AP Physics C: Electricity and Magnetism · Unit 11 · Topic 3 of 4 · 5 min lesson

Capacitors in Circuits

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

Quiz this topic · 5 questions

The big idea

Capacitors combine opposite to resistors: parallel adds, series adds reciprocals. In series they all carry the same charge; in parallel they share the same voltage. Once a circuit settles, no current flows into any capacitor.

1. Combining and sharing

ParallelC = C₁ + C₂; same ΔV

Series1/C = 1/C₁ + 1/C₂; same Q

Series voltagesΔV_i = Q/C_i: smaller C gets more ΔV

Connecting charged to unchargedQ conserved; ends at a common ΔV

When a charged capacitor is connected to an empty one, charge flows until their voltages match. The total charge stays the same; some energy is lost.

2. Steady state

After a long time, every capacitor is fully charged and carries no current, like a break in the wire. Find the currents through the resistors first, then each capacitor’s voltage from whatever it’s connected across.

In charge sharing, energy isn’t conserved: some is lost as heat (and radiation) while charge rushes between the capacitors, no matter how small the wire’s resistance.

3. Example, then try it

Example

2.0 µF and 6.0 µF capacitors, initially uncharged, are connected in series to a 12 V battery. Find the charge and the voltage on each.

C_eq = (2.0)(6.0)/(8.0) = 1.5 µF, so Q = (1.5)(12) = 18 µC on each.

ΔV₂ = 18/2.0 = 9.0 V; ΔV₆ = 18/6.0 = 3.0 V. The smaller capacitor gets the larger voltage.

Try it · work it on paper first

A 4.0 µF capacitor charged to 10 V is disconnected and then connected across an uncharged 1.0 µF capacitor, + plate to one plate and − plate to the other (the figure below). Find the final voltage, each charge, and the energy before and after.

Charge to share: Q = (4.0)(10) = 40 µC.

Common voltage: now in parallel, C = 5.0 µF, so ΔV = 40/5.0 = 8.0 V. Charges: 32 µC and 8.0 µC.

Energy before: ½(4.0 × 10⁻⁶)(10)² = 200 µJ.

Energy after: ½(5.0 × 10⁻⁶)(8.0)² = 160 µJ. 40 µJ (20%) is lost.

Answer: 8.0 V; 32 µC and 8.0 µC; 200 µJ → 160 µJ.

4. Common mistakes

Using resistor rules for capacitors.They’re reversed.

Giving series capacitors equal voltages.Equal charge, so ΔV = Q/C differs.

Conserving energy in charge sharing.Conserve charge; energy is lost.

Check yourself

A circuit with resistors and a capacitor has been connected to a battery for a long time. How much current flows through the capacitor’s branch?

Once the capacitor is fully charged, its voltage matches what it’s connected across and no more charge flows into it.

Ready to practice?

5 questions on Capacitors in Circuits, each with a hint and an explanation for every wrong answer.

Quiz this topic All of Unit 11