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Loop currents at t=0+ with two 1 F capacitors

Short answer

i1 = i2 = i3 = 20 A. Uncharged capacitors act as shorts, unfluxed inductor as open; all points have same potential so full 100 V across 5 Ω.

In the circuit shown in the following figure, find the loop currents i1, i2, i3 at t = 0⁺. [Figure: 100 V source in series with a 5 Ω resistor forming the outer loop (current i1). Inside, on the left, two 1 F capacitors stacked vertically between the top and bottom wires. A switch (t = 0) connects the junction of the two capacitors to the left end of a 10 Ω resistor; the right end of this 10 Ω resistor is a node joined to the top wire through a 1 H inductor and to the bottom wire through a 10 Ω resistor. Loop i2 is the upper loop (top 1 F capacitor, switch, 10 Ω, 1 H) and loop i3 is the lower loop (bottom 1 F capacitor, switch, 10 Ω, 10 Ω).]

Assumptions (figure description): the capacitors are uncharged and the inductor carries no current before t = 0. Take all mesh currents clockwise: i₁ in the source loop (100 V, 5 Ω), i₂ in the upper loop (top 1 F, switch, 10 Ω, 1 H) and i₃ in the lower loop (bottom 1 F, switch, 10 Ω, 10 Ω). Call the top wire T, the bottom wire B, the capacitor junction M and the node at the right end of the switched 10 Ω resistor X.

Equivalent circuit at t = 0⁺

  • Uncharged capacitors act as short circuits: T, M and B are at the same potential.
  • The inductor with zero initial current acts as an open circuit: i_L = 0.

Using the conditions

Current through the inductor (T to X): i_L = i₂ − i₁ = 0, so i₂ = i₁.

The two 10 Ω resistors are in series between M and B (through X). With M and B at the same potential, no voltage appears across the series pair, so

10(i₃ − i₂) + 10(i₃ − i₁) = 0  →  i₃ = (i₁ + i₂)/2 = i₁

So all three mesh currents are equal, and no current flows in the two 10 Ω resistors or the inductor.

The 100 V source now sees only the 5 Ω resistor, since the capacitors short the rest of the circuit:

i₁(0⁺) = 100/5 = 20 A

Answer: i₁ = i₂ = i₃ = 20 A at t = 0⁺ (the current circulates through the 5 Ω resistor and the two shorted capacitors).

✓ Answer checked by Claude Opus 5.5 (high effort)

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