Ohm's Law: How to Find Current, Voltage & Resistance

Ohm's law relates current, voltage, and resistance: I = ΔV/R, printed on the AP sheet in that form. To find current, divide the potential difference across a resistor by its resistance. Rearranged, ΔV = IR gives voltage and R = ΔV/I gives resistance. Power follows from P = IΔV.

AP Physics: Unit 11 (topics 11.3 Resistance, Resistivity, and Ohm's Law, 11.4 Electric Power). Ohm's law is topic 11.3 and electric power is topic 11.4 in AP Physics 2 Unit 11 (Electric Circuits, 15-18% exam weighting). The same circuit analysis appears in AP Physics C: Electricity and Magnetism Unit 11, weighted at 15-25%.

What Ohm's Law Says

Ohm's law says the current through a resistor equals the potential difference across it divided by its resistance. The AP Physics 2 equation sheet prints it as

I = \frac{\Delta V}{R}

where $I$ is current in amperes (A), $\Delta V$ is potential difference in volts (V), and $R$ is resistance in ohms ( $\Omega$ ). Most textbooks write the same relationship as $V = IR$ . The AP version uses $\Delta V$ to remind you that voltage is a difference in electric potential between two points, one on each side of the resistor.

The pairing matters. If you use the potential difference across one resistor, you get the current through that resistor. If you use the battery voltage with the total resistance of the circuit, you get the total current leaving the battery. Mixing a voltage from one part of the circuit with a resistance from another part is the most common Ohm's law error.

All Three Rearrangements

One equation, three uses. Solve for whichever quantity the problem hides:

Solve for	Equation	In words
Current	$I = \Delta V / R$	divide voltage by resistance
Voltage	$\Delta V = IR$	multiply current by resistance
Resistance	$R = \Delta V / I$	divide voltage by current

A quick unit check catches most algebra slips: one ohm is one volt per ampere, so amperes should come out as volts divided by ohms, and volts as amperes times ohms. If a problem gives current in milliamps, convert to amps first (1 mA = 0.001 A), or your resistance will be off by a factor of 1000.

You can practice all three forms with the Ohm's law calculator, which solves for the missing variable and shows the substitution step by step.

How to Find Current in a Circuit

To find current, work through three steps.

Identify the potential difference driving the current. For a simple circuit with one battery and one resistor, that is the battery voltage.
Find the resistance the current actually flows through. With one resistor, use its value directly. With several, reduce them to a single equivalent resistance first: series resistances add ( $R_{eq} = R_1 + R_2 + ...$ ), while parallel resistances combine through reciprocals. The series vs parallel circuits guide covers both reductions with examples.
Divide: $I = \Delta V / R$ .

Keep the voltage and resistance matched to the same piece of circuit. Battery voltage with equivalent resistance gives battery current. One resistor's voltage with that resistor's resistance gives that resistor's current. That single discipline prevents most circuit confusion before it starts.

Electric Power in a Resistor

The equation sheet gives electric power as $P = I \Delta V$ : power in watts equals current times potential difference. A resistor converts that electrical power into thermal energy, which is why chargers and laptops warm up while working.

Substituting Ohm's law into $P = I \Delta V$ produces two derived forms that save time on multiple-choice questions:

$P = I^2 R$ (substitute $\Delta V = IR$ ), useful when you know the current
$P = \frac{(\Delta V)^2}{R}$ (substitute $I = \Delta V / R$ ), useful when you know the voltage

Only $P = I \Delta V$ appears on the sheet; you derive the other two in one line. The squared forms explain a classic AP result: for resistors in series carrying the same current, the larger resistance dissipates more power, but for resistors in parallel sharing the same voltage, the smaller resistance dissipates more.

When a Resistor Is Ohmic

A resistor is ohmic when its resistance stays constant as the voltage across it changes. Plot current against potential difference for an ohmic resistor and you get a straight line through the origin with slope $1/R$ . Double the voltage, and the current doubles.

Plenty of real components are not ohmic. An incandescent bulb filament heats up as current increases, and hotter metal has higher resistivity, so its resistance climbs with voltage: the graph of $I$ versus $\Delta V$ bends over instead of staying straight. Diodes and LEDs are even further from linear.

This distinction shows up in lab-style AP questions. If an experiment gives you a table of voltage and current values, plot them. A straight line through the origin means the device is ohmic over that range, and the slope (or its reciprocal, depending on which axis holds which variable) gives the resistance.

Common Mistakes to Avoid

These errors account for most lost points on circuit problems.

Mismatched voltage and resistance. Using the full battery voltage with just one resistor's value in a multi-resistor circuit inflates the current. Reduce to an equivalent resistance, or use that one resistor's own voltage.
Milliamp slips. 250 mA is 0.250 A. Skipping the conversion makes resistances a thousand times too small.
Assuming everything is ohmic. A bulb's resistance at one operating point does not hold at other voltages. Apply a fixed $R$ only where the problem states the resistor is ohmic or gives a constant value.
Confusing through and across. Current flows through a resistor; voltage is measured across it. Ammeters go in series, voltmeters in parallel.
Dropping units. Carrying A, V, and $\Omega$ through every step catches errors early, and AP free-response graders expect them.

Current and Power in a Simple Circuit

A 9.0 V battery is connected across a single 45 $\Omega$ resistor. Ignore the battery's internal resistance. Find the current in the circuit and the power dissipated by the resistor.

Identify knowns: $\Delta V = 9.0\ \text{V}$ across the resistor and $R = 45\ \Omega$ .
Apply Ohm's law: $I = \Delta V / R = 9.0\ \text{V} / 45\ \Omega = 0.20\ \text{A}$ .
Apply the power equation from the sheet: $P = I \Delta V = (0.20\ \text{A})(9.0\ \text{V}) = 1.8\ \text{W}$ .
Check with a derived form: $P = I^2 R = (0.20\ \text{A})^2 (45\ \Omega) = (0.040)(45) = 1.8\ \text{W}$ . Both routes agree.

$I = 0.20$ A and $P = 1.8$ W.

Finding Resistance from Meter Readings

A voltmeter reads 6.0 V across a resistor while an ammeter in series with it reads 25 mA. Find the resistance and the power dissipated.

Convert the current to SI base units: $I = 25\ \text{mA} = 0.025\ \text{A}$ .
Rearrange Ohm's law for resistance: $R = \Delta V / I = 6.0\ \text{V} / 0.025\ \text{A} = 240\ \Omega$ .
Find the power: $P = I \Delta V = (0.025\ \text{A})(6.0\ \text{V}) = 0.15\ \text{W}$ .

$R = 240\ \Omega$ , dissipating $P = 0.15$ W as heat.

Current in a Two-Resistor Series Circuit

A 12 V battery drives current through a 4.0 $\Omega$ resistor and a 2.0 $\Omega$ resistor connected in series. Find the current and the potential difference across each resistor.

Series resistances add: $R_{eq} = 4.0\ \Omega + 2.0\ \Omega = 6.0\ \Omega$ .
Use the battery voltage with the equivalent resistance: $I = \Delta V / R_{eq} = 12\ \text{V} / 6.0\ \Omega = 2.0\ \text{A}$ . In series, this same 2.0 A flows through both resistors.
Voltage across the first resistor: $\Delta V_1 = I R_1 = (2.0\ \text{A})(4.0\ \Omega) = 8.0\ \text{V}$ .
Voltage across the second resistor: $\Delta V_2 = I R_2 = (2.0\ \text{A})(2.0\ \Omega) = 4.0\ \text{V}$ .
Check: $8.0\ \text{V} + 4.0\ \text{V} = 12\ \text{V}$ , matching the battery. Individual voltage drops in series must add to the total.

$I = 2.0$ A; $\Delta V_1 = 8.0$ V and $\Delta V_2 = 4.0$ V.

Frequently asked questions

Is V = IR the same as Ohm's law on the AP equation sheet?

Yes. The AP Physics 2 sheet prints Ohm's law as I = ΔV/R, which is V = IR solved for current. The delta emphasizes that voltage is a difference in electric potential between the two ends of the resistor. Use whichever arrangement isolates your unknown.

How do you find current in a circuit with more than one resistor?

Reduce the resistors to one equivalent resistance first. Series resistances add directly; parallel resistances combine through reciprocals. Then divide the battery voltage by the equivalent resistance to get the total current. The series vs parallel circuits guide walks through both reductions.

What does it mean for a resistor to be ohmic?

An ohmic resistor keeps a constant resistance as the voltage across it changes, so a graph of current versus voltage is a straight line through the origin. Non-ohmic devices such as incandescent bulb filaments, diodes, and LEDs change resistance with operating conditions, so a single fixed R only describes them at one point.

What units go with Ohm's law?

Current in amperes (A), potential difference in volts (V), and resistance in ohms. One ohm equals one volt per ampere. Watch for milliamps: convert 1 mA to 0.001 A before substituting, or your resistance will come out a factor of 1000 too small.

Where does Ohm's law show up on the AP exams?

In AP Physics 2 it is topic 11.3 (Resistance, Resistivity, and Ohm's Law) in Unit 11, Electric Circuits, which carries a 15-18% exam weighting; electric power is topic 11.4. AP Physics C: Electricity and Magnetism covers circuits in its own Unit 11, weighted at 15-25%.