AP Physics C: Electricity and Magnetism · Unit 11 · Topic 2 of 4 · 6 min lesson

Resistor Circuits

A short lesson with diagrams, a worked example, a problem to try, and a quick check. Then quiz yourself on 6 practice questions.

Quiz this topic · 6 questions

The big idea

Solve resistor circuits by combining series and parallel pieces, or with Kirchhoff’s rules when that doesn’t work. A real battery has internal resistance r: its terminal voltage drops to ε − Ir under load, and it delivers the most power when the load matches it, R = r.

1. Real batteries

I = ε/(R + r)battery with internal resistance r

ΔV_terminal = ε − Irdrops as current grows

P = I²Rpower to the load

Max power when R = rP_max = ε²/(4r)

Short circuit (R = 0)I = ε/r

Too little load resistance wastes power inside the battery; too much limits the current. The best match is R = r.

2. Tools for any circuit

Series: same I, R_eq = ΣR. Parallel: same ΔV, 1/R_eq = Σ1/R.

Kirchhoff: junction rule (charge conservation) and loop rule (energy conservation) handle circuits with several batteries.

Bulbs in series: same current, so P = I²R: the higher-resistance bulb glows brighter. A 40 W bulb (rated at 120 V, R = 360 Ω) outshines a 100 W bulb (R = 144 Ω) when they’re in series.

Power ratings assume the rated voltage; at other voltages, use the bulb’s resistance.

3. Example, then try it

Example

A battery with ε = 12 V and r = 1.0 Ω runs a 5.0 Ω resistor. Find the current, the terminal voltage, and the power to the resistor and lost inside the battery.

I = 12 ÷ 6.0 = 2.0 A; terminal voltage = 12 − (2.0)(1.0) = 10 V.

P_load = (2.0)²(5.0) = 20 W; P_internal = (2.0)²(1.0) = 4.0 W.

Try it · work it on paper first

A battery with ε = 12 V and r = 2.0 Ω powers an adjustable load R. Find the load power for R = 1.0, 2.0 and 4.0 Ω, then prove the maximum is at R = r.

R = 1.0 Ω: I = 12/3.0 = 4.0 A, P = 16 W. R = 2.0 Ω: I = 3.0 A, P = 18 W. R = 4.0 Ω: I = 2.0 A, P = 16 W.

General: P = ε²R/(R + r)².

Maximize: dP/dR = ε²[(R + r)² − 2R(R + r)]/(R + r)⁴ = 0 → R + r = 2R → R = r.

Max: P = ε²/(4r) = 144 ÷ 8.0 = 18 W. Only half the battery’s output reaches the load.

Answer: 16, 18 and 16 W; the maximum, 18 W, is at R = r = 2.0 Ω.

4. Common mistakes

Ignoring internal resistance.If r is given, the terminal voltage is less than ε.

“The 100 W bulb is always brighter.”In series, the higher-R (lower-wattage) bulb wins.

Maximum power = maximum efficiency.At R = r, efficiency is only 50%.

Check yourself

A battery with emf ε and internal resistance r is short-circuited by a thick wire (R ≈ 0). What current flows?

I = ε/(R + r) with R ≈ 0 gives ε/r. The internal resistance is the only thing limiting it, which is why shorted batteries get hot.

Ready to practice?

6 questions on Resistor Circuits, each with a hint and an explanation for every wrong answer.

Quiz this topic All of Unit 11