Unit 11 · Topic 11.8

11.8 Resistor-Capacitor (RC) Circuits

Capacitors combine the opposite way to resistors: they add directly in parallel and by reciprocals in series. In a circuit with resistors, a capacitor changes over time, acting like a wire when it's uncharged and like a break once it's fully charged. The time constant τ = RC tells you how fast that change happens.

Key terms

equivalent capacitance
RC circuit
time constant
charging and discharging
steady state

Capacitors in parallel and in series

Capacitors in parallel all have the same potential difference, and together they store more charge, like one capacitor with bigger plates:

$C_{\text{eq}} = C_1 + C_2 + \cdots \quad \text{(parallel)}$

Capacitors in series all hold the same charge Q. That follows from conservation of charge: the inner plates between two series capacitors are isolated, so whatever charge one gains, the other loses. Their potential differences add up to the total:

$\frac{1}{C_{\text{eq}}} = \frac{1}{C_1} + \frac{1}{C_2} + \cdots \quad \text{(series)}$

The equivalent capacitance of a series group is less than the smallest capacitor in it. These are the reverse of the resistor rules, so double-check which one you're using.

Charging and discharging

When an uncharged capacitor is first connected in a circuit, it acts like a plain wire: charge flows freely onto its plates, and there's no potential difference across it yet. As charge builds up, the potential difference across the capacitor grows and pushes back against the battery, so the current in its branch shrinks.

After a long time, the capacitor is fully charged. Its potential difference has reached its maximum and the current in its branch is zero, so the capacitor acts like a break in the circuit.

When a charged capacitor discharges through a resistor, its charge, its potential difference, its stored energy and the current all start at their largest values and fall toward zero.

Quantity	Charging (starts uncharged)	Discharging (starts charged)
Charge and ΔV on capacitor	start at 0, rise quickly, then level off	start at maximum, fall quickly, then level off near 0
Current in capacitor's branch	starts at its largest value, falls toward 0	starts at its largest value, falls toward 0
Capacitor acts like	a wire at first, a break at the end	a battery that runs down until no current flows

The time constant

How quickly a capacitor charges or discharges depends on the resistance R it charges through and its capacitance C. The time constant is:

$\tau = RC$

With R in ohms and C in farads, τ comes out in seconds. In one time constant, a charging capacitor reaches about 63% of its final charge, and a discharging capacitor falls to about 37% of its starting charge. Bigger R means less current to move the charge; bigger C means more charge to move. Either way it takes longer.

You'll describe the curves between the start and the end, but you won't calculate values at in-between times. The exam won't ask you to use exponential functions here.

Solving the start and the end

Two snapshots can be calculated exactly. Just after the switch closes, replace an uncharged capacitor with a wire. After a long time, remove the capacitor's branch entirely (no current there), find the potential difference across the capacitor's location, then use Q = CΔV and U = ½CΔV².

Worked examples

Try each one yourself first, then open the solution.

Example 1Calculator allowed
Series and parallel capacitors
A 3.0 μF and a 6.0 μF capacitor are connected to a 12 V battery. Find the equivalent capacitance, the charge on each, and the potential difference across each (a) in series and (b) in parallel.
Show the solutionHide the solution
1. Step 1: (a) Series: $\dfrac{1}{C_{\text{eq}}} = \dfrac{1}{3.0} + \dfrac{1}{6.0} = \dfrac{1}{2.0}$ , so $C_{\text{eq}}$ = 2.0 μF. Q = $C_{\text{eq}}\Delta V$ = (2.0 μF)(12 V) = 24 μC on each capacitor.
2. Step 2: Series potential differences: 3.0 μF: 24/3.0 = 8.0 V. 6.0 μF: 24/6.0 = 4.0 V. They add to 12 V. The smaller capacitor gets the larger share.
3. Step 3: (b) Parallel: $C_{\text{eq}}$ = 3.0 + 6.0 = 9.0 μF. Each has 12 V. Charges: (3.0)(12) = 36 μC and (6.0)(12) = 72 μC.
Answer: Series: 2.0 μF; 24 μC each; 8.0 V and 4.0 V. Parallel: 9.0 μF; 36 μC and 72 μC; 12 V each.
Example 2Calculator allowed
Just after and long after
A 12 V ideal battery, a switch and a 2.0 kΩ resistor R₁ are in series with a parallel pair: a 4.0 kΩ resistor R₂ and an uncharged 50 μF capacitor. Find the current through R₁ and R₂ just after the switch closes and a long time later. Then find the capacitor's final charge and stored energy.
Show the solutionHide the solution
1. Step 1: Just after: the uncharged capacitor acts like a wire, which shorts out R₂. So I₂ = 0, and the only resistance is R₁: I₁ = 12 V / 2000 Ω = 6.0 mA.
2. Step 2: Long after: the capacitor's branch has no current, so R₁ and R₂ are simply in series: I = 12 V / 6000 Ω = 2.0 mA through both.
3. Step 3: The capacitor is in parallel with R₂, so its potential difference is ΔV = (2.0 mA)(4000 Ω) = 8.0 V.
4. Step 4: Q = CΔV = (50 × 10⁻⁶)(8.0) = 4.0 × 10⁻⁴ C. U = ½QΔV = ½(4.0 × 10⁻⁴)(8.0) = 1.6 × 10⁻³ J.
Answer: Just after: I₁ = 6.0 mA, I₂ = 0. Long after: 2.0 mA in both. Q = 4.0 × 10⁻⁴ C, U = 1.6 × 10⁻³ J.
Example 3Calculator allowed
How charged after one time constant? (classic trap)
An uncharged 100 μF capacitor charges through a 10 kΩ resistor from a 9.0 V battery. Find the time constant. A student says the capacitor is fully charged after one time constant. Is that right? What is its charge after one time constant?
Show the solutionHide the solution
1. Step 1: τ = RC = (10 × 10³ Ω)(100 × 10⁻⁶ F) = 1.0 s.
2. Step 2: Final charge: $Q_{\text{max}}$ = CΔV = (100 × 10⁻⁶)(9.0) = 9.0 × 10⁻⁴ C.
3. Step 3: After one time constant, the capacitor has only about 63% of that: 0.63 × 9.0 × 10⁻⁴ ≈ 5.7 × 10⁻⁴ C. The student is wrong.
4. Step 4: The capacitor only approaches full charge. After several time constants it's so close that you can treat it as fully charged.
Answer: τ = 1.0 s. No: after 1.0 s it holds about 5.7 × 10⁻⁴ C, roughly 63% of the final 9.0 × 10⁻⁴ C.

Common mistakes

Using the resistor rules for capacitors. Capacitors add directly in parallel and by reciprocals in series.
Thinking a fully charged capacitor still carries current. Its branch has zero current in the long-time state.
Treating an uncharged capacitor as a break at the instant the switch closes. At first it acts like a wire.
Saying a capacitor is fully charged after one time constant. It's only about 63% charged.

On the exam

Expect to sketch qualitative graphs of charge, potential difference or current against time for charging and discharging, with the right starting values and shapes.
Start and end state calculations (just after the switch closes and after a long time) are the main quantitative skill here.
Lab questions may ask how to find a capacitance from a measured time constant, for example by timing how long the potential difference takes to fall to about 37% for several resistors and graphing τ against R (slope = C).

Connected topics

Videos

AP Physics 2 - Unit 11 - Lesson 11 - RC Circuits
Allen Tsao The STEM CoachWatch on YouTube (opens in a new tab)
RC Circuit Basics
Flipping PhysicsWatch on YouTube (opens in a new tab)
AP Physics - Capacitors in Steady State
Dan Fullerton (APlusPhysics)Watch on YouTube (opens in a new tab)
AP Physics 2 - Unit 11 - Lesson 10 - Series and Parallel Capacitance
Allen Tsao The STEM CoachWatch on YouTube (opens in a new tab)
Capacitors in series | Circuits | Physics | Khan Academy
khanacademymedicineWatch on YouTube (opens in a new tab)
Resistors and Capacitors
Bozeman ScienceWatch on YouTube (opens in a new tab)

Check yourself

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

Question 1 of 4Calculator allowed

A 2.0 μF capacitor and a 3.0 μF capacitor are first connected in parallel, and then in series. What are the two equivalent capacitances?

Question 2 of 4Calculator allowed

A 2.0 μF capacitor and a 6.0 μF capacitor are connected in series to a 12 V battery and fully charged. What is the potential difference across the 2.0 μF capacitor?

Question 3 of 4Calculator allowed

A 50 μF capacitor charges through a 10 kΩ resistor. What is the time constant of the circuit?

Question 4 of 4Calculator allowed

An uncharged capacitor is connected through a resistor to a battery. After one time constant, about what fraction of its final charge does the capacitor hold?

0 of 4 answered