Electric circuits practice problems AP Physics C: E&M, Unit 11
Current, resistance and power, series and parallel circuits, Kirchhoff's rules, internal resistance, and RC circuits. Try each question before you look: use the hint if you are stuck, then reveal the worked solution one step at a time.
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Key equations for electric circuits
| V = IR | Ohm's law |
|---|---|
| P = IV = I2R | power |
| τ = RC | time constant of an RC circuit |
| q(t) = Cε(1 − e−t/RC) | charging capacitor |
| q(t) = Q0e−t/RC | discharging capacitor |
Constants as on the AP reference sheet: k = 9.0 × 109 N·m²/C², ε0 = 8.85 × 10−12 C²/(N·m²), μ0 = 4π × 10−7 T·m/A, e = 1.6 × 10−19 C. You can type answers like 3.6e5 or 3.6x10^5. The full sheet is in our AP Physics C equation sheet.
Warm-up questions
Quick checks on the core ideas. If these feel easy, go straight to the exam-style questions.
Hint
Ohm's law.
Worked solution
- I = V/R = 12/4.0 = 3.0 A.
Hint
In series, resistances add.
Worked solution
- R = 2.0 + 3.0 + 5.0 = 10 Ω.
Hint
In parallel, the reciprocals add.
Worked solution
- 1/R = 1/6 + 1/3 = 1/2.
- R = 2.0 Ω.
Exam-style questions
Questions at the level of the AP Physics C: E&M exam. The ones marked Challenge combine more than one idea.
Hint
P = I2R.
Worked solution
- P = 2.02 × 6.0 = 24 W.
Hint
τ = RC, in base units.
Worked solution
- τ = 2000 × 50 × 10−6.
- τ = 0.10 s.
Hint
After one time constant, the capacitor has reached 1 − e−1 of its final value.
Worked solution
- VC = 12(1 − e−1) = 12 × 0.632.
- VC = 7.59 V.
Hint
Find the current first, using the total resistance.
Worked solution
- I = 12/(5.5 + 0.50) = 2.0 A.
- Terminal voltage = ε − Ir = 12 − 2.0 × 0.50.
- = 11 V.
Hint
Voltage falls as e−t/RC. Half the voltage means one “half-life”.
Worked solution
- RC = 104 × 10−4 = 1.0 s.
- 4.5 = 9.0e−t/RC, so t = RC ln 2.
- t = 0.693 s.
Free-response practice
One original question in the style of the exam's free-response section. Answer it in full, with working, before you open the scoring guide.
Mathematical routines style · 7 points · about 20 minutes
An uncharged capacitor C is connected in series with a resistor R, a battery of emf ε and a switch. The switch is closed at t = 0.
- (a) Use Kirchhoff's loop rule to write a differential equation for the charge q on the capacitor.
- (b) Solve it to find q(t).
- (c) Find the current immediately after the switch closes, and after a long time.
- (d) Show that the energy dissipated in the resistor while charging equals the energy finally stored in the capacitor.
Write your full answer on paper first, then open the scoring guide and tick each point you earned.
Scoring guide and model answer
Part (a)
ε − R(dq/dt) − q/C = 0.
Part (b)
Separating: dq/(Cε − q) = dt/(RC), giving q(t) = Cε(1 − e−t/RC).
Part (c)
At t = 0 the capacitor acts like a wire: I = ε/R. After a long time it is fully charged: I = 0.
Part (d)
The battery supplies Qε = Cε2; the capacitor stores ½Cε2; so the resistor dissipates the other ½Cε2.
Your mark: 0 of 7
Common mistakes in electric circuits
- Using e−t/RC for a charging capacitor's voltage instead of 1 − e−t/RC.
- Forgetting internal resistance when a battery's terminal voltage is asked for.
- Leaving kΩ and μF unconverted when calculating RC.
Keep going
Revise the ideas behind these questions:
Then try real exam questions: AP Physics C: E&M past papers, with official scoring guidelines. · All AP Physics C: E&M units