Capacitor Discharge in a Two-Loop DC Circuit
A 9.0 V battery with internal resistance 1.5 Ω is connected through a switch to a 10 Ω resistor R1 in series with a parallel pair R2 = 20 Ω and R3 = 30 Ω. A 47 μF capacitor is connected across R3. The switch has been closed for a long time. Then the switch is opened; it is in the battery branch, so the battery and R1 are disconnected from the capacitor, which remains connected only to R2 and R3. Assume ideal connecting wires. A student first predicts that no current flows anywhere in the steady state because the capacitor blocks direct current. The student then predicts that when the switch opens, the capacitor discharges through R1 as well as R2 and R3.
1. For the steady state with the switch closed, calculate the current through R1, the voltage across R3, and the charge on the capacitor.
Read the worked answer
In steady state the capacitor branch carries no current, but current flows through R1, R2, and R3. The parallel resistance is R_p = (20 × 30)/(20 + 30) = 12 Ω. Total resistance is R_tot = 1.5 + 10 + 12 = 23.5 Ω. Current I = 9.0/23.5 = 0.383 A.
The voltage across the parallel pair, and therefore across the capacitor, is V_c = I R_p = 0.383 × 12 = 4.60 V. Charge Q = C V_c = 47 × 10⁻⁶ × 4.60 = 2.16 × 10⁻⁴ C = 216 μC.
The student’s “no current anywhere” is wrong: only the capacitor branch has zero steady current; the resistor branches carry current.
2. Immediately after the switch opens, find the initial discharge current and the time constant. How long until the capacitor voltage falls to half its initial value?
Read the worked answer
After opening the switch, the battery and R1 are out of the circuit. The capacitor sees R2 and R3 in parallel: R_eq = 12 Ω. Initial discharge current I₀ = V_c/R_eq = 4.60/12 = 0.383 A. The time constant is τ = R_eq C = 12 × 47 × 10⁻⁶ = 5.64 × 10⁻⁴ s = 0.564 ms.
The voltage falls exponentially, so the half-voltage time is t₁/₂ = τ ln2 = 5.64 × 10⁻⁴ × 0.693 = 3.91 × 10⁻⁴ s = 0.391 ms.
3. Evaluate the student’s second prediction that R1 carries discharge current. What would change if the switch were placed so that R1 stayed connected across the capacitor?
Read the worked answer
The second prediction is wrong for the stated switch: opening the battery branch leaves R1 connected only to an open circuit, so no current can flow through R1. The discharge loop is the capacitor in parallel with R2 and R3.
If the switch were placed so that R1 remained connected across the capacitor, the discharge resistance would become R_eq′ = R1 + (R2 || R3) = 10 + 12 = 22 Ω. Then τ′ = 22 × 47 × 10⁻⁶ = 1.03 ms and the initial current would be 4.60/22 = 0.209 A. The circuit topology matters, not just the component values.