Ohm's Law Practice Problems

Solve for voltage, current, resistance and power from any two known values — the calculation underneath almost everything else on the exam.

Practise Ohm's Law

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What it is, and why electricians need it

Ohm's law states that the current through a conductor is directly proportional to the voltage across it and inversely proportional to its resistance: V = I × R. Combine it with the power equation P = V × I and you have four quantities, any two of which give you the other two.

This is not an academic exercise. Every day on the job you work out the current a nameplate implies, check whether a coil's resistance is what it should be, or turn a voltage drop across a connection into the power being wasted there. The exam tests the same skill, usually by giving you two values and asking for a third — so the real difficulty is recognising which of the twelve formula arrangements fits the two values you actually have.

One relationship repays particular attention: power varies with the square of voltage or current, not linearly. Double the voltage across a fixed resistance and the current doubles but the power quadruples. That is why a modest overvoltage destroys heating elements so quickly, and why a splice carrying twice the current gets four times as hot.

Formulas and principles

  • V = I × R

    V in volts, I in amperes, R in ohms. The original statement of the law.

    Rearranges to I = V ÷ R and R = V ÷ I.

  • P = V × I

    P in watts. Gives real power only for a resistive load.

    For a motor or transformer this gives volt-amperes; multiply by power factor for watts.

  • P = I² × R

    Use when you know current and resistance but not voltage.

    This is the form that explains why a high-resistance connection overheats.

  • P = V² ÷ R

    Use when you know voltage and resistance. Rearranges to R = V² ÷ P.

    R = V² ÷ P is how you find the expected resistance of a heating element from its nameplate.

  • I = √(P ÷ R) and V = √(P × R)

    Use when power and resistance are the only values given.

Now try it on 18 questions

One question at a time, marked the moment you answer, with the reasoning and the source behind it. Anything you miss is saved so you can come back to it.

Start the Ohm's Law drill

Worked examples

Scenario: A 4,500-watt water heater element is rated 240 volts. It produces no heat. What resistance should a healthy element read, and what does an infinite reading mean?

  1. Expected resistance comes from R = V² ÷ P.
  2. R = 240² ÷ 4,500 = 57,600 ÷ 4,500.
  3. R = 12.8 ohms for a sound element.
  4. An infinite reading means the element is open internally and must be replaced.
  5. A reading near zero ohms to the sheath would instead mean the element is shorted to the tank.

Answer: 12.8 ohms expected; infinite means an open element

Scenario: A terminal in a disconnect is running hot. You measure 3.1 volts across it while the circuit carries 24 amperes. How much power is being dissipated there?

  1. The connection behaves as a resistance in series with the load.
  2. P = V × I = 3.1 V × 24 A.
  3. P = 74.4 watts, concentrated in a single lug.
  4. Resistance is R = V ÷ I = 3.1 ÷ 24 = about 0.13 ohms.
  5. A sound termination drops only hundredths of a volt, so this is a fire hazard requiring immediate repair.

Answer: About 74 watts, from a connection of roughly 0.13 ohms

Common mistakes

  • Treating volts times amps as watts for every load. It is watts only for a resistive load. For a motor, a transformer or a ballasted luminaire, volts times amps gives volt-amperes, and real power needs the power factor.
  • Reaching for the wrong arrangement of the formula. Check your units cancel to the unit you want. Volts squared divided by watts yields ohms; watts divided by volts yields amperes. A unit check catches most formula errors in a few seconds.
  • Assuming power scales linearly with voltage. It scales with the square. Raising voltage 10 percent raises power 21 percent, which is enough to shorten an element's life considerably.
  • Using line-to-line voltage when the load is connected line-to-neutral. A load across 208 volts line-to-line draws very different current from the same load on 120 volts line-to-neutral. Read what the question says the load is connected to.
  • Measuring resistance on an energized circuit. An ohmmeter supplies its own test current, so outside voltage gives a false reading and can destroy the meter. De-energize, and isolate the component from parallel paths that would read as lower resistance.

Related practice topics

Topics that come up alongside this one on the exam.

Questions about this topic

Why do I only need two values to solve an Ohm's law problem?

Because voltage, current, resistance and power are all mathematically related. Once you know any two, the remaining two follow. That is what the Ohm's law wheel expresses: twelve formulas, all derived from just V = I × R and P = V × I.

Does Ohm's law apply to AC circuits?

It applies to both AC and DC, but for AC you must use RMS values — which is what a standard meter reads. Where a load is reactive rather than resistive, as motors, transformers and ballasted luminaires are, volts times amps gives volt-amperes rather than watts, and you need the power factor to get real power.

What is the most useful Ohm's law formula on the job?

I = P ÷ V, because nameplates give watts and voltage and what you need is current. A 1,500-watt heater on 120 volts draws 12.5 amperes, which tells you immediately whether it fits a 15-ampere circuit. Second most useful is R = V ÷ I for checking whether a coil or element is sound.

Why does my meter reading not match the calculation?

Usually because the load is not purely resistive, or there is voltage drop in the wiring, or the resistance has changed with temperature. An incandescent filament measures a few ohms cold and about 144 ohms hot at 120 volts and 100 watts, which is why such a lamp draws a large inrush current for the first instant after switching on.

How is Ohm's law used in troubleshooting?

To turn a measurement into a diagnosis. Measure 3.1 volts across a terminal carrying 24 amperes and you have 0.13 ohms of resistance dissipating 74 watts in a single lug — a fire risk, not a nuisance. A healthy termination drops a few hundredths of a volt. The same reasoning identifies open elements, shorted windings and high-resistance splices.

How precise do my answers need to be?

Each exercise states a tolerance beneath the input box, matched to the precision the calculation warrants. Sensible rounding is accepted; a wrong method is not. On the real exam the offered options are usually far enough apart that rounding is not the risk — choosing the wrong formula is.

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