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

RC Circuits

A short lesson with diagrams, a worked example, a problem to try, and a quick check. Then quiz yourself on 4 practice questions.

The big idea

A capacitor in a circuit changes over time. Just after the switch closes, an uncharged capacitor acts like a plain wire. After a long time, it’s fully charged and acts like a break in the wire: no current flows through it. The time constant τ = RC sets how fast it gets there.

1. Charging over time

Charge on C 63% τ t Qmax Current 37% τ t I0
The charge climbs quickly at first, then levels off. The current does the opposite.
t = 0, unchargedΔVC = 0: acts like a wire
t → ∞IC = 0: acts like an open switch
τ = RCtime constant (s)
After one τQ at 63% of max; I at 37% of start

2. Two snapshots solve most problems

At t = 0: replace each uncharged capacitor with a wire and solve the circuit normally.

After a long time: remove each capacitor’s branch (no current), solve the rest, then find the voltage across where the capacitor sits. Q = CΔV.

A bigger R slows the current; a bigger C needs more charge. Either one makes τ longer.

Discharging works the same way in reverse: the charge and current both decay, each falling to 37% after one τ.

3. Example, then try it

Example

A 12 V battery, a 4.0 Ω resistor and an uncharged capacitor are in series. Find the current just after the switch closes and after a long time.

At t = 0 the capacitor is like a wire: I = 12 ÷ 4.0 = 3.0 A.

After a long time: I = 0, and the capacitor has the full 12 V.

Try it · work it on paper first

A 12 V battery is connected in series with a 2.0 Ω resistor and then a parallel pair: a 4.0 Ω resistor and an uncharged 50 µF capacitor. Find the battery current just after the switch closes and after a long time, and the final charge on the capacitor.

1

t = 0: the capacitor acts like a wire and bypasses the 4.0 Ω resistor. I = 12 ÷ 2.0 = 6.0 A, all of it into the capacitor.

2

Long time: no current to the capacitor, so the 2.0 Ω and 4.0 Ω are in series: I = 12 ÷ 6.0 = 2.0 A.

3

Capacitor voltage: the same as the 4.0 Ω resistor beside it: (2.0)(4.0) = 8.0 V.

4

Charge: Q = CΔV = (50 × 10⁻⁶)(8.0) = 4.0 × 10⁻⁴ C (400 µC).

Answer: 6.0 A at first, 2.0 A later; 400 µC on the capacitor.

4. Common mistakes

Current through a charged capacitor.Once fully charged, no steady current flows through its branch.
“Fully charged after one τ.”Only 63%. It takes about 5τ to be essentially full.
Assuming the capacitor gets the full battery voltage.It gets the voltage of whatever it’s in parallel with.
Check yourself

A 2000 Ω resistor is in series with a 500 µF capacitor. What is the time constant?

τ = RC = (2000)(500 × 10⁻⁶) = 1.0 s.

Ready to practice?

4 questions on RC Circuits, 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.