Enter any two of voltage, current, resistance and power and the calculator gives the other two, with the formulas it used and the arithmetic written out the way an exam answer is. Units can be entered in millivolts, milliamps, kilohms or kilowatts. Below it, a second tool totals a list of resistors in series and in parallel, showing the reciprocal sum, and a third estimates the energy and cost of running a resistive load. All three are resistive DC arithmetic, which is also what the 309A formula sheet and the 310S electrical block ask for.
Ohm's law and power
Fill in any two of the four values. The other two are calculated. Change the unit beside a value if the question gives milliamps, kilohms or kilowatts.
Resistors in series and parallel
Energy cost of a resistive load
01Ohm's law in one paragraph
Ohm's law says that the current through a resistance is the voltage across it divided by that resistance: I = V / R. Double the voltage and the current doubles; double the resistance and the current halves. The same relation can be turned around to find whichever quantity is missing: V = I × R gives the voltage that a current develops across a resistance, and R = V / I gives the resistance from a measured voltage and current. The units are volts, amperes and ohms, and the law only works when all three are in those base units, so 250 mA goes in as 0.25 A and 4.7 kΩ as 4,700 Ω.
It is the first electrical relation both trades learn, and it is the one that every other electrical calculation on the exams rests on. Voltage drop, fuse and breaker sizing, heater and lamp ratings, series and parallel circuit analysis, the voltage-drop test on a starter cable and the resistance check on a sensor circuit are all Ohm's law with one extra step. The Red Seal formula sheet for Construction Electrician opens with it (as Current = E / R, with E for voltage), and the 310S standard's electrical block, at 23 of the 125 questions, is built on the same three quantities. If you can rearrange V = I × R without thinking, a large part of both exams becomes arithmetic.
02The power formulas and the twelve relations
Electrical power in a resistive circuit is voltage times current: P = V × I, in watts. Substitute Ohm's law into it and you get the two forms that are more useful in practice: P = I² × R, which gives the heat produced in a resistance from the current through it, and P = V² / R, which gives the power of a load from the voltage applied and its resistance. Each of the three power formulas can be rearranged for any of its quantities, which together with the three forms of Ohm's law gives twelve relations. Textbooks draw them as a wheel; the table below is the same twelve, grouped by what you are trying to find.
| To find | Given | Formula |
|---|---|---|
| Voltage (V) | I and R | V = I × R |
| P and I | V = P / I | |
| P and R | V = √(P × R) | |
| Current (I) | V and R | I = V / R |
| P and V | I = P / V | |
| P and R | I = √(P / R) | |
| Resistance (R) | V and I | R = V / I |
| V and P | R = V² / P | |
| P and I | R = P / I² | |
| Power (P) | V and I | P = V × I |
| I and R | P = I² × R | |
| V and R | P = V² / R |
You do not need to memorize all twelve. Learn the three base formulas, V = I × R, P = V × I and P = I² × R, and rearrange. The rule for picking one is simple: look at which two quantities the question gives you, find the row for the quantity it asks for, and use the formula whose Given column matches. The calculator does exactly that and prints the formula it chose, so it is a good way to check your choice before checking your arithmetic.
Two formulas on the 309A formula sheet are worth matching to this table. The sheet writes voltage as E rather than V, so Power = E² / R and Resistance = E² / P are the V² / R and V² / P rows above. The letters change between textbooks; the relations do not.
03Worked examples in exam style
Each of these is written the way a Red Seal question is: a situation, the numbers you need and one thing to find. Work them on paper first, then enter the given pair into the calculator and compare the working.
Example 1: current from voltage and resistance
Question. A 120 V circuit supplies a resistive load of 24 Ω. What current does the load draw?
Solution. Given V and R, find I: I = V / R = 120 / 24 = 5 A. The power of the load follows from P = V × I = 120 × 5 = 600 W, or from P = V² / R = 14,400 / 24 = 600 W; either formula gives the same answer, which is the check.
Example 2: resistance of a heater from its nameplate
Question. A baseboard heater is rated 4500 W at 240 V. What is the resistance of its element, and what current does it draw at rated voltage?
Solution. Given V and P, find R: R = V² / P = 240² / 4500 = 57,600 / 4500 = 12.8 Ω. Current: I = P / V = 4500 / 240 = 18.75 A. Check with Ohm's law: 18.75 × 12.8 = 240 V. A nameplate gives voltage and power, never resistance, so V² / P is the formula this kind of question is testing.
Example 3: power dissipated from current and resistance (310S)
Question. A blower motor circuit draws 20 A. The ground strap on the return side has a resistance of 0.05 Ω through a corroded bolt. How much voltage is lost across the strap, and how much power is turned into heat there?
Solution. Given I and R. Voltage: V = I × R = 20 × 0.05 = 1 V. Power: P = I² × R = 20² × 0.05 = 400 × 0.05 = 20 W. Check: P = V × I = 1 × 20 = 20 W. Twenty watts in a bolted joint is enough to warm it, and 1 V taken from a 12 V system is a large share of what the motor was supposed to get. The section on the automotive angle below takes this further.
Example 4: a series-parallel combination
Question. A 12 V source feeds R1 = 6 Ω in series with a parallel pair, R2 = 12 Ω and R3 = 6 Ω. Find the total resistance, the total current, the voltage across each resistor and the current through each.
Solution. Reduce the parallel pair first: R2 and R3 in parallel = (12 × 6) / (12 + 6) = 72 / 18 = 4 Ω. Total resistance = R1 + 4 = 6 + 4 = 10 Ω. Total current = V / R = 12 / 10 = 1.2 A. That whole current flows through R1, so the voltage across R1 = 1.2 × 6 = 7.2 V, and the voltage across the parallel pair = 1.2 × 4 = 4.8 V (check: 7.2 + 4.8 = 12 V). The parallel branches share that 4.8 V: current through R2 = 4.8 / 12 = 0.4 A, and through R3 = 4.8 / 6 = 0.8 A (check: 0.4 + 0.8 = 1.2 A). Total power = 12 × 1.2 = 14.4 W, which is also 1.2² × 6 + 0.4² × 12 + 0.8² × 6 = 8.64 + 1.92 + 3.84 = 14.4 W.
Example 5: milliamps
Question. A sensor circuit is supplied at 5 V and draws 25 mA. What is its resistance?
Solution. Convert first: 25 mA = 0.025 A. R = V / I = 5 / 0.025 = 200 Ω. Power: P = V × I = 5 × 0.025 = 0.125 W, or 125 mW. Leaving the current as 25 gives 0.2 Ω, which is wrong by a factor of a thousand and is usually one of the four options.
04Series and parallel rules, and the check that catches most errors
Series. Resistors in series carry the same current, and the total resistance is the sum: RT = R1 + R2 + R3 and so on. The source voltage divides across them in proportion to their resistance, so the biggest resistor drops the most voltage. The total is always larger than the largest resistor in the string.
Parallel. Resistors in parallel have the same voltage across them, and each branch draws its own current, I = V / R. The total current is the sum of the branch currents, which is why the reciprocals add: 1 / RT = 1 / R1 + 1 / R2 + 1 / R3, then RT = 1 divided by that sum. For exactly two resistors the product-over-sum shortcut is the same thing rearranged: RT = (R1 × R2) / (R1 + R2). For n equal resistors, RT = R / n.
The check. A parallel total is always less than the smallest branch, because adding a path can only let more current through. With 10, 20 and 30 Ω in parallel the total is 5.4545 Ω, less than 10. If your answer for a parallel group comes out larger than any resistor in it, you have added instead of taking reciprocals, or you forgot the final "1 divided by". It is the fastest sanity check in electrical arithmetic and it works on every parallel question on either exam.
Series-parallel. Reduce the circuit one group at a time: combine each parallel group into a single equivalent, then add the series parts. Find the total current, then work back outwards, using V = I × R for the voltage across each series element and I = V / R for the current in each parallel branch. Example 4 above shows the full sequence. Two checks at the end: the voltages across the series elements add up to the source, and the branch currents add up to the total.
05The automotive angle for 310S: why a small resistance matters at 12 V
Vehicle low-voltage systems run at a nominal 12 V, so there is very little voltage to spare. A resistance that would be invisible in a 120 V or 600 V circuit can take a noticeable share of 12 V, and Ohm's law shows why it depends on current, not only on the resistance itself. The drop across a connection is V = I × R: a fixed resistance drops ten times more voltage in a circuit carrying ten times the current.
Take a hypothetical corroded battery terminal with 0.02 Ω of resistance. In a lamp circuit drawing 2 A, the drop across it is 2 × 0.02 = 0.04 V, which no one would notice. Suppose the starter circuit through that same terminal draws 150 A while cranking. The drop is now 150 × 0.02 = 3 V, a quarter of the system voltage, and the heat in the joint is I² × R = 150² × 0.02 = 22,500 × 0.02 = 450 W. The starter gets 9 V instead of 12, cranks slowly or not at all, and the terminal gets hot. The lamp on the same terminal works normally. That pattern, a high-current circuit failing while low-current loads on the same connection work, is Ohm's law telling you where the resistance is.
It is also why an ohmmeter is the wrong tool for that fault. A meter pushes a very small test current through the connection, so 0.02 Ω reads as almost nothing and is lost in the meter's own lead resistance. Measuring the voltage across the connection while the circuit carries its working current shows the drop directly, which is the principle behind the voltage-drop test in the 310S standard's electrical diagnosis sub-tasks. On the exam, a question that gives a measured drop and a current is asking for R = V / I; one that gives a resistance and a current is asking for V = I × R; and one that asks why a connection is hot is asking for I² × R. No vehicle specification is needed for any of them, and the numbers above are illustrations, not specifications.
06The electrical angle for 309A: resistive arithmetic and where it stops
For a construction electrician, Ohm's law and the power formulas are the tools for resistive loads: electric heat, water heaters, incandescent and resistive lighting, and any DC circuit. A heater's nameplate power and voltage give its resistance and current (example 2); a known current through a known conductor resistance gives the voltage lost along it, which is what the voltage drop calculator does with a length and a conductor size instead of a resistance; and I² × R is why an undersized or loose connection heats up.
The Red Seal formula sheet for Construction Electrician lists exactly these relations, written with E for voltage: Current = E / R; Power = I² × R, E × I and E² / R; Resistance = E² / P. It is provided at the sitting, and the same page lists it in advance, so you do not have to memorize it, but you do have to know which line to use, and that comes from working problems. The sheet goes on to three-phase power, power factor and volt-amperes, which is where this tool stops.
That boundary matters. In an AC circuit with motors, transformers, ballasts or drives, current and voltage are not in step, the load has impedance rather than plain resistance, and true power is V × I × power factor rather than V × I; this calculator does not handle power factor, impedance or reactance, and it should not be used for them. For a resistive AC load, or for any DC circuit, the arithmetic here is complete and the answers match the formula sheet.
07Common mistakes
- Mixing milliamps and amps. The formulas need amperes. 25 mA is 0.025 A. Using 25 gives an answer a thousand times too small for resistance and a thousand times too large for voltage, and the wrong answer is usually among the options. The same goes for kΩ (multiply by 1,000) and kW (multiply by 1,000).
- Using P = V × I with the wrong pair. The V and I must belong to the same component. Multiplying the source voltage by a branch current gives the power of nothing in particular. In example 4, R2 dissipates 4.8 V × 0.4 A = 1.92 W, not 12 V × 0.4 A = 4.8 W.
- Forgetting to square. P = I² × R and P = V² / R. Doubling the current quadruples the heat. Writing I × R for power in example 3 gives 1 W instead of 20 W, which is the voltage drop, not the power.
- A parallel total larger than a branch. Adding parallel resistors instead of their reciprocals, or forgetting the final reciprocal, gives a total that is bigger than the smallest resistor. It never is. Check every parallel answer against the smallest branch.
- Series and parallel rules swapped. Current is the same everywhere in series; voltage is the same across every branch in parallel. A question that gives one branch's voltage in a parallel group has given you the voltage across all of them.
- Rounding too early. Keep at least four significant figures through the reciprocal sum. Rounding 1/30 to 0.03 turns 5.4545 Ω into 5.5556 Ω, which can move you to the wrong option on a question written to be close.
- Reading the wrong quantity from the question. A nameplate gives power and voltage, a meter gives voltage and current, a colour code or a data sheet gives resistance. Decide which two you have before choosing a formula, not after.
08What the 309A and 310S exams expect
Neither exam publishes topics at the question level, but both publish the block breakdown and the mix of question types, and both put electrical arithmetic squarely inside them.
309A Construction Electrician. Our 309A exam guide sets out the Red Seal Program's breakdown: 100 questions in four hours, 70 to pass, with generating, distribution and service systems at 28 questions, wiring systems at 30 and motors and control systems at 21. The program says 60 to 70% of the questions are procedural and application questions, which its preparation guide says can include calculations, and 20 to 30% are critical thinking questions that may need more than one step. A formula sheet listing the current, power and resistance relations is provided at the sitting, as is a calculator if one is needed; you cannot bring your own. Ohm's law questions on this exam tend to be one step inside a larger task: the current a heater draws before you size its circuit, the resistance of a load before you find its voltage drop, the power in a conductor before you decide whether it is overloaded.
310S Automotive Service Technician. Our 310S exam guide gives the breakdown: 125 questions in four hours, 88 correct to pass, with electrical and comfort control systems at 23 questions, tied for the heaviest block, and hybrid and electric vehicle systems at a further 9. The mix is recall 5 to 15%, procedural 40 to 50% and critical thinking 40 to 50%, and the preparation guide's examples of critical thinking questions give a symptom and ask for the cause. That is where Ohm's law appears on this exam: a slow-cranking starter with a measured drop, a lamp that is dim on one side, a circuit whose current is too high or too low for its resistance. The program provides the trade's acronym list at the sitting; whether a formula sheet is provided for the 310S is not stated, so know the three base formulas without one. A calculator is provided if needed.
For both, the preparation is the same: work problems until choosing the formula is automatic, convert units before you calculate, and run the checks (a second formula for the same answer, series voltages adding to the source, parallel totals below the smallest branch) every time. The free 309A questions and free 310S questions include electrical items in the same four-option format as the exams.
09Sources
- Red Seal Program: Construction Electrician trade page the trade the 309A exam certifies, with links to the occupational standard and the exam information page.
- Red Seal Program: Construction Electrician exam information the block breakdown and the formula sheet provided at the sitting, which lists the current, power and resistance relations used on this page.
- Red Seal Program: Automotive Service Technician trade page the trade the 310S exam certifies, with links to the 2023 occupational standard and the exam information page.
- TicketPrep: 309A Electrician Red Seal exam guide the block and task question counts, the question-type mix and what is provided in the exam room.
- TicketPrep: 310S Automotive Service Technician Red Seal exam guide the eight-block breakdown, the electrical block's sub-tasks, the question-type mix and what is provided in the exam room.
10Questions people ask
- What is Ohm's law?
- Ohm's law says the current through a resistance equals the voltage across it divided by the resistance: I = V / R. It can be rearranged as V = I x R to find voltage and R = V / I to find resistance. The units are volts, amperes and ohms, so convert milliamps to amps and kilohms to ohms before using it.
- How do I calculate electrical power from voltage and current?
- For a resistive load, power in watts is voltage times current: P = V x I. If you have current and resistance instead, use P = I squared x R. If you have voltage and resistance, use P = V squared / R. All three give the same answer for the same circuit, so using a second one is a good check.
- How do I find the resistance of a heater from its nameplate?
- Use R = V squared / P with the rated voltage and power. A 4500 W heater rated at 240 V has a resistance of 240 squared divided by 4500, which is 57,600 / 4500 = 12.8 ohms. Its current at rated voltage is P / V = 4500 / 240 = 18.75 A.
- Why is the total resistance of resistors in parallel less than the smallest one?
- Because every extra branch gives the current another path, so the group as a whole lets more current through than any single branch would. The total is 1 divided by the sum of the reciprocals: for 10, 20 and 30 ohms in parallel that is 1 / (0.1 + 0.05 + 0.0333) = 5.4545 ohms. If your parallel answer is larger than the smallest resistor, it is wrong.
- Does this calculator work for AC circuits?
- Only for resistive AC loads such as heaters and incandescent lighting, where voltage and current are in step. For motors, transformers, ballasts and drives the load has impedance and true power is V x I x power factor. The calculator does not handle power factor, impedance or reactance.
11Practice for this exam
Ohm's law is one step inside many questions on both exams. TicketPrep 309A and 310S practice tests follow the Red Seal Program's block-by-block breakdown, and every answer is explained, including why each wrong option is wrong. You can try free sample questions with no account.
Ten free questions per trade, no account: free practice questions. How our questions are written and checked: how our questions are made.