Unit 11 · Topic 11.8

11.8 Resistor-Capacitor (RC) Circuits

Capacitors in circuits change what happens over time. This topic covers combining capacitors in series and parallel, then solving RC circuits, where the loop rule gives a differential equation and the charge and current change exponentially with time constant $\tau = RC$ .

Key terms

equivalent capacitance
RC circuit
time constant
exponential charging and discharging
steady state

Capacitors in series and parallel

Capacitors in parallel share the same potential difference, and their charges add, so their capacitances add: $C_{eq} = C_1 + C_2 + \cdots$ . It's like making one capacitor with bigger plates.

Capacitors in series each hold the same amount of charge Q, because the charge on the inner plates is only separated, not supplied. Their potential differences add, so $\dfrac{1}{C_{eq}} = \dfrac{1}{C_1} + \dfrac{1}{C_2} + \cdots$ . The series equivalent is smaller than the smallest capacitor.

Notice that these rules are the reverse of the rules for resistors.

Charging an RC circuit

Connect an uncharged capacitor C in series with a resistor R and a battery of emf $\mathcal{E}$ , then close the switch. The loop rule gives $\mathcal{E} - IR - \dfrac{q}{C} = 0$ . Since $I = \dfrac{dq}{dt}$ , that's a differential equation for the charge: $\mathcal{E} - R\dfrac{dq}{dt} - \dfrac{q}{C} = 0$ .

Separating variables and integrating from q = 0 at t = 0 gives:

$q(t) = C\mathcal{E}\left(1 - e^{-t/RC}\right), \qquad I(t) = \frac{\mathcal{E}}{R}e^{-t/RC}$

The time constant is $\tau = RC$ , in seconds. After one time constant the capacitor has about 63% of its final charge ( $1 - e^{-1} \approx 0.632$ ) and the current has dropped to about 37% of its starting value. After about 5τ, the capacitor is essentially full.

Discharging

A capacitor with charge $Q_0$ discharging through a resistor follows $\dfrac{q}{C} + R\dfrac{dq}{dt} = 0$ , so $q(t) = Q_0e^{-t/RC}$ . Its potential difference and the current both decay by the same factor: after one τ, about 37% remains.

The time for any of these to fall to half is $t_{1/2} = \tau\ln 2 \approx 0.69\tau$ .

Right away and after a long time

These two limits answer many questions with no exponentials at all:

Just after the switch closes, an uncharged capacitor has no potential difference across it, so it acts like a plain wire.
A long time later (steady state), the capacitor is fully charged and no current flows in its branch, so it acts like a break in the circuit. Its potential difference equals whatever the rest of the circuit puts across that branch.
A capacitor's potential difference, and so its charge, can't jump suddenly; the current through its branch can.

Where the energy goes

While charging a capacitor from empty, the battery supplies $Q\mathcal{E} = C\mathcal{E}^2$ of energy. The capacitor ends up storing $\tfrac{1}{2}C\mathcal{E}^2$ , and the resistor turns the other $\tfrac{1}{2}C\mathcal{E}^2$ into thermal energy, no matter what R is. A bigger R just makes the charging slower.

Worked examples

Try each one yourself first, then open the solution.

Example 1Calculator allowed
Charging a capacitor
A 9.0 V battery, a 50 kΩ resistor and an uncharged 20 μF capacitor are connected in series, and the switch is closed at t = 0. Find the time constant, the starting current, the final charge, and the charge at t = 2.0 s.
Show the solutionHide the solution
1. Step 1: $\tau = RC = (50 \times 10^3)(20 \times 10^{-6}) = 1.0\text{ s}$ .
2. Step 2: At t = 0 the capacitor acts like a wire: $I_0 = \dfrac{\mathcal{E}}{R} = \dfrac{9.0}{50 \times 10^3} = 1.8 \times 10^{-4}\text{ A}$ .
3. Step 3: Final charge: $Q_{max} = C\mathcal{E} = (20 \times 10^{-6})(9.0) = 1.8 \times 10^{-4}\text{ C}$ , or 180 μC.
4. Step 4: At t = 2.0 s, which is 2τ: $q = 180\left(1 - e^{-2}\right) \approx 156\ \mu\text{C}$ , about 86% of the final charge.
Answer: τ = 1.0 s, I₀ = 0.18 mA, $Q_{max}$ = 180 μC, q(2.0 s) ≈ 156 μC
Example 2Calculator allowed
Deriving the discharge equation
A capacitor C with initial charge $Q_0$ is connected across a resistor R at t = 0. Derive q(t).
Show the solutionHide the solution
1. Step 1: Loop rule, walking from the capacitor's positive plate through the resistor: $\dfrac{q}{C} - IR = 0$ .
2. Step 2: The capacitor is losing charge, so the current through the resistor is $I = -\dfrac{dq}{dt}$ . Substituting: $\dfrac{q}{C} + R\dfrac{dq}{dt} = 0$ .
3. Step 3: Separate variables: $\dfrac{dq}{q} = -\dfrac{dt}{RC}$ .
4. Step 4: Integrate from $Q_0$ at time 0 to q at time t: $\ln\dfrac{q}{Q_0} = -\dfrac{t}{RC}$ , so $q(t) = Q_0e^{-t/RC}$ .
Answer: $q(t) = Q_0e^{-t/RC}$
Example 3Calculator allowed
A capacitor in parallel with a resistor (classic trap)
A 12 V ideal battery is connected in series with R₁ = 4.0 Ω. After R₁, the circuit splits into two parallel branches: one with R₂ = 8.0 Ω, the other with an uncharged 5.0 μF capacitor. Find the current through R₁ just after the switch closes and a long time later, and the capacitor's final charge.
Show the solutionHide the solution
1. Step 1: Just after closing, the uncharged capacitor acts like a wire, which shorts out R₂. All the current goes through R₁ and the capacitor: I = 12/4.0 = 3.0 A, and R₂ carries none.
2. Step 2: A long time later, the capacitor's branch carries no current. The circuit is R₁ and R₂ in series: I = 12/(4.0 + 8.0) = 1.0 A.
3. Step 3: The capacitor is in parallel with R₂, so it has the same potential difference: (1.0 A)(8.0 Ω) = 8.0 V. Its charge is Q = CΔV = (5.0 μF)(8.0 V) = 40 μC.
4. Step 4: The trap is charging the capacitor to the full 12 V. It only gets the potential difference of the branch it's parallel to.
Answer: 3.0 A just after closing, 1.0 A after a long time, final charge 40 μC (8.0 V across the capacitor)

Common mistakes

Using the resistor rules for capacitors. Capacitors in parallel add directly; in series, add reciprocals.
Assuming a capacitor always charges to the battery's emf. In steady state it gets the potential difference of whatever it's parallel with.
Writing the charging current as rising. In a charging RC circuit the current is biggest at the start and decays.
Mixing up the 63% and 37% points. After one τ, a charging capacitor is 63% full and a discharging one keeps 37%.

On the exam

Free-response questions often ask you to derive q(t) or I(t) from the loop rule. Show the loop equation, the substitution I = dq/dt, the separation of variables and the limits of integration.
Graph questions are common: sketch q, I or $\Delta V_C$ against t, or linearize data by graphing ln I against t, whose slope is $-\dfrac{1}{RC}$ .
For "just after" and "long after" questions, redraw the circuit with the capacitor as a wire or as a break, then use ordinary circuit rules.

Connected topics

Videos

AP Physics C E&M - Unit 11 - Lesson 11C - RC Circuits
Allen Tsao The STEM CoachWatch on YouTube (opens in a new tab)
RC Circuit Basics
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AP Physics C - RC Circuit Charging
Dan Fullerton (APlusPhysics)Watch on YouTube (opens in a new tab)
RC Circuits Physics Problems, Time Constant Explained, Capacitor Charging and Discharging
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RC Circuits | Physics with Professor Matt Anderson | M22-13
Physics with Professor Matt AndersonWatch on YouTube (opens in a new tab)

Check yourself

5 questions on 11.8 Resistor-Capacitor (RC) Circuits. Pick an answer to see if you got it, and why.

Question 1 of 5Calculator allowed

A 3.0 μF capacitor and a 6.0 μF capacitor are connected in series to an ideal 12 V battery and fully charged. What is the potential difference across the 3.0 μF capacitor?

Question 2 of 5Calculator allowed

Two identical capacitors, each with capacitance C, are connected in parallel. That pair is then connected in series with a third identical capacitor. What is the equivalent capacitance?

Question 3 of 5Calculator allowed

A 2.0 kΩ resistor is connected in series with a 50 μF capacitor. What is the time constant of the circuit?

An ideal 12 V battery, a switch and resistor R₁ = 2.0 kΩ are in series. After R₁ the circuit splits into two parallel branches that rejoin before the battery. One branch is resistor R₂ = 4.0 kΩ. The other branch is resistor R₃ = 1.0 kΩ in series with an initially uncharged 5.0 μF capacitor.

Described circuit

Question 4 of 5Calculator allowed

What is the current in R₁ immediately after the switch is closed?

Question 5 of 5Calculator allowed

After the switch has been closed for a long time, what is the charge on the capacitor?

0 of 5 answered