This AP Physics C lesson on RC circuits notes gives the key ideas first, then worked examples and diagrams. After that, practise with the worksheet on RC circuits and check yourself with the quiz.
Quick answer: Kirchhoff’s junction rule conserves charge and his loop rule conserves energy, giving equations for every branch current. In an RC circuit the loop rule ε − IR − q/C = 0 leads to q = Cε(1 − e^(−t/RC)) when charging and q = Q₀e^(−t/RC) when discharging, with time constant τ = RC.
Level: AP Physics C (Mechanics and E&M, calculus-based) · Curriculum: AP PC:E&M Unit 11
RC circuits notes: the key ideas
What are Kirchhoff’s rules and when do you use them?
Ohm’s law handles single resistors, but a circuit with several sources and branches needs two conservation laws. Also, one is about charge, the other about energy, and together they give enough equations to find every current.
Junction rule: the sum of currents into a junction equals the sum out of it (conservation of charge). Loop rule: the sum of potential changes around any closed loop is zero (conservation of energy).
Assign a current direction in every branch. Going round a loop, a battery contributes +ε when you cross from − to +, a resistor contributes −IR when you go with the current, and a capacitor contributes −q/C when you cross it from the positive plate to the negative plate. In addition, if a solved current comes out negative, it simply flows opposite to your guess.
Data: ε₁ = 12 V, R₁ = 2 Ω, ε₂ = 6 V, R₂ = 1 Ω and R₃ = 3 Ω as in the figure. Find the three currents.
Method: I₃ = I₁ + I₂. Left loop: 12 − 2I₁ − 3I₃ = 0, so 12 = 5I₁ + 3I₂. Right loop: 6 − I₂ − 3I₃ = 0, so 6 = 3I₁ + 4I₂. Solve the pair.
Answer: I₁ = 2.7 A, I₂ = −0.55 A (battery 2 is being charged) and I₃ = 2.2 A.
How does a capacitor charge through a resistor?
Close the switch in a series circuit with a battery ε, resistor R and capacitor C. At every instant the loop rule gives ε − IR − q/C = 0 with I = dq/dt. Moreover, this is a differential equation: dq/dt = (ε − q/C)/R. Separating variables and integrating from q = 0 gives an exponential approach to the final charge.
q(t) = Cε(1 − e−t/τ) I(t) = (ε/R) e−t/τ VC = ε(1 − e−t/τ) τ = RC
τ is the time constant (s) when R is in ohms and C is in farads. For discharge from Q₀: q = Q₀e−t/τ and I = (Q₀/τ)e−t/τ.
The time constant τ = RC is the time in which a charging capacitor reaches 63.2 % of its final charge, or a discharging one falls to 36.8 % of its initial charge.
Data: ε = 9.0 V, R = 20 kΩ and C = 47 μF, uncharged at t = 0. Find VC and I at t = 1.5 s.
Method: τ = RC = (2.0 × 10⁴)(47 × 10⁻⁶) = 0.94 s, so e−t/τ = e−1.596 = 0.203. Then VC = ε(1 − 0.203) and I = (ε/R)(0.203).
Answer: VC = 7.2 V and I = 91 μA.
How does a capacitor discharge and what is the time constant?
If the battery is removed and the capacitor discharges through R, the loop rule becomes q/C = IR with I = −dq/dt, so q decays as Q₀e−t/τ. After one time constant 36.8 % of the charge remains, and after 0.693τ half of it remains, because t½ = τ ln 2.
| Time | Charging: q/Cε | Charging: I/(ε/R) | Discharging: q/Q₀ |
|---|---|---|---|
| 0 | 0 | 1 | 1 |
| τ | 0.632 | 0.368 | 0.368 |
| 2τ | 0.865 | 0.135 | 0.135 |
| 3τ | 0.950 | 0.050 | 0.050 |
| 5τ | 0.993 | 0.007 | 0.007 |
Data: a 100 μF capacitor charged to 12 V discharges through 50 kΩ. Find the time for VC to fall to 3.0 V.
Method: τ = (5.0 × 10⁴)(1.0 × 10⁻⁴) = 5.0 s, and 3.0 = 12e−t/τ gives t = τ ln 4.
Answer: t = 6.9 s.
Energy accounting is neat. A battery supplies Cε² while charging, half is stored in the capacitor (½Cε²), and half is dissipated in the resistor, whatever the value of R. Integrating I²R from 0 to infinity confirms the resistor share.
Do not say the capacitor is fully charged after one time constant. At t = τ it holds only 63 % of its final charge. Treat it as essentially full after about 5τ.
An uncharged capacitor acts like a wire at t = 0, and a fully charged one acts like a break in the circuit. Solve those two limits first and the exponential fills the gap.
- Junction rule: ΣI in = ΣI out. Loop rule: Σ ΔV = 0 around any closed loop.
- Charging: q = Cε(1 − e^(−t/τ)), I = (ε/R)e^(−t/τ), τ = RC.
- Discharging: q = Q₀e^(−t/τ); 36.8 % remains at t = τ and half at t = 0.693τ.
- At t = 0 an uncharged capacitor is a wire; at t → ∞ it is an open circuit.
- Charging from a battery: Cε supplied, ½Cε² stored, ½Cε² lost as heat in R.
Kirchhoff's rules and RC transients: frequently asked questions
What does the time constant RC mean?
It is the time for a charging capacitor to reach 63.2 % of its final charge, or for a discharging one to fall to 36.8 % of its start. Similarly, a larger R or C makes the change slower.
Why does the current fall as a capacitor charges?
As q grows, the capacitor voltage q/C opposes the battery. The voltage across R, ε − q/C, shrinks, so I = (ε − q/C)/R decays exponentially to zero.
Where does the energy go when a capacitor is charged through a resistor?
The battery supplies Cε². Half is stored in the electric field as ½Cε² and the other half is dissipated as heat in the resistor, independent of the size of R.
How do I apply the loop rule with a capacitor?
Treat the capacitor as an element with potential difference q/C. Crossing it from its positive to its negative plate lowers the potential by q/C, so it enters like −q/C.
How long until a capacitor is considered fully charged?
After 5τ the charge is within 1 % of its final value, and after 7τ it is within 0.1 %. Engineers treat the circuit as steady after about five time constants.
