RC Circuits
A short lesson with diagrams, a worked example, a problem to try, and a quick check. Then quiz yourself on 9 practice questions.
Charging a capacitor through a resistor follows a differential equation from the loop rule: ε − R dQ/dt − Q/C = 0. Its solution is exponential, Q = Cε(1 − e−t/RC), with time constant τ = RC.
1. Solving the loop equation
2. Deriving it and using it
Separate variables: dQ/(Cε − Q) = dt/(RC). Integrate from 0: −ln[(Cε − Q)/Cε] = t/RC, so Q = Cε(1 − e−t/RC).
Energy: charging stores ½Cε², and the resistor dissipates another ½Cε², whatever R is. The battery supplies Cε² in all.
Complicated circuits: at t = 0 an uncharged capacitor acts like a wire; after a long time, like a break.
When the switch opens, a capacitor discharges through whatever resistors are still connected to it: use those for τ.
3. Example, then try it
The graph below plots ln V against t for a capacitor discharging as V = V0e−t/RC. Use its slope to find RC, and then C if R = 10 kΩ.
Slope = −1/RC, so RC = 2.0 s.
C = 2.0 ÷ (10 × 10³) = 200 µF.
A 9.0 V battery charges a 40 µF capacitor through a 5.0 kΩ resistor, starting uncharged. Find τ, the initial current, the final charge, the time to reach half the final charge, and the energy dissipated in the resistor.
τ = RC = (5.0 × 10³)(40 × 10⁻⁶) = 0.20 s. I0 = ε/R = 9.0 ÷ 5000 = 1.8 mA.
Final charge: Cε = (40 µF)(9.0) = 360 µC.
Half charged: 1 − e−t/τ = ½ → t = τ ln 2 = (0.20)(0.693) ≈ 0.14 s.
Heat in R: ½Cε² = ½(40 × 10⁻⁶)(81) ≈ 1.6 mJ, equal to the energy stored.
Answer: τ = 0.20 s, 1.8 mA, 360 µC, 0.14 s, 1.6 mJ.
4. Common mistakes
In an RC charging circuit, the resistance is doubled. What happens?
τ = RC doubles, so it takes twice as long. The final charge Cε doesn’t depend on R.
Ready to practice?
9 questions on RC Circuits, each with a hint and an explanation for every wrong answer.