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AP Physics 2 · Unit 11 · Topic 1 of 5 · 6 min lesson

Current & Resistance

A short lesson with diagrams, a worked example, a problem to try, and a quick check. Then quiz yourself on 5 practice questions.

The big idea

Current is the rate charge flows, I = ΔQ/Δt. Resistance is how hard a component makes it to push current through: R = ρL/A for a wire, and ΔV = IR. Current isn’t used up going around a circuit; energy is.

1. The key equations

I = ΔQ/Δtcurrent (A = C/s)
R = ρL/Alonger: more R; thicker: less R
ΔV = IROhm’s law
ΔVterminal = ε − Irreal battery with internal resistance r
ε r A R V Ammeter in series · voltmeter in parallel
Some of the battery’s emf is used up inside it, across r, so the voltage at its terminals is a little less than ε.

2. Wires, bulbs and meters

Resistance depends on area, not diameter: double the diameter and A is 4 times larger, so R is ¼ as large.

A resistor that obeys Ohm’s law has a straight-line I–V graph. A lightbulb doesn’t: its filament heats up, its resistance rises, and the graph curves.

An ammeter goes in series and has almost zero resistance, so it doesn’t change the current.

A voltmeter goes in parallel and has a huge resistance, so almost no current goes through it.

3. Example, then try it

Example

A nichrome wire (ρ = 1.1 × 10⁻⁶ Ω·m) is 2.0 m long with a cross-sectional area of 5.0 × 10⁻⁷ m². Find its resistance.

R = ρL/A = (1.1 × 10⁻⁶)(2.0) ÷ (5.0 × 10⁻⁷) = 4.4 Ω.

Try it · work it on paper first

A battery with ε = 9.0 V and internal resistance 0.50 Ω is connected to a 4.0 Ω resistor. Find the current, the terminal voltage, and the charge that passes in 30 s.

1

Current: the resistances are in series: I = ε ÷ (R + r) = 9.0 ÷ 4.5 = 2.0 A.

2

Terminal voltage: ε − Ir = 9.0 − (2.0)(0.50) = 8.0 V, the same as IR = (2.0)(4.0).

3

Charge: ΔQ = IΔt = (2.0)(30) = 60 C.

Answer: 2.0 A; 8.0 V at the terminals; 60 C in 30 s.

4. Common mistakes

“A thicker wire has more resistance.”Thicker means more area, so less resistance.
“Current gets used up.”The same current enters and leaves every component in series.
Mixing up the meters.Ammeter: series, tiny R. Voltmeter: parallel, huge R.
Check yourself

A wire of the same material and thickness is twice as long as another. How does its resistance compare?

R = ρL/A. Same ρ and A, double L, so double R.

Ready to practice?

5 questions on Current & Resistance, each with a hint and an explanation for every wrong answer.

Unit 11 topics

Every topic has a short lesson or quick review and its own quiz.