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Initial values with 20 Ω shorted by switch, 4 H, 2 F

Short answer

iL = 2 A, vC = 60 V, q = 120 C. 30 Ω: 2 A, 60 V; 10 Ω: 4 A, 40 V; 20 Ω: 0, 0; vL = 40 V. Switch shorts 20 Ω, using continuity of iL and vC.

The switch has been opened for a long time as shown in the figure below. At time t = 0, it is suddenly closed. At t = 0⁺, find current through inductor, voltage across capacitor, charge across capacitor, current and voltage across each resistor. [Figure: 100 V source in series with a 20 Ω resistor, which has a switch (t = 0) connected in parallel across it. After that, two parallel branches: a 30 Ω resistor in series with a 4 H inductor, and a 2 F capacitor in series with a 10 Ω resistor.]

Circuit: 100 V source in series with 20 Ω, which has a switch across it. The remaining circuit is two parallel branches: 30 Ω + 4 H, and 2 F + 10 Ω. The switch has been open for a long time, then closes at t = 0.

At t = 0⁻ (dc steady state, switch open)

  • Inductor = short, capacitor = open circuit.
  • Current: i_L = 100/(20 + 30) = 2 A.
  • No current flows in the 2 F + 10 Ω branch, so the capacitor voltage equals the voltage across the 30 Ω resistor: v_C = 30 × 2 = 60 V.

At t = 0⁺ (switch closed)

The inductor current and capacitor voltage cannot change suddenly. The closed switch short-circuits the 20 Ω resistor, so the full 100 V now appears across the two parallel branches.

QuantityValue
Inductor current i_L(0⁺)2 A
Capacitor voltage v_C(0⁺)60 V
Capacitor charge q = C v_C2 × 60 = 120 C
30 Ω resistori = 2 A, v = 60 V
Inductor voltage100 − 60 = 40 V
10 Ω resistori = (100 − 60)/10 = 4 A, v = 40 V
Capacitor current4 A
20 Ω resistorv = 0 V, i = 0 A (shorted by the switch)
Switch / source current2 + 4 = 6 A

Answer: i_L = 2 A, v_C = 60 V, q = 120 C; 30 Ω: 2 A, 60 V; 10 Ω: 4 A, 40 V; 20 Ω: 0 A, 0 V (shorted); v_L = 40 V.

✓ Answer checked by Claude Opus 5.5 (high effort)

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