be below absolute zero.');}\nfunction 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point in a conductor. Current tells you how quickly that charge passes. Divide the total charge by the elapsed time.\\n\\nConvert minutes to seconds before calculating. For varying current this gives the average over the stated interval, not necessarily its value at every instant.\",\"limits\":\"Average current; elapsed time must be positive.\",\"mistake\":\"Using minutes as though they were seconds.\",\"sourceIds\":[\"physics-dc\"]},{\"inputs\":[{\"key\":\"i\",\"label\":\"Current\",\"kind\":\"A\",\"example\":2,\"rule\":\"nonnegative\"},{\"key\":\"t\",\"label\":\"Time\",\"kind\":\"s\",\"example\":60,\"rule\":\"positive\"}],\"outputs\":[{\"label\":\"Charge transferred\",\"expression\":\"i*t\",\"unit\":\"C\"}],\"checks\":[],\"expected\":120,\"id\":\"core-02\",\"title\":\"Charge transferred\",\"section\":\"Foundations\",\"formula\":\"Q = I \u00d7 t\",\"explanation\":\"A steady flow transfers more charge when it lasts longer. Multiply amperes by seconds to obtain coulombs.\\n\\nYou can rearrange the relationship to t = Q\/I. Charge measures the amount transferred; it is not the same quantity as energy.\",\"limits\":\"Use steady current, or the average current for the interval.\",\"mistake\":\"Calling coulombs watts or joules.\",\"sourceIds\":[\"physics-dc\"]},{\"inputs\":[{\"key\":\"v\",\"label\":\"Voltage\",\"kind\":\"V\",\"example\":12,\"rule\":\"nonnegative\"},{\"key\":\"r\",\"label\":\"Resistance\",\"kind\":\"ohm\",\"example\":6,\"rule\":\"positive\"}],\"outputs\":[{\"label\":\"Ohm\u2019s law: current\",\"expression\":\"v\/r\",\"unit\":\"A\"}],\"checks\":[],\"expected\":2,\"id\":\"core-03\",\"title\":\"Ohm\u2019s law: current\",\"section\":\"Foundations\",\"formula\":\"I = V \/ R\",\"explanation\":\"Voltage provides the potential difference across the resistor. Resistance determines how much current flows for that voltage. Divide voltage by resistance.\\n\\nSketch the resistor and place the voltage across its two ends. Increasing resistance at fixed voltage reduces current; use that as a reasonableness check.\",\"limits\":\"A resistive element at the stated operating condition; general AC impedance needs Z.\",\"mistake\":\"Using an appliance power rating as its resistance.\",\"sourceIds\":[\"physics-dc\"]},{\"inputs\":[{\"key\":\"i\",\"label\":\"Current\",\"kind\":\"A\",\"example\":2,\"rule\":\"nonnegative\"},{\"key\":\"r\",\"label\":\"Resistance\",\"kind\":\"ohm\",\"example\":6,\"rule\":\"positive\"}],\"outputs\":[{\"label\":\"Ohm\u2019s law: voltage\",\"expression\":\"i*r\",\"unit\":\"V\"}],\"checks\":[],\"expected\":12,\"id\":\"core-04\",\"title\":\"Ohm\u2019s law: voltage\",\"section\":\"Foundations\",\"formula\":\"V = I \u00d7 R\",\"explanation\":\"The voltage drop across a resistor is current multiplied by resistance. This is also useful when assessing the drop along a known conductor resistance.\\n\\nA series circuit has one current, but each resistance can have a different voltage drop. Calculate each drop using that same current.\",\"limits\":\"DC resistance or a purely resistive AC element using consistent RMS quantities.\",\"mistake\":\"Applying the whole supply voltage to every series resistor.\",\"sourceIds\":[\"physics-dc\"]},{\"inputs\":[{\"key\":\"v\",\"label\":\"Voltage\",\"kind\":\"V\",\"example\":12,\"rule\":\"nonnegative\"},{\"key\":\"i\",\"label\":\"Current\",\"kind\":\"A\",\"example\":2,\"rule\":\"positive\"}],\"outputs\":[{\"label\":\"Ohm\u2019s law: resistance\",\"expression\":\"v\/i\",\"unit\":\"\u03a9\"}],\"checks\":[],\"expected\":6,\"id\":\"core-05\",\"title\":\"Ohm\u2019s law: resistance\",\"section\":\"Foundations\",\"formula\":\"R = V \/ I\",\"explanation\":\"Use the voltage across a component and current through it at the same operating point. Their ratio gives its resistance at that point.\\n\\nFor a filament lamp, resistance changes greatly as it heats. A hot operating resistance need not match a cold meter reading.\",\"limits\":\"Current must be nonzero; this does not imply all components have constant resistance.\",\"mistake\":\"Mixing a cold resistance with a hot operating current.\",\"sourceIds\":[\"physics-dc\"]},{\"inputs\":[{\"key\":\"area\",\"label\":\"Conductor area\",\"kind\":\"mm2\",\"example\":2.5,\"rule\":\"positive\"}],\"outputs\":[{\"label\":\"Cable area: mm\u00b2 to m\u00b2\",\"expression\":\"area\/1000000\",\"unit\":\"m\u00b2\"}],\"checks\":[],\"expected\":2.5e-06,\"id\":\"core-13\",\"title\":\"Cable area: mm\u00b2 to m\u00b2\",\"section\":\"Foundations\",\"formula\":\"A(m\u00b2) = A(mm\u00b2) \u00d7 10\u207b\u2076\",\"explanation\":\"Area is a squared length. A millimetre is one thousandth of a metre, so a square millimetre is one millionth of a square metre.\\n\\nUse m\u00b2 with resistivity expressed in \u03a9\u00b7m. If resistivity is supplied in \u03a9\u00b7mm\u00b2\/m, retain mm\u00b2 instead.\",\"limits\":\"Choose area units to match the resistivity units.\",\"mistake\":\"Dividing area by 1,000 instead of 1,000,000.\",\"sourceIds\":[\"physics-dc\"]},{\"inputs\":[{\"key\":\"d\",\"label\":\"Bare conductor diameter\",\"kind\":\"mm\",\"example\":2,\"rule\":\"positive\"}],\"outputs\":[{\"label\":\"Circular conductor area\",\"expression\":\"pi*d^2\/4\",\"unit\":\"mm\u00b2\"}],\"checks\":[],\"expected\":3.141592653589793,\"id\":\"circular-area\",\"title\":\"Circular conductor area\",\"section\":\"Foundations\",\"formula\":\"A = \u03c0d\u00b2 \/ 4\",\"explanation\":\"A circular conductor\u2019s area depends on the square of its diameter. Divide diameter by two to get radius, or use \u03c0d\u00b2\/4 directly.\\n\\nFor stranded conductors sum strand areas; the overall bundle diameter includes spaces and is not enough.\",\"limits\":\"Solid circular conductor, or diameter of one individual strand.\",\"mistake\":\"Using insulation diameter as copper diameter.\",\"sourceIds\":[\"physics-dc\"]},{\"inputs\":[{\"key\":\"v\",\"label\":\"Voltage\",\"kind\":\"V\",\"example\":230,\"rule\":\"nonnegative\"},{\"key\":\"i\",\"label\":\"Current\",\"kind\":\"A\",\"example\":10,\"rule\":\"nonnegative\"}],\"outputs\":[{\"label\":\"Electrical power: VI\",\"expression\":\"v*i\",\"unit\":\"W\"}],\"checks\":[],\"expected\":2300,\"id\":\"core-06\",\"title\":\"Electrical power: VI\",\"section\":\"Power and energy\",\"formula\":\"P = V \u00d7 I\",\"explanation\":\"Power is the rate of energy transfer. For a steady DC load, multiplying its voltage by current gives watts.\\n\\nRearrange to I = P\/V when estimating a DC load current. General AC real power also needs power factor, covered later.\",\"limits\":\"Steady DC or unity-power-factor AC using RMS values.\",\"mistake\":\"Using VI as real AC watts when power factor is below one.\",\"sourceIds\":[\"physics-dc\"]},{\"inputs\":[{\"key\":\"i\",\"label\":\"Current\",\"kind\":\"A\",\"example\":10,\"rule\":\"nonnegative\"},{\"key\":\"r\",\"label\":\"Resistance\",\"kind\":\"ohm\",\"example\":2,\"rule\":\"positive\"}],\"outputs\":[{\"label\":\"Resistive heating: I\u00b2R\",\"expression\":\"i^2*r\",\"unit\":\"W\"}],\"checks\":[],\"expected\":200,\"id\":\"core-07\",\"title\":\"Resistive heating: I\u00b2R\",\"section\":\"Power and energy\",\"formula\":\"P = I\u00b2 \u00d7 R\",\"explanation\":\"Current is squared before multiplying by resistance. If the same resistance carries twice the current, its heating power becomes four times as great.\\n\\nFor AC resistive heating use RMS current. This relationship explains why overcurrent and poor connections matter, but it does not predict temperature without thermal information.\",\"limits\":\"Resistance at the operating temperature; use RMS for AC.\",\"mistake\":\"Multiplying current by two instead of squaring it.\",\"sourceIds\":[\"physics-dc\"]},{\"inputs\":[{\"key\":\"v\",\"label\":\"Voltage\",\"kind\":\"V\",\"example\":12,\"rule\":\"nonnegative\"},{\"key\":\"r\",\"label\":\"Resistance\",\"kind\":\"ohm\",\"example\":6,\"rule\":\"positive\"}],\"outputs\":[{\"label\":\"Resistive power: V\u00b2\/R\",\"expression\":\"v^2\/r\",\"unit\":\"W\"}],\"checks\":[],\"expected\":24,\"id\":\"core-08\",\"title\":\"Resistive power: V\u00b2\/R\",\"section\":\"Power and energy\",\"formula\":\"P = V\u00b2 \/ R\",\"explanation\":\"Square the voltage across the resistor, then divide by resistance. This follows by replacing current in P = VI with V\/R.\\n\\nAt fixed voltage a smaller resistance dissipates more power. At fixed current the comparison is different, so always identify what is held constant.\",\"limits\":\"Pure resistance; voltage is DC or RMS as appropriate.\",\"mistake\":\"Squaring the answer after dividing V by R.\",\"sourceIds\":[\"physics-dc\"]},{\"inputs\":[{\"key\":\"p\",\"label\":\"Power\",\"kind\":\"W\",\"example\":1000,\"rule\":\"nonnegative\"},{\"key\":\"t\",\"label\":\"Time\",\"kind\":\"s\",\"example\":60,\"rule\":\"positive\"}],\"outputs\":[{\"label\":\"Energy in joules\",\"expression\":\"p*t\",\"unit\":\"J\"}],\"checks\":[],\"expected\":60000,\"id\":\"core-09\",\"title\":\"Energy in joules\",\"section\":\"Power and energy\",\"formula\":\"E = P \u00d7 t\",\"explanation\":\"A watt is one joule per second. Multiply constant power by time in seconds to obtain joules.\\n\\nPower describes the rate; energy describes the accumulated amount. A low-power device running for a long time can use substantial energy.\",\"limits\":\"Constant power, or average power for the period.\",\"mistake\":\"Using hours and labelling the result joules.\",\"sourceIds\":[\"physics-dc\"]},{\"inputs\":[{\"key\":\"v\",\"label\":\"Voltage\",\"kind\":\"V\",\"example\":12,\"rule\":\"nonnegative\"},{\"key\":\"i\",\"label\":\"Current\",\"kind\":\"A\",\"example\":2,\"rule\":\"nonnegative\"},{\"key\":\"t\",\"label\":\"Time\",\"kind\":\"s\",\"example\":60,\"rule\":\"positive\"}],\"outputs\":[{\"label\":\"Energy from voltage and current\",\"expression\":\"v*i*t\",\"unit\":\"J\"}],\"checks\":[],\"expected\":1440,\"id\":\"core-10\",\"title\":\"Energy from voltage and current\",\"section\":\"Power and energy\",\"formula\":\"E = V \u00d7 I \u00d7 t\",\"explanation\":\"Combine P = VI with E = Pt. Find the electrical power first, then multiply by elapsed seconds.\\n\\nFor varying loads use the appropriate average power or sum energy over separate time intervals.\",\"limits\":\"Steady DC, or resistive AC with RMS voltage\/current; time in seconds.\",\"mistake\":\"Treating apparent AC power as real energy for every load.\",\"sourceIds\":[\"physics-dc\"]},{\"inputs\":[{\"key\":\"p\",\"label\":\"Power\",\"kind\":\"W\",\"example\":1000,\"rule\":\"nonnegative\"},{\"key\":\"hours\",\"label\":\"Duration\",\"kind\":\"h\",\"example\":2,\"rule\":\"nonnegative\"}],\"outputs\":[{\"label\":\"Energy in kilowatt-hours\",\"expression\":\"p\/1000*hours\",\"unit\":\"kWh\"}],\"checks\":[],\"expected\":2,\"id\":\"core-11\",\"title\":\"Energy in kilowatt-hours\",\"section\":\"Power and energy\",\"formula\":\"E = P(kW) \u00d7 t(h)\",\"explanation\":\"Electricity meters commonly record kilowatt-hours. Divide watts by 1,000, then multiply by hours of operation.\\n\\nOne kWh is 3.6 million joules. Keep the chosen unit system consistent from start to finish.\",\"limits\":\"Constant or average real power.\",\"mistake\":\"Writing kW\/h when you mean kWh.\",\"sourceIds\":[\"physics-dc\"]},{\"inputs\":[{\"key\":\"v\",\"label\":\"Voltage\",\"kind\":\"V\",\"example\":230,\"rule\":\"nonnegative\"},{\"key\":\"i\",\"label\":\"Current\",\"kind\":\"A\",\"example\":10,\"rule\":\"nonnegative\"},{\"key\":\"hours\",\"label\":\"Duration\",\"kind\":\"h\",\"example\":2,\"rule\":\"nonnegative\"}],\"outputs\":[{\"label\":\"Kilowatt-hours from V and I\",\"expression\":\"v*i*hours\/1000\",\"unit\":\"kWh\"}],\"checks\":[],\"expected\":4.6,\"id\":\"core-12\",\"title\":\"Kilowatt-hours from V and I\",\"section\":\"Power and energy\",\"formula\":\"E(kWh) = V \u00d7 I \u00d7 t(h) \/ 1000\",\"explanation\":\"Calculate watts from voltage and current, convert to kilowatts and multiply by hours. This combines two familiar relationships.\\n\\nA motor\u2019s mechanical output is not its electrical input. AC loads may need power factor and efficiency information before using a rating.\",\"limits\":\"Steady DC or unity-power-factor AC.\",\"mistake\":\"Using seconds in this hours-based version.\",\"sourceIds\":[\"physics-dc\"]},{\"inputs\":[{\"key\":\"p\",\"label\":\"Power\",\"kind\":\"W\",\"example\":1000,\"rule\":\"nonnegative\"},{\"key\":\"hours\",\"label\":\"Duration\",\"kind\":\"h\",\"example\":2,\"rule\":\"nonnegative\"},{\"key\":\"tariff\",\"label\":\"Unit price (exclude standing charge)\",\"kind\":\"tariff\",\"example\":0.25,\"rule\":\"nonnegative\"}],\"outputs\":[{\"label\":\"Energy cost\",\"expression\":\"p\/1000*hours*tariff\",\"unit\":\"\u00a3\"}],\"checks\":[],\"expected\":0.5,\"id\":\"energy-cost\",\"title\":\"Energy cost\",\"section\":\"Power and energy\",\"formula\":\"Cost = P(kW) \u00d7 t(h) \u00d7 tariff\",\"explanation\":\"First find real energy consumption in kWh, then multiply by the supplied unit price. A tariff in pence must be divided by 100 for pounds.\\n\\nStanding charges, taxes or time-varying rates must be added separately when applicable.\",\"limits\":\"Use the actual tariff and account separately for standing charges and other fees.\",\"mistake\":\"Mixing pence and pounds.\",\"sourceIds\":[\"physics-dc\"]},{\"inputs\":[{\"key\":\"rho\",\"label\":\"Resistivity at operating temperature\",\"kind\":\"rho\",\"example\":1.72e-08,\"rule\":\"positive\"},{\"key\":\"l\",\"label\":\"Length\",\"kind\":\"m\",\"example\":10,\"rule\":\"positive\"},{\"key\":\"a\",\"label\":\"Area\",\"kind\":\"m2\",\"example\":2.5e-06,\"rule\":\"positive\"}],\"outputs\":[{\"label\":\"Resistance and resistivity\",\"expression\":\"rho*l\/a\",\"unit\":\"\u03a9\"}],\"checks\":[],\"expected\":0.0688,\"id\":\"core-14\",\"title\":\"Resistance and resistivity\",\"section\":\"Circuits and conductors\",\"formula\":\"R = \u03c1 \u00d7 l \/ A\",\"explanation\":\"Longer conductors have more resistance; greater cross-sectional area gives less resistance. Material and temperature affect resistivity.\\n\\nFor a two-wire circuit, distinguish one conductor\u2019s length from the full outgoing-and-return loop. This formula is not by itself a cable current-rating method.\",\"limits\":\"Uniform material\/area and stated temperature; dimensions must match \u03c1.\",\"mistake\":\"Forgetting the return conductor when finding loop resistance.\",\"sourceIds\":[\"physics-dc\"]},{\"inputs\":[{\"key\":\"r1\",\"label\":\"Resistance 1\",\"kind\":\"ohm\",\"example\":10,\"rule\":\"positive\"},{\"key\":\"r2\",\"label\":\"Resistance 2\",\"kind\":\"ohm\",\"example\":20,\"rule\":\"positive\"},{\"key\":\"r3\",\"label\":\"Resistance 3\",\"kind\":\"ohm\",\"example\":30,\"rule\":\"positive\"}],\"outputs\":[{\"label\":\"Series resistance\",\"expression\":\"r1+r2+r3\",\"unit\":\"\u03a9\"}],\"checks\":[],\"expected\":60,\"id\":\"series-r\",\"title\":\"Series resistance\",\"section\":\"Circuits and conductors\",\"formula\":\"R\u209c = R\u2081 + R\u2082 + R\u2083\",\"explanation\":\"In series there is one current path. Add the resistances because each contributes to the total opposition.\\n\\nThe total must exceed every individual positive resistance. Then find circuit current from the supply voltage and total resistance.\",\"limits\":\"Resistors in series, with no branching between them.\",\"mistake\":\"Using the parallel reciprocal rule for a single path.\",\"sourceIds\":[\"physics-circuits\"]},{\"inputs\":[{\"key\":\"r1\",\"label\":\"Resistance 1\",\"kind\":\"ohm\",\"example\":10,\"rule\":\"positive\"},{\"key\":\"r2\",\"label\":\"Resistance 2\",\"kind\":\"ohm\",\"example\":20,\"rule\":\"positive\"},{\"key\":\"r3\",\"label\":\"Resistance 3\",\"kind\":\"ohm\",\"example\":30,\"rule\":\"positive\"}],\"outputs\":[{\"label\":\"Parallel resistance\",\"expression\":\"1\/(1\/r1+1\/r2+1\/r3)\",\"unit\":\"\u03a9\"}],\"checks\":[],\"expected\":5.454545454545454,\"id\":\"parallel-r\",\"title\":\"Parallel resistance\",\"section\":\"Circuits and conductors\",\"formula\":\"1\/R\u209c = 1\/R\u2081 + 1\/R\u2082 + 1\/R\u2083\",\"explanation\":\"Parallel branches share the same two connection points. Add their conductances, then take the reciprocal to return to resistance.\\n\\nFor positive resistances the answer is smaller than the smallest branch resistance. Use this check before submitting.\",\"limits\":\"Three finite positive resistor branches across the same two nodes.\",\"mistake\":\"Stopping after adding reciprocals: that result is conductance, not resistance.\",\"sourceIds\":[\"physics-circuits\"]},{\"inputs\":[{\"key\":\"r1\",\"label\":\"Resistance 1\",\"kind\":\"ohm\",\"example\":10,\"rule\":\"positive\"},{\"key\":\"r2\",\"label\":\"Resistance 2\",\"kind\":\"ohm\",\"example\":20,\"rule\":\"positive\"}],\"outputs\":[{\"label\":\"Two parallel resistors\",\"expression\":\"r1*r2\/(r1+r2)\",\"unit\":\"\u03a9\"}],\"checks\":[],\"expected\":6.666666666666667,\"id\":\"two-parallel\",\"title\":\"Two parallel resistors\",\"section\":\"Circuits and conductors\",\"formula\":\"R\u209c = R\u2081R\u2082 \/ (R\u2081 + R\u2082)\",\"explanation\":\"For exactly two parallel resistors, product divided by sum is a convenient rearrangement of the reciprocal formula.\\n\\nDo not extend product-over-sum directly to three resistors. Reduce a pair first or use reciprocals.\",\"limits\":\"Exactly two positive resistors in parallel.\",\"mistake\":\"Using the product of three resistors divided by their sum.\",\"sourceIds\":[\"physics-circuits\"]},{\"inputs\":[{\"key\":\"r1\",\"label\":\"Resistance 1\",\"kind\":\"ohm\",\"example\":10,\"rule\":\"positive\"},{\"key\":\"r2\",\"label\":\"Resistance 2\",\"kind\":\"ohm\",\"example\":20,\"rule\":\"positive\"},{\"key\":\"r3\",\"label\":\"Resistance 3\",\"kind\":\"ohm\",\"example\":30,\"rule\":\"positive\"}],\"outputs\":[{\"label\":\"Series\u2013parallel resistance\",\"expression\":\"r1+r2*r3\/(r2+r3)\",\"unit\":\"\u03a9\"}],\"checks\":[],\"expected\":22,\"id\":\"combination-r\",\"title\":\"Series\u2013parallel resistance\",\"section\":\"Circuits and conductors\",\"formula\":\"R\u209c = R\u2081 + R\u2082R\u2083 \/ (R\u2082 + R\u2083)\",\"explanation\":\"Reduce the parallel part to a single equivalent resistance before adding the series part. Work from the inside of the grouping outward.\\n\\nR\u2081 carries total current. After it, current splits between R\u2082 and R\u2083 and recombines on return.\",\"limits\":\"R\u2081 in series with the parallel pair R\u2082 and R\u2083.\",\"mistake\":\"Adding all three resistances although two are parallel.\",\"sourceIds\":[\"physics-circuits\"]},{\"inputs\":[{\"key\":\"v\",\"label\":\"Voltage\",\"kind\":\"V\",\"example\":12,\"rule\":\"nonnegative\"},{\"key\":\"r1\",\"label\":\"Resistance 1\",\"kind\":\"ohm\",\"example\":10,\"rule\":\"positive\"},{\"key\":\"r2\",\"label\":\"Resistance 2\",\"kind\":\"ohm\",\"example\":20,\"rule\":\"positive\"}],\"outputs\":[{\"label\":\"Voltage divider\",\"expression\":\"v*r2\/(r1+r2)\",\"unit\":\"V\"}],\"checks\":[],\"expected\":8,\"id\":\"voltage-divider\",\"title\":\"Voltage divider\",\"section\":\"Circuits and conductors\",\"formula\":\"V\u2082 = V\u209b \u00d7 R\u2082 \/ (R\u2081 + R\u2082)\",\"explanation\":\"Series resistors share voltage in proportion to their resistance. Find the fraction of total resistance represented by R\u2082 and multiply by supply voltage.\\n\\nAttaching a load across R\u2082 changes the effective resistance. Recalculate the parallel combination before applying this formula.\",\"limits\":\"Unloaded divider and ideal voltage source.\",\"mistake\":\"Ignoring loading by another component.\",\"sourceIds\":[\"physics-circuits\"]},{\"inputs\":[{\"key\":\"i\",\"label\":\"Current\",\"kind\":\"A\",\"example\":3,\"rule\":\"nonnegative\"},{\"key\":\"r1\",\"label\":\"Resistance 1\",\"kind\":\"ohm\",\"example\":10,\"rule\":\"positive\"},{\"key\":\"r2\",\"label\":\"Resistance 2\",\"kind\":\"ohm\",\"example\":20,\"rule\":\"positive\"}],\"outputs\":[{\"label\":\"Current divider\",\"expression\":\"i*r2\/(r1+r2)\",\"unit\":\"A\"},{\"label\":\"Current through R2\",\"expression\":\"i*r1\/(r1+r2)\",\"unit\":\"A\"}],\"checks\":[],\"expected\":2,\"id\":\"current-divider\",\"title\":\"Current divider\",\"section\":\"Circuits and conductors\",\"formula\":\"I\u2081 = I\u209c \u00d7 R\u2082 \/ (R\u2081 + R\u2082)\",\"explanation\":\"The smaller resistance takes the larger current. For two parallel branches, current in R\u2081 uses the other resistance R\u2082 in the numerator.\\n\\nCheck by calculating the common branch voltage: I\u2081R\u2081 should equal I\u2082R\u2082, and the two currents must add to total current.\",\"limits\":\"Two positive parallel resistances.\",\"mistake\":\"Putting R\u2081 in the numerator when finding its own branch current.\",\"sourceIds\":[\"physics-circuits\"]},{\"inputs\":[{\"key\":\"total\",\"label\":\"Total entering current\",\"kind\":\"A\",\"example\":10,\"rule\":\"nonnegative\"},{\"key\":\"known\",\"label\":\"Sum of known outgoing currents\",\"kind\":\"A\",\"example\":6,\"rule\":\"nonnegative\"}],\"outputs\":[{\"label\":\"Missing outgoing current (negative means entering)\",\"expression\":\"total-known\",\"unit\":\"A\"}],\"checks\":[],\"expected\":4,\"id\":\"kcl\",\"title\":\"Kirchhoff: current balance\",\"section\":\"Circuits and conductors\",\"formula\":\"\u03a3I\u1d62\u2099 = \u03a3I\u2092\u1d64\u209c\",\"explanation\":\"A junction does not continually accumulate charge in an ordinary steady circuit. Total current entering therefore equals total current leaving.\\n\\nDraw arrows and label known branch currents. Subtract known outgoing currents from the total to find a missing branch.\",\"limits\":\"Steady currents; directions must be assigned consistently.\",\"mistake\":\"Saying current is used up by a component.\",\"sourceIds\":[\"physics-circuits\"]},{\"inputs\":[{\"key\":\"known\",\"label\":\"Sum of known signed voltage rises and drops\",\"kind\":\"V\",\"example\":-12,\"rule\":\"signed\"}],\"outputs\":[{\"label\":\"Missing signed voltage change\",\"expression\":\"-known\",\"unit\":\"V\"}],\"checks\":[],\"expected\":12,\"id\":\"kvl\",\"title\":\"Kirchhoff: voltage balance\",\"section\":\"Circuits and conductors\",\"formula\":\"\u03a3V around a loop = 0\",\"explanation\":\"Add voltage rises and drops with consistent signs around a closed loop. In a simple resistive loop the supply rise equals the total drops.\\n\\nThis is an energy balance per unit charge. Do not add magnitudes without noting whether each term is a rise or a drop.\",\"limits\":\"Lumped DC circuit with no unmodelled induced emf.\",\"mistake\":\"Using the same sign for all rises and drops.\",\"sourceIds\":[\"physics-circuits\"]},{\"inputs\":[{\"key\":\"r1\",\"label\":\"Resistance at reference temperature\",\"kind\":\"ohm\",\"example\":10,\"rule\":\"positive\"},{\"key\":\"alpha\",\"label\":\"Temperature coefficient at reference temperature\",\"kind\":\"alpha\",\"example\":0.00393,\"rule\":\"signed\"},{\"key\":\"t1\",\"label\":\"Reference temperature\",\"kind\":\"temp\",\"example\":20,\"rule\":\"signed\"},{\"key\":\"t2\",\"label\":\"Final temperature\",\"kind\":\"temp\",\"example\":70,\"rule\":\"signed\"}],\"outputs\":[{\"label\":\"Resistance and temperature\",\"expression\":\"r1*(1+alpha*(t2-t1))\",\"unit\":\"\u03a9\"}],\"checks\":[{\"expression\":\"1+alpha*(t2-t1)>0\",\"message\":\"Temperature model must give a positive resistance factor.\"}],\"expected\":11.965,\"id\":\"resistance-temperature\",\"title\":\"Resistance and temperature\",\"section\":\"Circuits and conductors\",\"formula\":\"R\u2082 = R\u2081[1 + \u03b1(T\u2082 \u2212 T\u2081)]\",\"explanation\":\"Within an appropriate range a metal conductor\u2019s resistance rises approximately linearly with temperature. Calculate the temperature change, multiply by the supplied coefficient and add one.\\n\\nMultiply this dimensionless factor by reference resistance. The coefficient must correspond to the reference temperature used.\",\"limits\":\"Linear approximation; supplied material coefficient, not a universal constant.\",\"mistake\":\"Multiplying by temperature instead of temperature change.\",\"sourceIds\":[\"physics-dc\",\"d-temp\"]},{\"inputs\":[{\"key\":\"i\",\"label\":\"Current\",\"kind\":\"A\",\"example\":10,\"rule\":\"nonnegative\"},{\"key\":\"area\",\"label\":\"Conductor area\",\"kind\":\"mm2\",\"example\":2.5,\"rule\":\"positive\"}],\"outputs\":[{\"label\":\"Current density\",\"expression\":\"i\/area\",\"unit\":\"A\/mm\u00b2\"}],\"checks\":[],\"expected\":4,\"id\":\"current-density\",\"title\":\"Current density\",\"section\":\"Circuits and conductors\",\"formula\":\"J = I \/ A\",\"explanation\":\"Current density relates current to conductor cross-section. It helps compare loading but is not a complete current-carrying-capacity calculation.\\n\\nThermal conditions, insulation, grouping and installation method still matter for cable selection.\",\"limits\":\"Average uniform current density; area units define output units.\",\"mistake\":\"Using a fixed A\/mm\u00b2 rule as a substitute for cable design.\",\"sourceIds\":[\"physics-dc\"]},{\"inputs\":[{\"key\":\"c\",\"label\":\"Capacitance\",\"kind\":\"F\",\"example\":0.0001,\"rule\":\"positive\"},{\"key\":\"v\",\"label\":\"Voltage\",\"kind\":\"V\",\"example\":12,\"rule\":\"nonnegative\"}],\"outputs\":[{\"label\":\"Capacitance and charge\",\"expression\":\"c*v\",\"unit\":\"C\"}],\"checks\":[],\"expected\":0.0012,\"id\":\"charge-cap\",\"title\":\"Capacitance and charge\",\"section\":\"Capacitance and inductance\",\"formula\":\"Q = C \u00d7 V\",\"explanation\":\"A capacitor stores separated charge. Its capacitance relates stored charge to the voltage across it.\\n\\nConvert microfarads to farads before multiplying. Capacitor voltage can remain after the supply is removed; a calculation is not proof of discharge.\",\"limits\":\"Ideal fixed capacitor.\",\"mistake\":\"Leaving capacitance in \u03bcF but labelling the answer C.\",\"sourceIds\":[\"physics-cap\"]},{\"inputs\":[{\"key\":\"c\",\"label\":\"Capacitance\",\"kind\":\"F\",\"example\":0.0001,\"rule\":\"positive\"},{\"key\":\"v\",\"label\":\"Voltage\",\"kind\":\"V\",\"example\":12,\"rule\":\"nonnegative\"}],\"outputs\":[{\"label\":\"Capacitor energy\",\"expression\":\"c*v^2\/2\",\"unit\":\"J\"}],\"checks\":[],\"expected\":0.0072,\"id\":\"cap-energy\",\"title\":\"Capacitor energy\",\"section\":\"Capacitance and inductance\",\"formula\":\"E = \u00bdCV\u00b2\",\"explanation\":\"Stored capacitor energy increases with the square of voltage. Double the voltage and the stored energy becomes four times as large at unchanged capacitance.\\n\\nUse the actual terminal voltage and capacitance in farads. Stored energy must be managed through an appropriate safe procedure.\",\"limits\":\"Ideal capacitor; not a discharge safety assessment.\",\"mistake\":\"Forgetting the half or failing to square voltage.\",\"sourceIds\":[\"physics-cap\"]},{\"inputs\":[{\"key\":\"c1\",\"label\":\"Capacitance 1\",\"kind\":\"F\",\"example\":0.0001,\"rule\":\"positive\"},{\"key\":\"c2\",\"label\":\"Capacitance 2\",\"kind\":\"F\",\"example\":0.0002,\"rule\":\"positive\"},{\"key\":\"c3\",\"label\":\"Capacitance 3\",\"kind\":\"F\",\"example\":0.0003,\"rule\":\"positive\"}],\"outputs\":[{\"label\":\"Capacitors in parallel\",\"expression\":\"c1+c2+c3\",\"unit\":\"F\"}],\"checks\":[],\"expected\":0.0006,\"id\":\"caps-parallel\",\"title\":\"Capacitors in parallel\",\"section\":\"Capacitance and inductance\",\"formula\":\"C\u209c = C\u2081 + C\u2082 + C\u2083\",\"explanation\":\"Capacitors in parallel share voltage and their charge capacities add. Sum their capacitances using one consistent unit.\\n\\nUnlike parallel resistors, parallel capacitors increase the total capacitance. Component voltage ratings still matter.\",\"limits\":\"Ideal capacitors across the same two nodes.\",\"mistake\":\"Applying the parallel-resistance reciprocal rule.\",\"sourceIds\":[\"physics-cap\"]},{\"inputs\":[{\"key\":\"c1\",\"label\":\"Capacitance 1\",\"kind\":\"F\",\"example\":0.0001,\"rule\":\"positive\"},{\"key\":\"c2\",\"label\":\"Capacitance 2\",\"kind\":\"F\",\"example\":0.0002,\"rule\":\"positive\"}],\"outputs\":[{\"label\":\"Capacitors in series\",\"expression\":\"c1*c2\/(c1+c2)\",\"unit\":\"F\"}],\"checks\":[],\"expected\":6.666666666666667e-05,\"id\":\"caps-series\",\"title\":\"Capacitors in series\",\"section\":\"Capacitance and inductance\",\"formula\":\"C\u209c = C\u2081C\u2082 \/ (C\u2081 + C\u2082)\",\"explanation\":\"Two series capacitors have a total capacitance smaller than either capacitor. Product over sum gives the two-component result.\\n\\nVoltage sharing depends on capacitance and practical leakage; putting capacitors in series does not automatically ensure safe voltage sharing.\",\"limits\":\"Two ideal capacitors in series.\",\"mistake\":\"Assuming equal voltage across unequal capacitors.\",\"sourceIds\":[\"physics-cap\"]},{\"inputs\":[{\"key\":\"r\",\"label\":\"Resistance\",\"kind\":\"ohm\",\"example\":1000,\"rule\":\"positive\"},{\"key\":\"c\",\"label\":\"Capacitance\",\"kind\":\"F\",\"example\":0.0001,\"rule\":\"positive\"}],\"outputs\":[{\"label\":\"RC time constant\",\"expression\":\"r*c\",\"unit\":\"s\"}],\"checks\":[],\"expected\":0.1,\"id\":\"rc-time\",\"title\":\"RC time constant\",\"section\":\"Capacitance and inductance\",\"formula\":\"\u03c4 = R \u00d7 C\",\"explanation\":\"The time constant sets the pace of a first-order RC response. After one time constant, an initially uncharged capacitor reaches about 63.2% of its final charging voltage.\\n\\nAfter one discharging time constant about 36.8% remains. Five time constants is an approximation, never proof that a real capacitor is safe to touch.\",\"limits\":\"Ideal first-order RC network; equivalent resistance seen by capacitor.\",\"mistake\":\"Calling one time constant a full charge or discharge.\",\"sourceIds\":[\"physics-circuits\"]},{\"inputs\":[{\"key\":\"v\",\"label\":\"Voltage\",\"kind\":\"V\",\"example\":12,\"rule\":\"nonnegative\"},{\"key\":\"r\",\"label\":\"Resistance\",\"kind\":\"ohm\",\"example\":1000,\"rule\":\"positive\"},{\"key\":\"c\",\"label\":\"Capacitance\",\"kind\":\"F\",\"example\":0.0001,\"rule\":\"positive\"},{\"key\":\"t\",\"label\":\"Elapsed time\",\"kind\":\"s\",\"example\":0.1,\"rule\":\"nonnegative\"}],\"outputs\":[{\"label\":\"Capacitor discharge voltage\",\"expression\":\"v*exp(-t\/(r*c))\",\"unit\":\"V\"}],\"checks\":[],\"expected\":4.414553294057308,\"id\":\"rc-discharge\",\"title\":\"Capacitor discharge voltage\",\"section\":\"Capacitance and inductance\",\"formula\":\"V(t) = V\u2080e^(\u2212t\/RC)\",\"explanation\":\"Voltage falls exponentially through a fixed discharge resistance. Work out RC in seconds, divide elapsed time by it, change the sign and evaluate the exponential.\\n\\nThe model approaches zero asymptotically. Real leakage, dielectric absorption and failed discharge components require practical verification.\",\"limits\":\"Ideal isolated capacitor discharging through R; no additional supply.\",\"mistake\":\"Using base 10 instead of e.\",\"sourceIds\":[\"physics-circuits\"]},{\"inputs\":[{\"key\":\"h\",\"label\":\"Inductance\",\"kind\":\"H\",\"example\":0.1,\"rule\":\"positive\"},{\"key\":\"i\",\"label\":\"Current\",\"kind\":\"A\",\"example\":2,\"rule\":\"nonnegative\"}],\"outputs\":[{\"label\":\"Inductor energy\",\"expression\":\"h*i^2\/2\",\"unit\":\"J\"}],\"checks\":[],\"expected\":0.2,\"id\":\"inductor-energy\",\"title\":\"Inductor energy\",\"section\":\"Capacitance and inductance\",\"formula\":\"E = \u00bdLI\u00b2\",\"explanation\":\"An inductor stores energy in its magnetic field. Square its current, multiply by inductance and divide by two.\\n\\nInterrupting current can create a large induced voltage. Suppression and safe switching need component-specific design.\",\"limits\":\"Linear inductor; saturation and losses excluded.\",\"mistake\":\"Using mH directly as H.\",\"sourceIds\":[\"physics-ind\"]},{\"inputs\":[{\"key\":\"h\",\"label\":\"Inductance\",\"kind\":\"H\",\"example\":0.1,\"rule\":\"positive\"},{\"key\":\"r\",\"label\":\"Resistance\",\"kind\":\"ohm\",\"example\":10,\"rule\":\"positive\"}],\"outputs\":[{\"label\":\"RL time constant\",\"expression\":\"h\/r\",\"unit\":\"s\"}],\"checks\":[],\"expected\":0.01,\"id\":\"rl-time\",\"title\":\"RL time constant\",\"section\":\"Capacitance and inductance\",\"formula\":\"\u03c4 = L \/ R\",\"explanation\":\"Current in a first-order RL circuit changes gradually. Divide inductance by the total series resistance to find the time constant.\\n\\nInclude coil resistance in the total. The relationship describes the transient pace, not the final DC current.\",\"limits\":\"Linear inductor and fixed resistance.\",\"mistake\":\"Multiplying L by R by analogy with RC.\",\"sourceIds\":[\"physics-ind\"]},{\"inputs\":[{\"key\":\"t\",\"label\":\"Time\",\"kind\":\"s\",\"example\":0.02,\"rule\":\"positive\"}],\"outputs\":[{\"label\":\"Frequency from period\",\"expression\":\"1\/t\",\"unit\":\"Hz\"}],\"checks\":[],\"expected\":50,\"id\":\"core-15\",\"title\":\"Frequency from period\",\"section\":\"AC foundations\",\"formula\":\"f = 1 \/ T\",\"explanation\":\"Frequency counts complete cycles per second. If each cycle takes T seconds, divide one second by T.\\n\\nConvert milliseconds to seconds first. A half-cycle is not a whole period.\",\"limits\":\"Periodic waveform; T must represent a complete cycle.\",\"mistake\":\"Entering milliseconds directly and losing a factor of 1,000.\",\"sourceIds\":[\"physics-ac\"]},{\"inputs\":[{\"key\":\"f\",\"label\":\"Frequency\",\"kind\":\"Hz\",\"example\":50,\"rule\":\"positive\"}],\"outputs\":[{\"label\":\"Period from frequency\",\"expression\":\"1\/f\",\"unit\":\"s\"}],\"checks\":[],\"expected\":0.02,\"id\":\"core-16\",\"title\":\"Period from frequency\",\"section\":\"AC foundations\",\"formula\":\"T = 1 \/ f\",\"explanation\":\"Higher frequency means less time per cycle. Divide one by the frequency to obtain seconds.\\n\\nMultiply seconds by 1,000 for milliseconds. At 50 Hz a full cycle is 20 ms and a half-cycle is 10 ms.\",\"limits\":\"Periodic waveform and positive frequency.\",\"mistake\":\"Confusing half-cycle time with period.\",\"sourceIds\":[\"physics-ac\"]},{\"inputs\":[{\"key\":\"f\",\"label\":\"Frequency\",\"kind\":\"Hz\",\"example\":100,\"rule\":\"positive\"},{\"key\":\"wave\",\"label\":\"Wavelength\",\"kind\":\"m\",\"example\":3,\"rule\":\"positive\"}],\"outputs\":[{\"label\":\"Wave speed\",\"expression\":\"f*wave\",\"unit\":\"m\/s\"}],\"checks\":[],\"expected\":300,\"id\":\"core-18\",\"title\":\"Wave speed\",\"section\":\"AC foundations\",\"formula\":\"v = f \u00d7 \u03bb\",\"explanation\":\"A wave travels one wavelength in one period. Multiplying cycles per second by metres per cycle gives metres per second.\\n\\nWave propagation and electron drift are different motions. Use this relationship for the specified wave and medium.\",\"limits\":\"Wave speed depends on medium; not an electron-drift formula.\",\"mistake\":\"Treating signal speed as the speed of individual electrons.\",\"sourceIds\":[\"physics-wave\"]},{\"inputs\":[{\"key\":\"speed\",\"label\":\"Wave speed in medium\",\"kind\":\"speed\",\"example\":300000000,\"rule\":\"positive\"},{\"key\":\"f\",\"label\":\"Frequency\",\"kind\":\"Hz\",\"example\":100000000,\"rule\":\"positive\"}],\"outputs\":[{\"label\":\"Wavelength\",\"expression\":\"speed\/f\",\"unit\":\"m\"}],\"checks\":[],\"expected\":3,\"id\":\"core-19\",\"title\":\"Wavelength\",\"section\":\"AC foundations\",\"formula\":\"\u03bb = v \/ f\",\"explanation\":\"At a fixed propagation speed, higher frequency gives shorter wavelength. Divide the supplied speed by frequency.\\n\\nUse hertz rather than megahertz unless you make the corresponding conversion.\",\"limits\":\"Use the speed for the stated medium, not automatically vacuum speed.\",\"mistake\":\"Forgetting that 1 MHz = 1,000,000 Hz.\",\"sourceIds\":[\"physics-wave\"]},{\"inputs\":[{\"key\":\"v\",\"label\":\"RMS voltage\",\"kind\":\"V\",\"example\":230,\"rule\":\"nonnegative\"}],\"outputs\":[{\"label\":\"Peak voltage\",\"expression\":\"sqrt(2)*v\",\"unit\":\"V\"}],\"checks\":[],\"expected\":325.2691193458119,\"id\":\"core-20\",\"title\":\"Peak voltage\",\"section\":\"AC foundations\",\"formula\":\"V\u209a = \u221a2 \u00d7 V\u1d63\u2098\u209b\",\"explanation\":\"For a sinusoidal voltage centred on zero, the peak is \u221a2 times RMS. RMS indicates equivalent resistive heating.\\n\\nThis factor is tied to the sine-wave shape. A square wave or distorted waveform needs its own RMS calculation.\",\"limits\":\"Sinusoidal waveform with zero DC offset.\",\"mistake\":\"Using \u221a2 for every waveform.\",\"sourceIds\":[\"physics-ac\"]},{\"inputs\":[{\"key\":\"v\",\"label\":\"Peak voltage\",\"kind\":\"V\",\"example\":100,\"rule\":\"nonnegative\"}],\"outputs\":[{\"label\":\"RMS voltage\",\"expression\":\"v\/sqrt(2)\",\"unit\":\"V\"}],\"checks\":[],\"expected\":70.71067811865474,\"id\":\"core-21\",\"title\":\"RMS voltage\",\"section\":\"AC foundations\",\"formula\":\"V\u1d63\u2098\u209b = V\u209a \/ \u221a2\",\"explanation\":\"Divide a sine wave\u2019s peak voltage by \u221a2 to find RMS voltage. RMS is not the arithmetic average over a complete AC cycle.\\n\\nA symmetric sine wave has zero full-cycle mean while still delivering heating power into a resistor.\",\"limits\":\"Sine wave centred on zero.\",\"mistake\":\"Confusing zero average with zero RMS.\",\"sourceIds\":[\"physics-ac\"]},{\"inputs\":[{\"key\":\"v\",\"label\":\"Peak voltage\",\"kind\":\"V\",\"example\":100,\"rule\":\"nonnegative\"}],\"outputs\":[{\"label\":\"Peak-to-peak voltage\",\"expression\":\"2*v\",\"unit\":\"V\"}],\"checks\":[],\"expected\":200,\"id\":\"core-22\",\"title\":\"Peak-to-peak voltage\",\"section\":\"AC foundations\",\"formula\":\"V\u209a\u209a = 2 \u00d7 V\u209a\",\"explanation\":\"Peak-to-peak measures from the most negative point to the most positive. For a symmetrical waveform those magnitudes are equal.\\n\\nFor unequal peaks use Vmax \u2212 Vmin instead. Do not double an RMS reading unless you also apply the appropriate peak conversion.\",\"limits\":\"Equal positive and negative peak magnitudes for the 2V\u209a form.\",\"mistake\":\"Doubling RMS and calling the result peak-to-peak.\",\"sourceIds\":[\"physics-ac\"]},{\"inputs\":[{\"key\":\"i\",\"label\":\"Peak current\",\"kind\":\"A\",\"example\":10,\"rule\":\"nonnegative\"}],\"outputs\":[{\"label\":\"RMS current\",\"expression\":\"i\/sqrt(2)\",\"unit\":\"A\"}],\"checks\":[],\"expected\":7.071067811865475,\"id\":\"core-23\",\"title\":\"RMS current\",\"section\":\"AC foundations\",\"formula\":\"I\u1d63\u2098\u209b = I\u209a \/ \u221a2\",\"explanation\":\"For a sinusoidal current, divide its peak by \u221a2. The RMS current is the useful quantity for resistive heating calculations.\\n\\nElectronic loads can draw non-sinusoidal current even from a sinusoidal voltage supply. The \u221a2 conversion then need not apply.\",\"limits\":\"Sinusoidal current centred on zero.\",\"mistake\":\"Assuming sinusoidal voltage guarantees sinusoidal current.\",\"sourceIds\":[\"physics-ac\"]},{\"inputs\":[{\"key\":\"f\",\"label\":\"Frequency\",\"kind\":\"Hz\",\"example\":50,\"rule\":\"positive\"},{\"key\":\"h\",\"label\":\"Inductance\",\"kind\":\"H\",\"example\":0.1,\"rule\":\"positive\"}],\"outputs\":[{\"label\":\"Inductive reactance\",\"expression\":\"2*pi*f*h\",\"unit\":\"\u03a9\"}],\"checks\":[],\"expected\":31.415926535897935,\"id\":\"inductive-reactance\",\"title\":\"Inductive reactance\",\"section\":\"AC foundations\",\"formula\":\"X\u2097 = 2\u03c0fL\",\"explanation\":\"An ideal inductor opposes changes in current. In sinusoidal steady state its reactance increases with frequency and inductance.\\n\\nReactance is not resistance: an ideal inductor stores and returns energy rather than dissipating average real power.\",\"limits\":\"Sinusoidal steady state, linear inductor.\",\"mistake\":\"Using mH without conversion to H.\",\"sourceIds\":[\"physics-ac\"]},{\"inputs\":[{\"key\":\"f\",\"label\":\"Frequency\",\"kind\":\"Hz\",\"example\":50,\"rule\":\"positive\"},{\"key\":\"c\",\"label\":\"Capacitance\",\"kind\":\"F\",\"example\":0.0001,\"rule\":\"positive\"}],\"outputs\":[{\"label\":\"Capacitive reactance\",\"expression\":\"1\/(2*pi*f*c)\",\"unit\":\"\u03a9\"}],\"checks\":[],\"expected\":31.830988618379067,\"id\":\"capacitive-reactance\",\"title\":\"Capacitive reactance\",\"section\":\"AC foundations\",\"formula\":\"X\ua700 = 1 \/ (2\u03c0fC)\",\"explanation\":\"A capacitor\u2019s reactance falls as frequency or capacitance increases. Calculate the entire denominator before taking its reciprocal.\\n\\nThis is a positive magnitude. In complex impedance, the capacitor contributes a negative imaginary term.\",\"limits\":\"Sinusoidal steady state, ideal capacitor.\",\"mistake\":\"Dividing by 2\u03c0 then multiplying by fC.\",\"sourceIds\":[\"physics-ac\"]},{\"inputs\":[{\"key\":\"r\",\"label\":\"Resistance\",\"kind\":\"ohm\",\"example\":3,\"rule\":\"nonnegative\"},{\"key\":\"xl\",\"label\":\"Inductive reactance magnitude\",\"kind\":\"ohm\",\"example\":5,\"rule\":\"nonnegative\"},{\"key\":\"xc\",\"label\":\"Capacitive reactance magnitude\",\"kind\":\"ohm\",\"example\":1,\"rule\":\"nonnegative\"}],\"outputs\":[{\"label\":\"Series RLC impedance\",\"expression\":\"sqrt(r^2+(xl-xc)^2)\",\"unit\":\"\u03a9\"}],\"checks\":[],\"expected\":5,\"id\":\"series-impedance\",\"title\":\"Series RLC impedance\",\"section\":\"AC foundations\",\"formula\":\"|Z| = \u221a[R\u00b2 + (X\u2097 \u2212 X\ua700)\u00b2]\",\"explanation\":\"Subtract capacitive reactance from inductive reactance to find the net reactive component. Resistance and net reactance are perpendicular components, so use Pythagoras.\\n\\nIf X\ua700 is greater, the net reactance is negative, but impedance magnitude is still positive.\",\"limits\":\"Series RLC at one sinusoidal frequency.\",\"mistake\":\"Adding R and reactance magnitudes arithmetically.\",\"sourceIds\":[\"physics-ac\"]},{\"inputs\":[{\"key\":\"v\",\"label\":\"Voltage\",\"kind\":\"V\",\"example\":230,\"rule\":\"nonnegative\"},{\"key\":\"z\",\"label\":\"Impedance magnitude\",\"kind\":\"ohm\",\"example\":23,\"rule\":\"positive\"}],\"outputs\":[{\"label\":\"AC current from impedance\",\"expression\":\"v\/z\",\"unit\":\"A\"}],\"checks\":[],\"expected\":10,\"id\":\"ac-current-z\",\"title\":\"AC current from impedance\",\"section\":\"AC foundations\",\"formula\":\"I = V \/ |Z|\",\"explanation\":\"Once the circuit impedance magnitude is known, divide RMS voltage by it to obtain RMS current magnitude.\\n\\nThe phase angle needs separate information about resistance and reactance. A current magnitude alone does not state power factor.\",\"limits\":\"Sinusoidal steady state; use matching RMS quantities.\",\"mistake\":\"Using only resistance in an inductive load.\",\"sourceIds\":[\"physics-ac\"]},{\"inputs\":[{\"key\":\"r\",\"label\":\"Resistance\",\"kind\":\"ohm\",\"example\":3,\"rule\":\"positive\"},{\"key\":\"xl\",\"label\":\"Inductive reactance magnitude\",\"kind\":\"ohm\",\"example\":5,\"rule\":\"nonnegative\"},{\"key\":\"xc\",\"label\":\"Capacitive reactance magnitude\",\"kind\":\"ohm\",\"example\":1,\"rule\":\"nonnegative\"}],\"outputs\":[{\"label\":\"Series impedance phase angle\",\"expression\":\"atan((xl-xc)\/r)\",\"unit\":\"\u00b0\"}],\"checks\":[],\"expected\":53.13010235415598,\"id\":\"phase-angle\",\"title\":\"Series impedance phase angle\",\"section\":\"AC foundations\",\"formula\":\"\u03c6 = atan[(X\u2097 \u2212 X\ua700) \/ R]\",\"explanation\":\"For positive resistance, arctangent of net reactance over resistance gives the impedance angle. Positive means inductive: current lags voltage. Negative means capacitive: current leads voltage.\\n\\nSet a calculator to degrees or convert radians explicitly. A displayed negative angle is meaningful, not automatically an error.\",\"limits\":\"Positive R in a series sinusoidal circuit.\",\"mistake\":\"Confusing voltage-relative-to-current with current-relative-to-voltage.\",\"sourceIds\":[\"physics-ac\"]},{\"inputs\":[{\"key\":\"h\",\"label\":\"Inductance\",\"kind\":\"H\",\"example\":0.1,\"rule\":\"positive\"},{\"key\":\"c\",\"label\":\"Capacitance\",\"kind\":\"F\",\"example\":0.0001,\"rule\":\"positive\"}],\"outputs\":[{\"label\":\"LC resonant frequency\",\"expression\":\"1\/(2*pi*sqrt(h*c))\",\"unit\":\"Hz\"}],\"checks\":[],\"expected\":50.329212104487034,\"id\":\"resonance\",\"title\":\"LC resonant frequency\",\"section\":\"AC foundations\",\"formula\":\"f\u2080 = 1 \/ (2\u03c0\u221aLC)\",\"explanation\":\"At ideal resonance the inductive and capacitive reactance magnitudes match. Convert both quantities to base units before taking their product and square root.\\n\\nSeries resonance leaves only the series resistance in the ideal model. Practical resonance may create large currents or component voltages.\",\"limits\":\"Ideal LC resonance; real losses and parasitics alter behaviour.\",\"mistake\":\"Forgetting the square root covers LC together.\",\"sourceIds\":[\"physics-ac\"]},{\"inputs\":[{\"key\":\"er\",\"label\":\"Relative permittivity\",\"kind\":\"number\",\"example\":4,\"rule\":\"positive\"},{\"key\":\"mr\",\"label\":\"Relative permeability\",\"kind\":\"number\",\"example\":1,\"rule\":\"positive\"}],\"outputs\":[{\"label\":\"Wave speed in a medium\",\"expression\":\"299792458\/sqrt(er*mr)\",\"unit\":\"m\/s\"}],\"checks\":[],\"expected\":149896229,\"id\":\"medium-wave\",\"title\":\"Wave speed in a medium\",\"section\":\"AC foundations\",\"formula\":\"v = 299792458 \/ \u221a(\u03b5r \u00d7 \u03bcr)\",\"explanation\":\"In the simple lossless-medium model, electrical and magnetic material properties set electromagnetic wave speed. The equivalent absolute-property formula is v = 1\/\u221a(\u03b5\u03bc).\\n\\nRelative values have no units. Do not mix relative and absolute properties in the same formula.\",\"limits\":\"Homogeneous, linear, lossless model; real cable propagation depends on geometry and frequency. Uses the exact vacuum light speed, 299,792,458 m\/s, rather than the rounded teaching value.\",\"mistake\":\"Assuming this is electron drift speed.\",\"sourceIds\":[\"physics-wave\"]},{\"inputs\":[{\"key\":\"v\",\"label\":\"Peak voltage\",\"kind\":\"V\",\"example\":100,\"rule\":\"nonnegative\"},{\"key\":\"theta\",\"label\":\"Angle\",\"kind\":\"deg\",\"example\":30,\"rule\":\"signed\"}],\"outputs\":[{\"label\":\"Instantaneous sine-wave voltage\",\"expression\":\"v*sin(theta)\",\"unit\":\"V\"}],\"checks\":[],\"expected\":50,\"id\":\"instant-sine\",\"title\":\"Instantaneous sine-wave voltage\",\"section\":\"AC foundations\",\"formula\":\"v = V\u209a \u00d7 sin \u03b8\",\"explanation\":\"RMS, peak and instantaneous values answer different questions. At a specific angle, multiply the peak by the sine of that angle.\\n\\nNegative values mean the opposite polarity. An instantaneous zero does not mean the RMS supply is zero.\",\"limits\":\"Pure sine wave with zero phase at its rising zero crossing; degree mode.\",\"mistake\":\"Putting RMS voltage in place of peak.\",\"sourceIds\":[\"physics-ac\"]},{\"inputs\":[{\"key\":\"v\",\"label\":\"Sine-wave peak voltage\",\"kind\":\"V\",\"example\":100,\"rule\":\"nonnegative\"}],\"outputs\":[{\"label\":\"Mean of a rectified sine wave\",\"expression\":\"2*v\/pi\",\"unit\":\"V\"}],\"checks\":[],\"expected\":63.66197723675813,\"id\":\"rectified-mean\",\"title\":\"Mean of a rectified sine wave\",\"section\":\"AC foundations\",\"formula\":\"Vmean = 2V\u209a \/ \u03c0\",\"explanation\":\"An ideal full-wave rectifier turns negative half-cycles positive. The average of that unsmoothed waveform is about 0.637 times its peak.\\n\\nThe mean of the original signed sine wave over a complete cycle is zero. A smoothing capacitor changes the waveform and its average.\",\"limits\":\"Ideal diodes, no smoothing capacitor and a sinusoidal input.\",\"mistake\":\"Confusing rectified average with RMS or smoothed DC.\",\"sourceIds\":[\"physics-ac\"]},{\"inputs\":[{\"key\":\"b\",\"label\":\"Flux density\",\"kind\":\"T\",\"example\":0.5,\"rule\":\"nonnegative\"},{\"key\":\"a\",\"label\":\"Area\",\"kind\":\"m2\",\"example\":0.02,\"rule\":\"positive\"},{\"key\":\"theta\",\"label\":\"Angle\",\"kind\":\"deg\",\"example\":60,\"rule\":\"signed\"}],\"outputs\":[{\"label\":\"Magnetic flux\",\"expression\":\"b*a*cos(theta)\",\"unit\":\"Wb\"}],\"checks\":[],\"expected\":0.005,\"id\":\"flux\",\"title\":\"Magnetic flux\",\"section\":\"Magnetism\",\"formula\":\"\u03a6 = B \u00d7 A \u00d7 cos \u03b8\",\"explanation\":\"Flux describes the magnetic field passing through a surface. Use the angle to the normal, a line perpendicular to the surface.\\n\\nFor a field perpendicular to the surface, \u03b8 is zero and cos \u03b8 is one. Convert square centimetres into square metres before calculating.\",\"limits\":\"Uniform field over a flat surface.\",\"mistake\":\"Using the angle to the surface instead of the normal.\",\"sourceIds\":[\"physics-flux\"]},{\"inputs\":[{\"key\":\"n\",\"label\":\"Number of turns\",\"kind\":\"count\",\"example\":100,\"rule\":\"integer\"},{\"key\":\"flux\",\"label\":\"Magnitude of flux change per turn\",\"kind\":\"Wb\",\"example\":0.02,\"rule\":\"nonnegative\"},{\"key\":\"t\",\"label\":\"Time\",\"kind\":\"s\",\"example\":0.1,\"rule\":\"positive\"}],\"outputs\":[{\"label\":\"Induced voltage: changing flux\",\"expression\":\"n*flux\/t\",\"unit\":\"V\"}],\"checks\":[],\"expected\":20,\"id\":\"faraday\",\"title\":\"Induced voltage: changing flux\",\"section\":\"Magnetism\",\"formula\":\"|e| = N \u00d7 |\u0394\u03a6| \/ \u0394t\",\"explanation\":\"A changing magnetic flux induces a voltage in a coil. More turns or a faster change gives a larger induced voltage.\\n\\nThe magnitude formula does not state polarity. Lenz\u2019s law gives the opposing direction, represented by a minus sign in the signed Faraday equation.\",\"limits\":\"Magnitude of the average emf for the stated flux change; each turn links the same flux. It is not the average of the absolute emf if polarity reverses during the interval.\",\"mistake\":\"Forgetting that flux change is per turn.\",\"sourceIds\":[\"physics-flux\"]},{\"inputs\":[{\"key\":\"b\",\"label\":\"Flux density\",\"kind\":\"T\",\"example\":0.5,\"rule\":\"nonnegative\"},{\"key\":\"l\",\"label\":\"Length\",\"kind\":\"m\",\"example\":2,\"rule\":\"positive\"},{\"key\":\"speed\",\"label\":\"Perpendicular velocity\",\"kind\":\"speed\",\"example\":3,\"rule\":\"nonnegative\"}],\"outputs\":[{\"label\":\"Voltage from conductor motion\",\"expression\":\"b*l*speed\",\"unit\":\"V\"}],\"checks\":[],\"expected\":3,\"id\":\"motional-emf\",\"title\":\"Voltage from conductor motion\",\"section\":\"Magnetism\",\"formula\":\"e = B \u00d7 l \u00d7 v\",\"explanation\":\"A moving conductor cuts magnetic flux and develops an emf. Only the length within the field contributes.\\n\\nThis simple form needs mutually perpendicular field, motion and conductor. Other angles require the appropriate perpendicular component.\",\"limits\":\"Uniform field and mutually perpendicular geometry.\",\"mistake\":\"Using total wire length rather than active length.\",\"sourceIds\":[\"physics-flux\"]},{\"inputs\":[{\"key\":\"b\",\"label\":\"Flux density\",\"kind\":\"T\",\"example\":0.5,\"rule\":\"nonnegative\"},{\"key\":\"i\",\"label\":\"Current\",\"kind\":\"A\",\"example\":2,\"rule\":\"nonnegative\"},{\"key\":\"l\",\"label\":\"Length\",\"kind\":\"m\",\"example\":3,\"rule\":\"positive\"},{\"key\":\"theta\",\"label\":\"Angle between current and field (0\u2013180\u00b0)\",\"kind\":\"deg\",\"example\":90,\"rule\":\"angle180\"}],\"outputs\":[{\"label\":\"Force on a conductor\",\"expression\":\"b*i*l*sin(theta)\",\"unit\":\"N\"}],\"checks\":[],\"expected\":3,\"id\":\"magnetic-force\",\"title\":\"Force on a conductor\",\"section\":\"Magnetism\",\"formula\":\"F = B \u00d7 I \u00d7 l \u00d7 sin \u03b8\",\"explanation\":\"A current-carrying conductor in a magnetic field experiences a force. The effect is greatest at right angles to the field.\\n\\nThis relationship underpins motor force. Direction is separate from the magnitude calculation.\",\"limits\":\"Straight conductor in a uniform field.\",\"mistake\":\"Using cos \u03b8 instead of sin \u03b8 for this angle.\",\"sourceIds\":[\"physics-magnetic\"]},{\"inputs\":[{\"key\":\"v\",\"label\":\"Voltage\",\"kind\":\"V\",\"example\":230,\"rule\":\"nonnegative\"},{\"key\":\"i\",\"label\":\"Current\",\"kind\":\"A\",\"example\":10,\"rule\":\"nonnegative\"},{\"key\":\"pf\",\"label\":\"Power factor\",\"kind\":\"factor\",\"example\":0.8,\"rule\":\"fraction0\"}],\"outputs\":[{\"label\":\"Single-phase active power\",\"expression\":\"v*i*pf\",\"unit\":\"W\"}],\"checks\":[],\"expected\":1840,\"id\":\"single-phase-active-power\",\"title\":\"Single-phase active power\",\"section\":\"AC power\",\"formula\":\"P = V \u00d7 I \u00d7 PF\",\"explanation\":\"Active power is the rate at which electrical energy becomes work or heat. Voltage multiplied by current gives apparent power, and power factor identifies the active proportion.\\n\\nWrite the unknown first. To find current, rearrange I = P\/(V \u00d7 PF); convert kW into W before using volts and amperes.\",\"limits\":\"Use RMS voltage and current. The cos \u03c6 form assumes sinusoidal voltage and current; for distorted waveforms use measured true power factor, P\/S.\",\"mistake\":\"Do not treat shaft output kW as electrical input kW or use PF as a percentage.\",\"sourceIds\":[\"ac-schneider\"]},{\"inputs\":[{\"key\":\"v\",\"label\":\"Voltage\",\"kind\":\"V\",\"example\":230,\"rule\":\"nonnegative\"},{\"key\":\"i\",\"label\":\"Current\",\"kind\":\"A\",\"example\":10,\"rule\":\"nonnegative\"}],\"outputs\":[{\"label\":\"Single-phase apparent power\",\"expression\":\"v*i\",\"unit\":\"VA\"}],\"checks\":[],\"expected\":2300,\"id\":\"single-phase-apparent-power\",\"title\":\"Single-phase apparent power\",\"section\":\"AC power\",\"formula\":\"S = V \u00d7 I\",\"explanation\":\"Apparent power is useful when discussing transformer and supply loading. It includes the effect of current that does not contribute to active power.\\n\\nKeep VA separate from W. A 2 kVA load uses 2 kW only at unity power factor; current is found from I = S\/V.\",\"limits\":\"Use RMS voltage and current. The cos \u03c6 form assumes sinusoidal voltage and current; for distorted waveforms use measured true power factor, P\/S.\",\"mistake\":\"Do not call a VA result watts without accounting for power factor.\",\"sourceIds\":[\"ac-schneider\"]},{\"inputs\":[{\"key\":\"p\",\"label\":\"Power\",\"kind\":\"W\",\"example\":800,\"rule\":\"nonnegative\"},{\"key\":\"s\",\"label\":\"Apparent power\",\"kind\":\"VA\",\"example\":1000,\"rule\":\"positive\"}],\"outputs\":[{\"label\":\"Power factor from measured power\",\"expression\":\"p\/s\",\"unit\":\"factor\"}],\"checks\":[{\"expression\":\"p<=s\",\"message\":\"Active power must not exceed apparent power (use matching units).\"}],\"expected\":0.8,\"id\":\"power-factor\",\"title\":\"Power factor from measured power\",\"section\":\"AC power\",\"formula\":\"PF = P \/ S\",\"explanation\":\"Power factor compares active power with apparent power. An importing load with 8 kW and 10 kVA has a power factor of 0.8.\\n\\nWith harmonics present, true power factor includes distortion as well as phase displacement. A cos \u03c6 reading alone may not describe the complete load.\",\"limits\":\"Use RMS voltage and current. The cos \u03c6 form assumes sinusoidal voltage and current; for distorted waveforms use measured true power factor, P\/S.\",\"mistake\":\"Use matching prefixes: kW\/kVA or W\/VA, never W\/kVA.\",\"sourceIds\":[\"ac-schneider\"]},{\"inputs\":[{\"key\":\"p\",\"label\":\"Power\",\"kind\":\"W\",\"example\":800,\"rule\":\"nonnegative\"},{\"key\":\"s\",\"label\":\"Apparent power\",\"kind\":\"VA\",\"example\":1000,\"rule\":\"positive\"}],\"outputs\":[{\"label\":\"Reactive power and the power triangle\",\"expression\":\"sqrt(s^2-p^2)\",\"unit\":\"var\"}],\"checks\":[{\"expression\":\"p<=s\",\"message\":\"Active power must not exceed apparent power.\"}],\"expected\":600,\"id\":\"reactive-power-triangle\",\"title\":\"Reactive power and the power triangle\",\"section\":\"AC power\",\"formula\":\"Q = \u221a(S\u00b2 \u2212 P\u00b2)\",\"explanation\":\"The right-angled power triangle relates active, reactive and apparent power for sinusoidal conditions. Squaring matching units keeps the calculation consistent.\\n\\nThis equation gives a magnitude. State separately whether reactive power is inductive (lagging) or capacitive (leading); the square root does not identify its sign.\",\"limits\":\"Sinusoidal loads only for S\u00b2 = P\u00b2 + Q\u00b2. Harmonic distortion can add a distortion contribution; do not infer kvar from this triangle in that case.\",\"mistake\":\"Do not subtract S \u2212 P without first squaring both terms.\",\"sourceIds\":[\"ac-schneider\"]},{\"inputs\":[{\"key\":\"v\",\"label\":\"Line-to-line RMS voltage\",\"kind\":\"V\",\"example\":400,\"rule\":\"nonnegative\"},{\"key\":\"i\",\"label\":\"Line RMS current\",\"kind\":\"A\",\"example\":10,\"rule\":\"nonnegative\"},{\"key\":\"pf\",\"label\":\"Power factor\",\"kind\":\"factor\",\"example\":0.8,\"rule\":\"fraction0\"}],\"outputs\":[{\"label\":\"Balanced three-phase active power\",\"expression\":\"sqrt(3)*v*i*pf\",\"unit\":\"W\"}],\"checks\":[],\"expected\":5542.562584220408,\"id\":\"three-phase-active-power\",\"title\":\"Balanced three-phase active power\",\"section\":\"AC power\",\"formula\":\"P = \u221a3 \u00d7 V_L \u00d7 I_L \u00d7 PF\",\"explanation\":\"For a balanced sinusoidal load, one line current and the line-to-line voltage describe all three phases. The factor \u221a3 accounts for the three-phase relationship.\\n\\nThe result is already the total power; do not multiply it by three again. To find current, use I_L = P\/(\u221a3 \u00d7 V_L \u00d7 PF).\",\"limits\":\"Balanced sinusoidal load. For unbalanced loads calculate or measure each phase appropriately and sum active powers. Do not use a simple average current for a badly unbalanced installation.\",\"mistake\":\"Use line-to-line voltage here, not line-to-neutral voltage.\",\"sourceIds\":[\"ac-schneider\"]},{\"inputs\":[{\"key\":\"v\",\"label\":\"Line-to-line RMS voltage\",\"kind\":\"V\",\"example\":400,\"rule\":\"nonnegative\"},{\"key\":\"i\",\"label\":\"Line RMS current\",\"kind\":\"A\",\"example\":10,\"rule\":\"nonnegative\"}],\"outputs\":[{\"label\":\"Balanced three-phase apparent power\",\"expression\":\"sqrt(3)*v*i\",\"unit\":\"VA\"}],\"checks\":[],\"expected\":6928.203230275509,\"id\":\"three-phase-apparent-power\",\"title\":\"Balanced three-phase apparent power\",\"section\":\"AC power\",\"formula\":\"S = \u221a3 \u00d7 V_L \u00d7 I_L\",\"explanation\":\"A transformer or balanced three-phase load can be described in kVA. This equation calculates the apparent power associated with its line voltage and current.\\n\\nTo estimate rated line current from kVA, multiply kVA by 1000 and divide by \u221a3 \u00d7 V_L. This is a steady-state calculation, not a protective-device selection.\",\"limits\":\"Balanced sinusoidal three-phase conditions. Equipment overload, harmonics, starting and ambient ratings require separate checks.\",\"mistake\":\"Do not insert kVA directly when the formula expects VA.\",\"sourceIds\":[\"ac-schneider\"]},{\"inputs\":[{\"key\":\"v\",\"label\":\"Star winding RMS voltage\",\"kind\":\"V\",\"example\":230,\"rule\":\"nonnegative\"},{\"key\":\"i\",\"label\":\"Winding RMS current\",\"kind\":\"A\",\"example\":10,\"rule\":\"nonnegative\"}],\"outputs\":[{\"label\":\"Star connection: line and winding voltage\",\"expression\":\"sqrt(3)*v\",\"unit\":\"V\"},{\"label\":\"Line current\",\"expression\":\"i\",\"unit\":\"A\"}],\"checks\":[],\"expected\":398.37168574084177,\"id\":\"star-voltage\",\"title\":\"Star connection: line and winding voltage\",\"section\":\"AC power\",\"formula\":\"V_L = \u221a3 \u00d7 V_ph; I_L = I_ph\",\"explanation\":\"Each star-connected winding runs between a line and the star point. For a balanced supply and load, its voltage is the line voltage divided by \u221a3.\\n\\nLine current equals winding current in star. Winding voltage is especially important when interpreting motor nameplates; the correct connection must match the stated supply and rating.\",\"limits\":\"Balanced three-phase star load. A floating star point can move under unbalance; do not assume V_L\/\u221a3 then.\",\"mistake\":\"Never assume every 400 V motor should run in star. Read its actual winding and connection ratings.\",\"sourceIds\":[\"star-delta\"]},{\"inputs\":[{\"key\":\"i\",\"label\":\"Delta winding RMS current\",\"kind\":\"A\",\"example\":10,\"rule\":\"nonnegative\"},{\"key\":\"v\",\"label\":\"Winding RMS voltage\",\"kind\":\"V\",\"example\":400,\"rule\":\"nonnegative\"}],\"outputs\":[{\"label\":\"Delta connection: line and winding current\",\"expression\":\"sqrt(3)*i\",\"unit\":\"A\"},{\"label\":\"Line-to-line voltage\",\"expression\":\"v\",\"unit\":\"V\"}],\"checks\":[],\"expected\":17.32050807568877,\"id\":\"delta-current\",\"title\":\"Delta connection: line and winding current\",\"section\":\"AC power\",\"formula\":\"I_L = \u221a3 \u00d7 I_ph; V_L = V_ph\",\"explanation\":\"Each delta winding is connected across two lines, so it receives line-to-line voltage. The line current is a vector combination of winding currents.\\n\\nFor a balanced load, divide line current by \u221a3 to find winding current. The relationship is not a plain arithmetic addition of two winding currents.\",\"limits\":\"Balanced sinusoidal three-phase delta load. Switching star\/delta requires a correctly rated motor and starter; this equation alone is not a wiring instruction.\",\"mistake\":\"Do not multiply line current by \u221a3 when finding the smaller winding current.\",\"sourceIds\":[\"star-delta\"]},{\"inputs\":[{\"key\":\"p\",\"label\":\"Power\",\"kind\":\"W\",\"example\":10000,\"rule\":\"nonnegative\"},{\"key\":\"pf1\",\"label\":\"Initial lagging power factor\",\"kind\":\"factor\",\"example\":0.8,\"rule\":\"fraction\"},{\"key\":\"pf2\",\"label\":\"Target lagging power factor\",\"kind\":\"factor\",\"example\":1,\"rule\":\"fraction\"}],\"outputs\":[{\"label\":\"Power factor correction: required kvar\",\"expression\":\"p*(sqrt(1-pf1^2)\/pf1-sqrt(1-pf2^2)\/pf2)\",\"unit\":\"var\"}],\"checks\":[{\"expression\":\"pf2>=pf1\",\"message\":\"Target power factor must be at least the initial value for this correction model.\"}],\"expected\":7500,\"id\":\"power-factor-correction\",\"title\":\"Power factor correction: required kvar\",\"section\":\"AC power\",\"formula\":\"Q_c = P \u00d7 (tan \u03c6\u2081 \u2212 tan \u03c6\u2082)\",\"explanation\":\"Correction offsets part of an inductive load's reactive demand, reducing upstream current for the same active power. Calculate the original and desired reactive powers and subtract.\\n\\nUse acos to recover each angle before using tan, or calculate tan \u03c6 = \u221a(1\/PF\u00b2 \u2212 1). This is a sizing principle, not a complete capacitor-bank design.\",\"limits\":\"Sinusoidal lagging PF improved towards unity at constant kW. Harmonics, resonance, variable load, switching, discharge and overcompensation require specialist\/manufacturer assessment. Results are in var (divide by 1000 for kvar).\",\"mistake\":\"Do not subtract the power factors themselves and multiply by kW.\",\"sourceIds\":[\"pfc-schneider\"]},{\"inputs\":[{\"key\":\"f\",\"label\":\"Frequency\",\"kind\":\"Hz\",\"example\":50,\"rule\":\"positive\"},{\"key\":\"poles\",\"label\":\"Total poles (not pole pairs)\",\"kind\":\"count\",\"example\":4,\"rule\":\"poles\"}],\"outputs\":[{\"label\":\"Synchronous speed\",\"expression\":\"120*f\/poles\",\"unit\":\"rpm\"}],\"checks\":[],\"expected\":1500,\"id\":\"core-17\",\"title\":\"Synchronous speed\",\"section\":\"Motors and transformers\",\"formula\":\"n\u209b = 120 \u00d7 f \/ p\",\"explanation\":\"The rotating magnetic field speed depends on frequency and total pole count. More poles at the same frequency give lower field speed.\\n\\nAn induction motor normally runs below this speed in motoring operation. The difference is accounted for by slip, introduced later.\",\"limits\":\"Use total poles, not pole pairs; this is field speed, not automatically rotor speed.\",\"mistake\":\"Entering two pole pairs as two poles.\",\"sourceIds\":[\"physics-ac\",\"motor-abb\"]},{\"inputs\":[{\"key\":\"ns\",\"label\":\"Synchronous field speed\",\"kind\":\"rpm\",\"example\":1500,\"rule\":\"positive\"},{\"key\":\"nr\",\"label\":\"Rotor speed\",\"kind\":\"rpm\",\"example\":1440,\"rule\":\"signed\"}],\"outputs\":[{\"label\":\"Induction-motor slip\",\"expression\":\"(ns-nr)\/ns*100\",\"unit\":\"%\"}],\"checks\":[],\"expected\":4,\"id\":\"motor-slip\",\"title\":\"Induction-motor slip\",\"section\":\"Motors and transformers\",\"formula\":\"s% = (n_s \u2212 n_r) \/ n_s \u00d7 100\",\"explanation\":\"Slip measures how far rotor speed falls below field speed, relative to field speed. In normal induction motoring it is positive.\\n\\nTo work backwards, use n_r = n_s \u00d7 (1 \u2212 s). Convert a stated percentage such as 4% into 0.04 before using the reverse equation.\",\"limits\":\"Steady operating point of an induction machine. Slip varies with load; generating and braking conditions can have different signs or ranges.\",\"mistake\":\"Divide by synchronous speed, not rotor speed.\",\"sourceIds\":[\"motor-abb\"]},{\"inputs\":[{\"key\":\"power\",\"label\":\"Mechanical output power\",\"kind\":\"W\",\"example\":7500,\"rule\":\"nonnegative\"},{\"key\":\"speed\",\"label\":\"Actual shaft speed\",\"kind\":\"rpm\",\"example\":1500,\"rule\":\"positive\"}],\"outputs\":[{\"label\":\"Shaft torque from power and speed\",\"expression\":\"9550*(power\/1000)\/speed\",\"unit\":\"N\u00b7m\"}],\"checks\":[],\"expected\":47.75,\"id\":\"shaft-torque\",\"title\":\"Shaft torque from power and speed\",\"section\":\"Motors and transformers\",\"formula\":\"T \u2248 9550 \u00d7 P_out \/ n\",\"explanation\":\"Power combines turning force with rotational speed. For a given output power, a slower rotating shaft requires a greater torque.\\n\\nUse mechanical output kW and actual shaft speed. If only electrical input is known, account for motor efficiency before calculating torque.\",\"limits\":\"Steady rotation, n > 0. The constant 9550 is rounded. The equation cannot predict starting torque at zero speed; use the motor\/drive torque-speed data.\",\"mistake\":\"Do not use watts in a formula expecting kW or use synchronous speed instead of shaft speed.\",\"sourceIds\":[\"torque-abb\"]},{\"inputs\":[{\"key\":\"output\",\"label\":\"Mechanical output power\",\"kind\":\"W\",\"example\":800,\"rule\":\"nonnegative\"},{\"key\":\"input\",\"label\":\"Active electrical input power\",\"kind\":\"W\",\"example\":1000,\"rule\":\"positive\"}],\"outputs\":[{\"label\":\"Motor efficiency and losses\",\"expression\":\"output\/input\",\"unit\":\"factor\"},{\"label\":\"Motor losses\",\"expression\":\"input-output\",\"unit\":\"W\"}],\"checks\":[{\"expression\":\"output<=input\",\"message\":\"Output power must not exceed input power.\"}],\"expected\":0.8,\"id\":\"motor-efficiency\",\"title\":\"Motor efficiency and losses\",\"section\":\"Motors and transformers\",\"formula\":\"\u03b7 = P_out \/ P_in; P_loss = P_in \u2212 P_out\",\"explanation\":\"Efficiency compares useful shaft power with active electrical input. The difference becomes losses in the motor, including winding, iron and mechanical losses.\\n\\nUse the same power units on both sides. Power factor and efficiency are separate: power factor links active and apparent input power, while efficiency links input and output.\",\"limits\":\"Values must refer to the same operating point. Nameplate efficiency at rated load is not necessarily part-load efficiency.\",\"mistake\":\"Do not divide shaft kW by input kVA and call it efficiency.\",\"sourceIds\":[\"motor-current\"]},{\"inputs\":[{\"key\":\"power\",\"label\":\"Mechanical shaft output\",\"kind\":\"W\",\"example\":7500,\"rule\":\"nonnegative\"},{\"key\":\"v\",\"label\":\"Line-to-line RMS voltage\",\"kind\":\"V\",\"example\":400,\"rule\":\"positive\"},{\"key\":\"pf\",\"label\":\"Power factor\",\"kind\":\"factor\",\"example\":0.8,\"rule\":\"fraction\"},{\"key\":\"eta\",\"label\":\"Efficiency\",\"kind\":\"factor\",\"example\":0.9,\"rule\":\"fraction\"}],\"outputs\":[{\"label\":\"Three-phase motor current from output power\",\"expression\":\"power\/(sqrt(3)*v*pf*eta)\",\"unit\":\"A\"}],\"checks\":[],\"expected\":15.035163260146504,\"id\":\"motor-line-current\",\"title\":\"Three-phase motor current from output power\",\"section\":\"Motors and transformers\",\"formula\":\"I_L = P_out \u00d7 1000 \/ (\u221a3 \u00d7 V_L \u00d7 PF \u00d7 \u03b7)\",\"explanation\":\"Motor shaft ratings do not directly describe electrical current. First allow for efficiency to obtain electrical input power, then allow for power factor and the three-phase relationship.\\n\\nThis combined equation is a useful check against nameplate current. Retain unrounded values through the calculation and round only the final current.\",\"limits\":\"Balanced sinusoidal three-phase operation at the stated load. This is running current, not starting current. Do not use it alone to choose overloads, cables or protective devices; drive input current also differs from motor current.\",\"mistake\":\"Convert efficiency percentages to decimals and shaft kW to W.\",\"sourceIds\":[\"motor-current\"]},{\"inputs\":[{\"key\":\"vp\",\"label\":\"Primary winding RMS voltage\",\"kind\":\"V\",\"example\":230,\"rule\":\"nonnegative\"},{\"key\":\"np\",\"label\":\"Primary turns\",\"kind\":\"count\",\"example\":1000,\"rule\":\"integer\"},{\"key\":\"ns\",\"label\":\"Secondary turns\",\"kind\":\"count\",\"example\":100,\"rule\":\"integer\"}],\"outputs\":[{\"label\":\"Transformer voltage and turns ratio\",\"expression\":\"vp*ns\/np\",\"unit\":\"V\"}],\"checks\":[],\"expected\":23,\"id\":\"transformer-voltage-ratio\",\"title\":\"Transformer voltage and turns ratio\",\"section\":\"Motors and transformers\",\"formula\":\"V_s \/ V_p = N_s \/ N_p\",\"explanation\":\"In the ideal model, voltage follows the turns ratio. A secondary with fewer turns than the primary gives a lower voltage.\\n\\nRearrange before substituting: V_s = V_p \u00d7 N_s\/N_p, or N_s = N_p \u00d7 V_s\/V_p. Keep winding quantities distinct from three-phase line quantities.\",\"limits\":\"Ideal transformer approximation with alternating flux; real winding drops and regulation affect loaded voltage. Turns ratio does not establish safe isolation; an autotransformer can have the same ratio without isolation.\",\"mistake\":\"Do not assume winding voltage ratio always equals three-phase line voltage ratio for unlike connections.\",\"sourceIds\":[\"transformer-openstax\"]},{\"inputs\":[{\"key\":\"vp\",\"label\":\"Primary winding RMS voltage\",\"kind\":\"V\",\"example\":230,\"rule\":\"positive\"},{\"key\":\"vs\",\"label\":\"Secondary winding RMS voltage\",\"kind\":\"V\",\"example\":23,\"rule\":\"nonnegative\"},{\"key\":\"is\",\"label\":\"Secondary current\",\"kind\":\"A\",\"example\":10,\"rule\":\"nonnegative\"}],\"outputs\":[{\"label\":\"Ideal transformer current ratio\",\"expression\":\"vs*is\/vp\",\"unit\":\"A\"}],\"checks\":[],\"expected\":1,\"id\":\"transformer-current-ratio\",\"title\":\"Ideal transformer current ratio\",\"section\":\"Motors and transformers\",\"formula\":\"I_p \/ I_s = V_s \/ V_p = N_s \/ N_p\",\"explanation\":\"For an ideal transformer, a reduction in voltage is accompanied by a greater secondary current for the same transferred apparent power. Current ratio is inverse to voltage ratio when comparing the same sides.\\n\\nUse I_p = V_s \u00d7 I_s\/V_p to estimate the primary load-current component. Real transformers also take magnetising current and incur losses.\",\"limits\":\"Ideal single-phase model, negligible losses and magnetising current. It does not predict actual no-load current or permission to exceed winding ratings.\",\"mistake\":\"Do not multiply primary current by the voltage step-down ratio to find secondary current.\",\"sourceIds\":[\"transformer-openstax\"]},{\"inputs\":[{\"key\":\"flux\",\"label\":\"Received luminous flux\",\"kind\":\"lm\",\"example\":10000,\"rule\":\"nonnegative\"},{\"key\":\"a\",\"label\":\"Area\",\"kind\":\"m2\",\"example\":20,\"rule\":\"positive\"}],\"outputs\":[{\"label\":\"Illuminance from light reaching a surface\",\"expression\":\"flux\/a\",\"unit\":\"lx\"}],\"checks\":[],\"expected\":500,\"id\":\"average-illuminance\",\"title\":\"Illuminance from light reaching a surface\",\"section\":\"Lighting\",\"formula\":\"E_avg = \u03a6_received \/ A\",\"explanation\":\"Illuminance describes light arriving on a surface. Dividing the total received lumens by the area gives average illuminance, even though individual positions may be brighter or darker.\\n\\nStart by identifying whether a question gives emitted or received light. A luminaire's total emitted lumens do not all reach the chosen working plane.\",\"limits\":\"Use received flux over the stated area. Average illuminance alone does not establish uniformity, glare, task suitability or compliance.\",\"mistake\":\"Do not divide a lamp's rated emitted lumens by floor area without accounting for utilisation and maintenance.\",\"sourceIds\":[\"light-erco\"]},{\"inputs\":[{\"key\":\"iv\",\"label\":\"Luminous intensity in ray direction\",\"kind\":\"cd\",\"example\":1000,\"rule\":\"nonnegative\"},{\"key\":\"d\",\"label\":\"Source-to-surface distance\",\"kind\":\"m\",\"example\":2,\"rule\":\"positive\"}],\"outputs\":[{\"label\":\"Point-source illuminance and distance\",\"expression\":\"iv\/d^2\",\"unit\":\"lx\"}],\"checks\":[],\"expected\":250,\"id\":\"inverse-square-light\",\"title\":\"Point-source illuminance and distance\",\"section\":\"Lighting\",\"formula\":\"E = I_v \/ d\u00b2\",\"explanation\":\"For a sufficiently small source illuminating a perpendicular surface, illuminance falls with the square of distance. Doubling the distance reduces the direct illuminance to a quarter.\\n\\nUse candela in the relevant direction from the luminaire's photometric data. Distance must be squared; for an unknown distance, use d = \u221a(I_v\/E).\",\"limits\":\"Point-source approximation, direct light, surface perpendicular to incident ray, negligible absorption. Extended sources at close distances, reflections and tilted surfaces need fuller calculations.\",\"mistake\":\"Do not substitute lumens for candela, and do not use mounting height if the ray travels a longer slant distance.\",\"sourceIds\":[\"light-erco\"]},{\"inputs\":[{\"key\":\"lux\",\"label\":\"Target maintained illuminance\",\"kind\":\"lx\",\"example\":500,\"rule\":\"positive\"},{\"key\":\"a\",\"label\":\"Area\",\"kind\":\"m2\",\"example\":20,\"rule\":\"positive\"},{\"key\":\"flux\",\"label\":\"Initial luminaire output\",\"kind\":\"lm\",\"example\":4000,\"rule\":\"positive\"},{\"key\":\"uf\",\"label\":\"Utilisation factor\",\"kind\":\"factor\",\"example\":0.6,\"rule\":\"fraction\"},{\"key\":\"mf\",\"label\":\"Maintenance factor\",\"kind\":\"factor\",\"example\":0.8,\"rule\":\"fraction\"}],\"outputs\":[{\"label\":\"Lumen method: estimating luminaire numbers\",\"expression\":\"ceil(lux*a\/(flux*uf*mf))\",\"unit\":\"luminaires\"},{\"label\":\"Unrounded luminaire count\",\"expression\":\"lux*a\/(flux*uf*mf)\",\"unit\":\"count\"}],\"checks\":[],\"expected\":6,\"id\":\"lumen-method\",\"title\":\"Lumen method: estimating luminaire numbers\",\"section\":\"Lighting\",\"formula\":\"N = E_m \u00d7 A \/ (\u03a6 \u00d7 UF \u00d7 MF)\",\"explanation\":\"This estimates how many luminaires are needed to achieve a stated maintained average. The utilisation factor accounts for how emitted light reaches the working plane; maintenance allows for reductions with age and dirt.\\n\\nCalculate the unrounded count, then round upward. A designer must still check layout and distribution using actual photometric data.\",\"limits\":\"Preliminary general-lighting estimate. Use the project illuminance target and manufacturer utilisation data on the same lumen basis. Do not double-count optical losses. Check layout, uniformity, glare and spacing separately.\",\"mistake\":\"Do not round the number downward or use 80 instead of 0.80 for an 80% factor.\",\"sourceIds\":[\"lighting-eaton\"]},{\"inputs\":[{\"key\":\"lux\",\"label\":\"Initial illuminance\",\"kind\":\"lx\",\"example\":500,\"rule\":\"nonnegative\"},{\"key\":\"mf\",\"label\":\"Maintenance factor\",\"kind\":\"factor\",\"example\":0.8,\"rule\":\"fraction0\"}],\"outputs\":[{\"label\":\"Lighting maintenance factor\",\"expression\":\"lux*mf\",\"unit\":\"lx\"}],\"checks\":[],\"expected\":400,\"id\":\"lighting-maintenance\",\"title\":\"Lighting maintenance factor\",\"section\":\"Lighting\",\"formula\":\"E_m = E_initial \u00d7 MF\",\"explanation\":\"Initial light output declines over time and dirt affects the luminaire and room. Maintenance factor connects the initial design calculation with the maintained value expected under a stated maintenance plan.\\n\\nTo find the initial illuminance needed, divide the maintained target by MF. The result is greater than the target when MF is below one.\",\"limits\":\"Choose factors from actual product data, environment, replacement and cleaning plan. MF is not a universal constant. This does not establish a statutory target.\",\"mistake\":\"Do not multiply the maintained target by MF when finding the greater initial value required.\",\"sourceIds\":[\"mf-erco\"]},{\"inputs\":[{\"key\":\"flux\",\"label\":\"Luminaire luminous flux\",\"kind\":\"lm\",\"example\":4000,\"rule\":\"nonnegative\"},{\"key\":\"p\",\"label\":\"Power\",\"kind\":\"W\",\"example\":40,\"rule\":\"positive\"}],\"outputs\":[{\"label\":\"Luminaire luminous efficacy\",\"expression\":\"flux\/p\",\"unit\":\"lm\/W\"}],\"checks\":[],\"expected\":100,\"id\":\"luminaire-efficacy\",\"title\":\"Luminaire luminous efficacy\",\"section\":\"Lighting\",\"formula\":\"K = \u03a6 \/ P\",\"explanation\":\"Luminous efficacy measures emitted light per watt of electrical input. Compare complete-luminaire values on the same basis when evaluating products.\\n\\nA high lm\/W value does not guarantee good lighting on the task: distribution, glare, colour quality, controls and maintenance also matter. To recover flux, multiply efficacy by power.\",\"limits\":\"Use total luminaire output and total input, including driver losses. Do not compare LED-chip efficacy directly with complete-fitting efficacy.\",\"mistake\":\"Do not confuse luminous efficacy in lm\/W with efficiency expressed as a percentage.\",\"sourceIds\":[\"efficacy-thorn\"]},{\"inputs\":[{\"key\":\"length\",\"label\":\"Room length\",\"kind\":\"m\",\"example\":8,\"rule\":\"positive\"},{\"key\":\"width\",\"label\":\"Room width\",\"kind\":\"m\",\"example\":6,\"rule\":\"positive\"},{\"key\":\"height\",\"label\":\"Luminaire height above working plane\",\"kind\":\"m\",\"example\":2.5,\"rule\":\"positive\"}],\"outputs\":[{\"label\":\"Room index for utilisation data\",\"expression\":\"length*width\/(height*(length+width))\",\"unit\":\"number\"}],\"checks\":[],\"expected\":1.3714285714285714,\"id\":\"room-index\",\"title\":\"Room index for utilisation data\",\"section\":\"Lighting\",\"formula\":\"K_r = L \u00d7 W \/ [H_m \u00d7 (L + W)]\",\"explanation\":\"Room index summarises rectangular room geometry for selecting utilisation-factor data. A wide shallow space behaves differently from a narrow deep one.\\n\\nMeasure mounting height from the working plane, not automatically from the floor. Use the resulting index with the manufacturer's reflectance and utilisation tables.\",\"limits\":\"Rectangular-room lumen-method geometry. Correct reflectances and manufacturer photometric data are still required; the index alone does not give illuminance.\",\"mistake\":\"Do not substitute ceiling height unless the luminaires are at the ceiling and the working plane is at floor level.\",\"sourceIds\":[\"lighting-eaton\"]},{\"inputs\":[{\"key\":\"p\",\"label\":\"Power\",\"kind\":\"W\",\"example\":2300,\"rule\":\"nonnegative\"},{\"key\":\"v\",\"label\":\"Supply RMS voltage\",\"kind\":\"V\",\"example\":230,\"rule\":\"positive\"},{\"key\":\"pf\",\"label\":\"Power factor\",\"kind\":\"factor\",\"example\":0.8,\"rule\":\"fraction\"}],\"outputs\":[{\"label\":\"Design current from electrical input power\",\"expression\":\"p\/(v*pf)\",\"unit\":\"A\"}],\"checks\":[],\"expected\":12.5,\"id\":\"design-current\",\"title\":\"Design current from electrical input power\",\"section\":\"Installation design\",\"formula\":\"Ib = P \/ (U \u00d7 cos \u03c6)\",\"explanation\":\"Find the current that the load is expected to draw before selecting protection and conductors. Divide the electrical input power by voltage and power factor.\\n\\nFor a motor nameplate giving mechanical output, first divide output power by efficiency to obtain electrical input. Do not apply efficiency twice.\",\"limits\":\"Sinusoidal single-phase model. Non-linear loads require true power factor and manufacturer data. Current alone does not select a cable.\",\"mistake\":\"Putting kW into a formula expecting W, or using mechanical output as electrical input.\",\"sourceIds\":[\"d-power\"]},{\"inputs\":[{\"key\":\"rating\",\"label\":\"Protective-device rated current\",\"kind\":\"A\",\"example\":32,\"rule\":\"nonnegative\"},{\"key\":\"ca\",\"label\":\"Ambient correction factor\",\"kind\":\"number\",\"example\":0.94,\"rule\":\"positive\"},{\"key\":\"cg\",\"label\":\"Grouping correction factor\",\"kind\":\"number\",\"example\":0.8,\"rule\":\"positive\"}],\"outputs\":[{\"label\":\"Minimum tabulated current rating\",\"expression\":\"rating\/(ca*cg)\",\"unit\":\"A\"}],\"checks\":[],\"expected\":42.5531914893617,\"id\":\"cable-correction\",\"title\":\"Cable rating correction factors\",\"section\":\"Installation design\",\"formula\":\"It \u2265 In \/ (Ca \u00d7 Cg)\",\"explanation\":\"A tabulated cable rating applies to stated reference conditions. Correction factors reduce the usable rating in the actual installation.\\n\\nHere only ambient and grouping apply. Divide the protective-device rating by their product to obtain a minimum tabulated rating, then check the other design constraints.\",\"limits\":\"Ambient and grouping factors only. Enter applicable verified factors; other factors and fuse provisions may apply. Confirm the complete method, Ib \u2264 In \u2264 Iz, voltage drop and protective-device requirements. This does not select a cable size.\",\"mistake\":\"Multiplying In by the factors when finding the required tabulated rating.\",\"sourceIds\":[\"d-cable\"]},{\"inputs\":[{\"key\":\"mv\",\"label\":\"Circuit voltage-drop coefficient\",\"kind\":\"mvam\",\"example\":18,\"rule\":\"nonnegative\"},{\"key\":\"i\",\"label\":\"Current\",\"kind\":\"A\",\"example\":10,\"rule\":\"nonnegative\"},{\"key\":\"l\",\"label\":\"Length\",\"kind\":\"m\",\"example\":20,\"rule\":\"positive\"}],\"outputs\":[{\"label\":\"Voltage drop using mV\/A\/m\",\"expression\":\"mv*i*l\/1000\",\"unit\":\"V\"}],\"checks\":[],\"expected\":3.6,\"id\":\"tabulated-voltage-drop\",\"title\":\"Voltage drop using mV\/A\/m\",\"section\":\"Installation design\",\"formula\":\"\u0394U = m \u00d7 Ib \u00d7 L \/ 1000\",\"explanation\":\"A circuit voltage-drop table can already include the outgoing and return conductor effects. Multiply its applicable mV\/A\/m figure by current and one-way length.\\n\\nConvert millivolts to volts by dividing by 1000. Compare against the allocated project limit after including upstream circuits.\",\"limits\":\"Use the applicable published circuit mV\/A\/m value, temperature, phase arrangement and power-factor treatment. The coefficient must already include circuit conductor effects; enter one-way length.\",\"mistake\":\"Doubling route length when the tabulated figure already covers the circuit.\",\"sourceIds\":[\"d-drop\"]},{\"inputs\":[{\"key\":\"drop\",\"label\":\"Voltage drop\",\"kind\":\"V\",\"example\":4.6,\"rule\":\"nonnegative\"},{\"key\":\"v\",\"label\":\"Nominal voltage\",\"kind\":\"V\",\"example\":230,\"rule\":\"positive\"}],\"outputs\":[{\"label\":\"Voltage drop as a percentage\",\"expression\":\"100*drop\/v\",\"unit\":\"%\"}],\"checks\":[],\"expected\":2,\"id\":\"voltage-drop-percent\",\"title\":\"Voltage drop as a percentage\",\"section\":\"Installation design\",\"formula\":\"\u0394U% = 100 \u00d7 \u0394U \/ Un\",\"explanation\":\"Percentage drop expresses lost voltage relative to nominal circuit voltage. A 4.6 V drop on 230 V is 2%.\\n\\nA design may allocate part of its total voltage-drop allowance to a submain and the remainder to a final circuit. Use the same voltage basis throughout.\",\"limits\":\"Compare the result with the applicable project and regulatory allowance, including upstream drop. Equipment starting and operating requirements also matter.\",\"mistake\":\"Using 230 V for a line-to-line 400 V drop.\",\"sourceIds\":[\"d-drop\"]},{\"inputs\":[{\"key\":\"i\",\"label\":\"Current\",\"kind\":\"A\",\"example\":10,\"rule\":\"nonnegative\"},{\"key\":\"l\",\"label\":\"Length\",\"kind\":\"m\",\"example\":100,\"rule\":\"positive\"},{\"key\":\"r\",\"label\":\"Resistance per conductor\",\"kind\":\"ohmkm\",\"example\":1,\"rule\":\"nonnegative\"},{\"key\":\"x\",\"label\":\"Reactance per conductor\",\"kind\":\"ohmkm\",\"example\":0.1,\"rule\":\"nonnegative\"},{\"key\":\"pf\",\"label\":\"Power factor\",\"kind\":\"factor\",\"example\":0.8,\"rule\":\"fraction0\"}],\"outputs\":[{\"label\":\"Voltage drop with resistance and reactance\",\"expression\":\"2*i*(l\/1000)*(r*pf+x*sqrt(1-pf^2))\",\"unit\":\"V\"}],\"checks\":[],\"expected\":1.72,\"id\":\"rx-voltage-drop\",\"title\":\"Voltage drop with resistance and reactance\",\"section\":\"Installation design\",\"formula\":\"\u0394U = 2 I L (r cos \u03c6 + x sin \u03c6)\",\"explanation\":\"For a single-phase two-wire circuit, both conductors contribute impedance. Resolve resistance and reactance using the load power factor, then multiply by twice the one-way length and current.\\n\\nWith cos \u03c6 = 0.8, sin \u03c6 = 0.6 for a lagging load. A balanced three-phase circuit uses \u221a3 instead of 2 and gives line-to-line drop.\",\"limits\":\"Steady sinusoidal lagging-load approximation; r must match conductor temperature. Do not combine per-metre data with kilometres.\",\"mistake\":\"Adding r and x directly without the power-factor terms.\",\"sourceIds\":[\"d-drop\"]},{\"inputs\":[{\"key\":\"u\",\"label\":\"Nominal voltage to earth\",\"kind\":\"V\",\"example\":230,\"rule\":\"positive\"},{\"key\":\"cmin\",\"label\":\"Minimum-voltage factor\",\"kind\":\"factor\",\"example\":0.95,\"rule\":\"fraction\"},{\"key\":\"ia\",\"label\":\"Operating current at required disconnection time\",\"kind\":\"A\",\"example\":160,\"rule\":\"positive\"}],\"outputs\":[{\"label\":\"Loop impedance and operating current\",\"expression\":\"u*cmin\/ia\",\"unit\":\"\u03a9\"}],\"checks\":[],\"expected\":1.365625,\"id\":\"loop-disconnection\",\"title\":\"Loop impedance and operating current\",\"section\":\"Installation design\",\"formula\":\"Zs,max = U0 \u00d7 Cmin \/ Ia\",\"explanation\":\"Automatic disconnection depends on enough fault current flowing to operate the device within the required time. Obtain Ia from the applicable characteristic or accepted data.\\n\\nCmin allows for lower supply voltage in this calculation. It is distinct from a temperature allowance for a measured cold loop value.\",\"limits\":\"Obtain Cmin and Ia from the applicable standard and device data for the required disconnection time. A numerical limit does not demonstrate full compliance.\",\"mistake\":\"Using the device rated current In in place of Ia.\",\"sourceIds\":[\"d-looplimit\"]},{\"inputs\":[{\"key\":\"v\",\"label\":\"Voltage\",\"kind\":\"V\",\"example\":230,\"rule\":\"nonnegative\"},{\"key\":\"z\",\"label\":\"Complete fault-loop impedance\",\"kind\":\"ohm\",\"example\":0.2,\"rule\":\"positive\"}],\"outputs\":[{\"label\":\"Prospective fault current estimate\",\"expression\":\"v\/z\",\"unit\":\"A\"}],\"checks\":[],\"expected\":1150,\"id\":\"fault-current\",\"title\":\"Prospective fault current estimate\",\"section\":\"Installation design\",\"formula\":\"If = U \/ Zloop\",\"explanation\":\"Fault current is limited by the entire supply-and-return loop. In this simplified model divide the fault-driving voltage by that loop impedance.\\n\\nFor a line-to-earth fault use voltage to earth and the earth loop. A line-neutral fault uses its own return path; its impedance need not equal the earth loop.\",\"limits\":\"Simplified steady RMS estimate, not a breaking-capacity assessment. Maximum and minimum fault studies need appropriate voltage factors, sources and impedance data.\",\"mistake\":\"Using only the resistance of the damaged section as the complete loop.\",\"sourceIds\":[\"d-fault\"]},{\"inputs\":[{\"key\":\"i\",\"label\":\"Current\",\"kind\":\"A\",\"example\":1000,\"rule\":\"nonnegative\"},{\"key\":\"t\",\"label\":\"Time\",\"kind\":\"s\",\"example\":0.1,\"rule\":\"positive\"},{\"key\":\"k\",\"label\":\"Material\/temperature factor k\",\"kind\":\"k\",\"example\":115,\"rule\":\"positive\"}],\"outputs\":[{\"label\":\"Minimum thermal conductor area\",\"expression\":\"sqrt(i^2*t)\/k\",\"unit\":\"mm\u00b2\"}],\"checks\":[{\"expression\":\"t<=5\",\"message\":\"The adiabatic calculation is limited to fault durations of 5 seconds or less.\"}],\"expected\":2.749806661015982,\"id\":\"adiabatic-cpc\",\"title\":\"Adiabatic protective conductor calculation\",\"section\":\"Installation design\",\"formula\":\"S \u2265 \u221a(I\u00b2t) \/ k\",\"explanation\":\"The adiabatic model checks whether a conductor can absorb fault energy during a short fault without exceeding the temperature basis of k. Square current, multiply by time, take the square root and divide by k.\\n\\nThe calculated area is a minimum thermal value. Actual selection also needs the available conductor sizes, mechanical minima and all other applicable requirements.\",\"limits\":\"Use the correct k for conductor material, insulation and temperature limits within the adiabatic method validity. Current-limiting devices may require manufacturer let-through energy. Round up to an available size and check mechanical minima separately. Valid for fault durations up to 5 seconds; for current-limiting operation below 0.1 s, use verified manufacturer let-through data rather than assumed constant RMS current.\",\"mistake\":\"Using I \u00d7 t instead of I\u00b2t, or rounding conductor size down.\",\"sourceIds\":[\"d-adiabatic\"]},{\"inputs\":[{\"key\":\"k\",\"label\":\"Material\/temperature factor k\",\"kind\":\"k\",\"example\":115,\"rule\":\"positive\"},{\"key\":\"area\",\"label\":\"Conductor area\",\"kind\":\"mm2\",\"example\":2.5,\"rule\":\"positive\"},{\"key\":\"energy\",\"label\":\"Device let-through energy\",\"kind\":\"A2s\",\"example\":50000,\"rule\":\"nonnegative\"}],\"outputs\":[{\"label\":\"Conductor thermal withstand\",\"expression\":\"k^2*area^2\",\"unit\":\"A\u00b2s\"},{\"label\":\"Thermal withstand minus let-through\",\"expression\":\"k^2*area^2-energy\",\"unit\":\"A\u00b2s\"}],\"checks\":[],\"expected\":82656.25,\"id\":\"let-through\",\"title\":\"Protective-device let-through energy\",\"section\":\"Installation design\",\"formula\":\"I\u00b2t \u2264 k\u00b2S\u00b2\",\"explanation\":\"A current-limiting device changes the current waveform during interruption. Its published let-through energy can be compared directly with conductor withstand.\\n\\nUse manufacturer data for the actual device, voltage and prospective fault current. The simple constant-current I\u00b2 \u00d7 t estimate is not a substitute for those data.\",\"limits\":\"This is a thermal comparison only; device breaking capacity, selectivity and disconnection requirements remain separate.\",\"mistake\":\"Comparing an A\u00b2s energy figure with a current in A.\",\"sourceIds\":[\"d-energy\"]},{\"inputs\":[{\"key\":\"ra\",\"label\":\"Electrode and protective-conductor resistance\",\"kind\":\"ohm\",\"example\":100,\"rule\":\"positive\"},{\"key\":\"idn\",\"label\":\"RCD rated residual operating current\",\"kind\":\"A\",\"example\":0.03,\"rule\":\"positive\"},{\"key\":\"limit\",\"label\":\"Applicable touch-voltage limit\",\"kind\":\"V\",\"example\":50,\"rule\":\"positive\"}],\"outputs\":[{\"label\":\"Calculated RA \u00d7 I\u0394n\",\"expression\":\"ra*idn\",\"unit\":\"V\"},{\"label\":\"Arithmetic maximum resistance\",\"expression\":\"limit\/idn\",\"unit\":\"\u03a9\"},{\"label\":\"Touch-voltage limit minus calculated voltage\",\"expression\":\"limit-ra*idn\",\"unit\":\"V\"}],\"checks\":[],\"expected\":3,\"id\":\"tt-rcd\",\"title\":\"TT resistance and residual-current relationship\",\"section\":\"Installation design\",\"formula\":\"RA \u00d7 I\u0394n \u2264 UL\",\"explanation\":\"This relationship coordinates a TT protective arrangement with the applicable touch-voltage limit. Convert milliamps to amps before multiplying resistance by residual operating current.\\n\\nA numerical result below the limit is only one part of the assessment. Reliable electrode performance, RCD selection and required disconnection time must also be addressed.\",\"limits\":\"Enter the applicable touch-voltage limit. The arithmetic maximum RA is not a recommended electrode target or a complete compliance check. Verify RCD operation and disconnection time separately.\",\"mistake\":\"Treating 30 mA as 30 A, or assuming a high calculated maximum is a good electrode design.\",\"sourceIds\":[\"d-tt\"]},{\"inputs\":[{\"key\":\"i1\",\"label\":\"Connected group 1 current\",\"kind\":\"A\",\"example\":20,\"rule\":\"nonnegative\"},{\"key\":\"f1\",\"label\":\"Group 1 coincidence factor\",\"kind\":\"factor\",\"example\":0.8,\"rule\":\"fraction0\"},{\"key\":\"i2\",\"label\":\"Connected group 2 current\",\"kind\":\"A\",\"example\":30,\"rule\":\"nonnegative\"},{\"key\":\"f2\",\"label\":\"Group 2 coincidence factor\",\"kind\":\"factor\",\"example\":0.5,\"rule\":\"fraction0\"},{\"key\":\"i3\",\"label\":\"Connected group 3 current\",\"kind\":\"A\",\"example\":10,\"rule\":\"nonnegative\"},{\"key\":\"f3\",\"label\":\"Group 3 coincidence factor\",\"kind\":\"factor\",\"example\":1,\"rule\":\"fraction0\"}],\"outputs\":[{\"label\":\"Maximum demand with an agreed coincidence factor\",\"expression\":\"i1*f1+i2*f2+i3*f3\",\"unit\":\"A\"}],\"checks\":[],\"expected\":41,\"id\":\"demand-assumptions\",\"title\":\"Maximum demand with an agreed coincidence factor\",\"section\":\"Installation design\",\"formula\":\"Idemand = \u03a3(Igroup \u00d7 fgroup)\",\"explanation\":\"Not every load necessarily operates at full load at the same time. A documented demand model can apply an agreed coincidence factor to each group and sum the resulting currents.\\n\\nThe factor must reflect the actual use, controls and relevant guidance. It is not permission to apply an arbitrary reduction to a protective device or an individual continuously loaded circuit.\",\"limits\":\"Single-phase grouped-current model. Use justified project coincidence factors, not arbitrary diversity reductions. Phase imbalance and correlated loads need separate consideration. Enter zero current for unused groups.\",\"mistake\":\"Confusing a coincidence factor below 1 with a diversity factor defined as its reciprocal.\",\"sourceIds\":[\"d-diversity\"]},{\"inputs\":[{\"key\":\"ze\",\"label\":\"External loop impedance\",\"kind\":\"ohm\",\"example\":0.35,\"rule\":\"nonnegative\"},{\"key\":\"r\",\"label\":\"Circuit line + CPC resistance at design temperature\",\"kind\":\"ohm\",\"example\":0.5,\"rule\":\"nonnegative\"}],\"outputs\":[{\"label\":\"Calculated earth fault loop impedance\",\"expression\":\"ze+r\",\"unit\":\"\u03a9\"}],\"checks\":[],\"expected\":0.85,\"id\":\"earth-loop\",\"title\":\"Calculated earth fault loop impedance\",\"section\":\"Testing and verification\",\"formula\":\"Zs \u2248 Ze + (R1 + R2)\",\"explanation\":\"The fault loop includes the supply path and the circuit line and protective conductors. For a small circuit where reactance is negligible, add the external impedance to the circuit resistance.\\n\\nUse values on a consistent temperature basis. Compare with the relevant protective-device requirement only after confirming the applicable disconnection time and conditions.\",\"limits\":\"Approximation; larger cables can require complex impedance. Measured parallel paths can reduce Zs: do not subtract measured Ze from measured Zs to claim a continuity-test result.\",\"mistake\":\"Treating a cold R1 + R2 as the hot design value.\",\"sourceIds\":[\"d-loop\"]},{\"inputs\":[{\"key\":\"r1\",\"label\":\"Line resistance per metre\",\"kind\":\"mohmm\",\"example\":7.41,\"rule\":\"nonnegative\"},{\"key\":\"r2\",\"label\":\"CPC resistance per metre\",\"kind\":\"mohmm\",\"example\":12.1,\"rule\":\"nonnegative\"},{\"key\":\"l\",\"label\":\"Length\",\"kind\":\"m\",\"example\":20,\"rule\":\"positive\"}],\"outputs\":[{\"label\":\"Radial line and CPC resistance\",\"expression\":\"(r1+r2)*l\/1000\",\"unit\":\"\u03a9\"}],\"checks\":[],\"expected\":0.3902,\"id\":\"radial-continuity\",\"title\":\"Radial line and CPC resistance\",\"section\":\"Testing and verification\",\"formula\":\"R1 + R2 = (r1 + r2)L \/ 1000\",\"explanation\":\"Add the per-metre resistances of the line and circuit protective conductor, then multiply by the one-way circuit length. Both conductors are already included in the sum.\\n\\nThe calculated value helps interpret a continuity test. The test itself requires safe isolation and competent procedures; an expected value does not prove continuity.\",\"limits\":\"Use conductor resistances at the same stated temperature. Joints, test leads, spurs and cable variation affect readings. This calculation does not prove continuity.\",\"mistake\":\"Doubling the length after already adding line and CPC resistance.\",\"sourceIds\":[\"d-loop\"]},{\"inputs\":[{\"key\":\"r1\",\"label\":\"Line end-to-end resistance\",\"kind\":\"ohm\",\"example\":0.4,\"rule\":\"positive\"},{\"key\":\"rn\",\"label\":\"Neutral end-to-end resistance\",\"kind\":\"ohm\",\"example\":0.4,\"rule\":\"positive\"}],\"outputs\":[{\"label\":\"Ring continuity: equal line and neutral\",\"expression\":\"(r1+rn)\/4\",\"unit\":\"\u03a9\"}],\"checks\":[{\"expression\":\"abs(r1-rn)<=0.000000001\",\"message\":\"This constant-reading model requires equal line and neutral resistances.\"}],\"expected\":0.2,\"id\":\"ring-continuity\",\"title\":\"Ring continuity: equal line and neutral\",\"section\":\"Testing and verification\",\"formula\":\"Rcross = (r1 + rn) \/ 4\",\"explanation\":\"For an intact uniform ring with equal line and neutral resistances, opposite-end cross-connections give two equal resistance paths in parallel at every point. Their combined resistance is one quarter of the sum of the end-to-end values.\\n\\nThis lesson concerns line-neutral readings with r1 = rn. CPC combinations with unequal conductor resistance do not generally give a perfectly constant reading around the ring.\",\"limits\":\"Ideal ring without spurs or parallel paths; dead testing only. This calculation does not replace the complete ring test procedure and interpretation.\",\"mistake\":\"Claiming the same exact quarter-sum at every line-CPC point when conductor sizes differ.\",\"sourceIds\":[\"d-ring\"]},{\"inputs\":[{\"key\":\"v\",\"label\":\"Nominal battery voltage\",\"kind\":\"V\",\"example\":12,\"rule\":\"nonnegative\"},{\"key\":\"capacity\",\"label\":\"Battery capacity\",\"kind\":\"Ah\",\"example\":100,\"rule\":\"positive\"},{\"key\":\"usable\",\"label\":\"Usable fraction\",\"kind\":\"factor\",\"example\":0.8,\"rule\":\"fraction0\"}],\"outputs\":[{\"label\":\"Battery nominal and usable energy\",\"expression\":\"v*capacity*usable\",\"unit\":\"Wh\"}],\"checks\":[],\"expected\":960,\"id\":\"battery-energy\",\"title\":\"Battery nominal and usable energy\",\"section\":\"Energy systems\",\"formula\":\"Eusable \u2248 Vnom \u00d7 Q \u00d7 fusable\",\"explanation\":\"Nominal voltage multiplied by ampere-hours gives a nominal watt-hour estimate. Multiply by the stated usable fraction to estimate the DC energy available within the chosen operating window.\\n\\nCells in series increase voltage while ampere-hours stay the same; matched parallel strings increase ampere-hours. Actual usable energy depends on battery conditions and controls.\",\"limits\":\"First-order estimate. Manufacturer discharge curves, temperature, ageing, discharge rate, BMS limits and system losses determine delivered energy.\",\"mistake\":\"Calling Ah an energy unit without including battery voltage.\",\"sourceIds\":[\"d-battery\"]},{\"inputs\":[{\"key\":\"energy\",\"label\":\"Usable DC energy\",\"kind\":\"Wh\",\"example\":1000,\"rule\":\"positive\"},{\"key\":\"eta\",\"label\":\"Inverter efficiency\",\"kind\":\"factor\",\"example\":0.9,\"rule\":\"fraction\"},{\"key\":\"load\",\"label\":\"Steady AC load\",\"kind\":\"W\",\"example\":100,\"rule\":\"positive\"}],\"outputs\":[{\"label\":\"Estimated battery-backed runtime\",\"expression\":\"energy*eta\/load\",\"unit\":\"h\"}],\"checks\":[],\"expected\":9,\"id\":\"ups-runtime\",\"title\":\"Estimated battery-backed runtime\",\"section\":\"Energy systems\",\"formula\":\"t \u2248 EDC \u00d7 \u03b7 \/ Pload\",\"explanation\":\"Multiply usable battery energy by conversion efficiency to estimate the AC energy available. Divide by the steady load to estimate hours of operation.\\n\\nFor an actual UPS, use the manufacturer runtime curve for the battery configuration and load. A simple energy calculation is useful for understanding the relationship, not guaranteeing autonomy.\",\"limits\":\"Ignores changing efficiency, idle consumption and discharge-rate effects unless included in supplied energy\/efficiency. No guaranteed emergency-system runtime.\",\"mistake\":\"Using nominal energy when the question gives a smaller usable amount, or returning hours as minutes.\",\"sourceIds\":[\"d-ups\"]},{\"inputs\":[{\"key\":\"power\",\"label\":\"Useful output power\",\"kind\":\"W\",\"example\":900,\"rule\":\"nonnegative\"},{\"key\":\"eta\",\"label\":\"Conversion efficiency\",\"kind\":\"factor\",\"example\":0.9,\"rule\":\"fraction\"}],\"outputs\":[{\"label\":\"Conversion efficiency and losses\",\"expression\":\"power*(1\/eta-1)\",\"unit\":\"W\"},{\"label\":\"Input power\",\"expression\":\"power\/eta\",\"unit\":\"W\"}],\"checks\":[],\"expected\":100,\"id\":\"system-loss\",\"title\":\"Conversion efficiency and losses\",\"section\":\"Energy systems\",\"formula\":\"Ploss = Pout(1\/\u03b7 \u2212 1)\",\"explanation\":\"Efficiency compares useful output with input. Divide the required output by efficiency to obtain input power, then subtract output to obtain the loss.\\n\\nLoss appears mainly as heat and can affect enclosure cooling and energy use. Efficiency varies with operating point, so use data for the actual load.\",\"limits\":\"Steady-state supplied efficiency; not a full thermal design. Auxiliary loads must be accounted for consistently.\",\"mistake\":\"Subtracting the efficiency percentage directly from watts.\",\"sourceIds\":[\"d-efficiency\"]}],\"units\":[{\"kind\":\"V\",\"options\":[{\"label\":\"V\",\"factor\":1},{\"label\":\"mV\",\"factor\":0.001},{\"label\":\"kV\",\"factor\":1000}]},{\"kind\":\"A\",\"options\":[{\"label\":\"A\",\"factor\":1},{\"label\":\"mA\",\"factor\":0.001},{\"label\":\"kA\",\"factor\":1000}]},{\"kind\":\"ohm\",\"options\":[{\"label\":\"\u03a9\",\"factor\":1},{\"label\":\"m\u03a9\",\"factor\":0.001},{\"label\":\"k\u03a9\",\"factor\":1000},{\"label\":\"M\u03a9\",\"factor\":1000000.0}]},{\"kind\":\"s\",\"options\":[{\"label\":\"s\",\"factor\":1},{\"label\":\"ms\",\"factor\":0.001},{\"label\":\"min\",\"factor\":60},{\"label\":\"h\",\"factor\":3600}]},{\"kind\":\"h\",\"options\":[{\"label\":\"h\",\"factor\":1},{\"label\":\"min\",\"factor\":0.016666666666666666},{\"label\":\"s\",\"factor\":0.0002777777777777778}]},{\"kind\":\"W\",\"options\":[{\"label\":\"W\",\"factor\":1},{\"label\":\"kW\",\"factor\":1000},{\"label\":\"MW\",\"factor\":1000000.0}]},{\"kind\":\"Hz\",\"options\":[{\"label\":\"Hz\",\"factor\":1},{\"label\":\"kHz\",\"factor\":1000},{\"label\":\"MHz\",\"factor\":1000000.0}]},{\"kind\":\"m\",\"options\":[{\"label\":\"m\",\"factor\":1},{\"label\":\"mm\",\"factor\":0.001},{\"label\":\"km\",\"factor\":1000}]},{\"kind\":\"mm\",\"options\":[{\"label\":\"mm\",\"factor\":1},{\"label\":\"m\",\"factor\":1000}]},{\"kind\":\"m2\",\"options\":[{\"label\":\"m\u00b2\",\"factor\":1},{\"label\":\"mm\u00b2\",\"factor\":1e-06},{\"label\":\"cm\u00b2\",\"factor\":0.0001}]},{\"kind\":\"mm2\",\"options\":[{\"label\":\"mm\u00b2\",\"factor\":1},{\"label\":\"m\u00b2\",\"factor\":1000000.0}]},{\"kind\":\"F\",\"options\":[{\"label\":\"F\",\"factor\":1},{\"label\":\"\u00b5F\",\"factor\":1e-06},{\"label\":\"nF\",\"factor\":1e-09},{\"label\":\"mF\",\"factor\":0.001}]},{\"kind\":\"H\",\"options\":[{\"label\":\"H\",\"factor\":1},{\"label\":\"mH\",\"factor\":0.001},{\"label\":\"\u00b5H\",\"factor\":1e-06}]},{\"kind\":\"factor\",\"options\":[{\"label\":\"decimal\",\"factor\":1},{\"label\":\"%\",\"factor\":0.01}]},{\"kind\":\"temp\",\"options\":[{\"label\":\"\u00b0C\",\"factor\":1}]},{\"kind\":\"deg\",\"options\":[{\"label\":\"\u00b0\",\"factor\":1},{\"label\":\"rad\",\"factor\":57.29577951308232}]},{\"kind\":\"count\",\"options\":[{\"label\":\"count\",\"factor\":1}]},{\"kind\":\"number\",\"options\":[{\"label\":\"no unit\",\"factor\":1}]},{\"kind\":\"C\",\"options\":[{\"label\":\"C\",\"factor\":1},{\"label\":\"mC\",\"factor\":0.001}]},{\"kind\":\"rho\",\"options\":[{\"label\":\"\u03a9\u00b7m\",\"factor\":1},{\"label\":\"\u03a9\u00b7mm\u00b2\/m\",\"factor\":1e-06}]},{\"kind\":\"alpha\",\"options\":[{\"label\":\"per 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Expects window.TF_CALC = {name, h1, sub, calcs, run}. *\/\n(function(){\n'use strict';\nconst D=window.TF_CALC, root=document.getElementById('tfc');\nif(!D||!root)return;\nconst esc=s=>String(s==null?'':s).replace(\/[&<>\"']\/g,c=>({'&':'&amp;','<':'&lt;','>':'&gt;','\"':'&quot;',\"'\":'&#39;'}[c]));\nconst byId=Object.fromEntries(D.calcs.map(c=>[c.id,c]));\nconst groups=[];D.calcs.forEach(c=>{if(!groups.includes(c.group))groups.push(c.group)});\nconst 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';\nconst MX=window.TF_MATH||{};\nfunction cardMath(c){const m=MX[c.id];return m&&m.m&&m.m.length?`<span class=\"f tfc-cm\" role=\"math\" aria-label=\"${esc(c.formula)}\">${m.m[0]}<\/span>`:`<span class=\"f\">${esc(c.formula)}<\/span>`}\nfunction mathHTML(c){const m=MX[c.id];if(!m||!m.m||!m.m.length)return `<div class=\"tfc-formula\">${esc(c.formula)}<\/div>`;\n const w=(m.w||[]).length?`<dl class=\"tfc-where\"><dt class=\"tfc-where-h\">where<\/dt>${m.w.map(x=>`<div><dt><math>${x[0].startsWith('<')?x[0]:`<mi>${esc(x[0])}<\/mi>`}<\/math><\/dt><dd>${esc(x[1])}<\/dd><\/div>`).join('')}<\/dl>`:'';\n return `<div class=\"tfc-math\" role=\"math\" aria-label=\"${esc(c.formula)}\">${m.m.join('')}${w}<\/div>`}\nlet filterGroup='All', query='';\nconst TRADES=[['Electrical','https:\/\/tradefoxapp.com\/calculators\/electrical\/'],['Gas and plumbing','https:\/\/tradefoxapp.com\/calculators\/gas-plumbing\/'],['Bricklaying','https:\/\/tradefoxapp.com\/calculators\/bricklaying\/'],['Carpentry','https:\/\/tradefoxapp.com\/calculators\/carpentry\/']];\nfunction localNav(){\n const others=TRADES.filter(t=>t[0].toLowerCase()!==D.name.toLowerCase());\n return `<nav class=\"tfc-local\" aria-label=\"Calculators\"><div class=\"tfc-wrap tfc-local-in\"><a class=\"tfc-local-title\" href=\"#\/\">${esc(D.name)} calculators<\/a><div class=\"tfc-local-links\"><a href=\"https:\/\/tradefoxapp.com\/calculators\/\">All trades<\/a>${others.map(t=>`<a href=\"${t[1]}\">${esc(t[0])}<\/a>`).join('')}<\/div><\/div><\/nav>`;\n}\n\nfunction fmt(v){\n if(!Number.isFinite(v))return '\u2014';\n if(v===0)return '0';\n const a=Math.abs(v);\n if(a>=1e9||a<1e-4){const [m,e]=v.toExponential(4).split('e');return Number(m)+' \u00d7 10'+String(Number(e)).replace(\/-\/,'\u207b').replace(\/\\d\/g,d=>'\u2070\u00b9\u00b2\u00b3\u2074\u2075\u2076\u2077\u2078\u2079'[d]);}\n return Number(v.toPrecision(6)).toLocaleString('en-GB',{maximumFractionDigits:6});\n}\nfunction title(t){try{document.title=t}catch(e){}}\n\nfunction listView(){\n const q=query.trim().toLowerCase();\n const hits=D.calcs.filter(c=>(filterGroup==='All'||c.group===filterGroup)&&(!q||(c.title+' '+c.group+' '+c.formula+' '+(c.intro||[]).join(' ')).toLowerCase().includes(q)));\n const sections=groups.map(g=>{const cs=hits.filter(c=>c.group===g);if(!cs.length)return '';return `<section class=\"tfc-group\"><h2 class=\"tfc-h2\">${esc(g)}<\/h2><div class=\"tfc-grid\">${cs.map(c=>`<a class=\"tfc-card\" href=\"#\/${encodeURIComponent(c.id)}\"><span class=\"k\">${esc(c.group)}<\/span><span class=\"t\">${esc(c.title)}<\/span>${cardMath(c)}<\/a>`).join('')}<\/div><\/section>`}).join('');\n return sections||`<p class=\"tfc-empty\">No calculators match \u201c${esc(query)}\u201d. 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