- Asked 2/12
- 2068 Shrawan · 8 marks
Initial values of all currents and voltages at t=0+
Short answer
iL = 2 A, vC = 6 V, qC = 12 C; i3Ω = 2 A, v3Ω = 6 V; i1Ω = iC = 4 A, v1Ω = 4 V; vL = 4 V; i2Ω = 0, v2Ω = 0.
The switch has been open for a long time in the circuit shown below and at t = 0 it is suddenly closed. Find iL, vC, qC, i2Ω, i3Ω, i1Ω, iC, vL, v3Ω, v1Ω, v2Ω at t = 0⁺. [Figure: 10 V source in series with a 2 Ω resistor that has a switch (t = 0) connected in parallel across it. After that, two parallel branches: a 3 Ω resistor in series with a 4 H inductor, and a 2 F capacitor in series with a 1 Ω resistor.]
Circuit: 10 V source in series with 2 Ω, which has a switch connected across it. Then two parallel branches: 3 Ω + 4 H, and 2 F + 1 Ω. The switch has been open for a long time and closes at t = 0.
At t = 0⁻ (steady state: L short, C open)
i_L = 10/(2 + 3) = 2 A
v_C = voltage across the 3 Ω resistor = 3 × 2 = 6 V (no current in the C branch)
At t = 0⁺
The inductor current and capacitor voltage are continuous. The closed switch shorts the 2 Ω resistor, so the full 10 V is across the two parallel branches.
| Quantity | Value at 0⁺ | Working |
|---|---|---|
| i_L | 2 A | continuity |
| v_C | 6 V | continuity |
| q_C | 12 C | C·v_C = 2 × 6 |
| i₃Ω | 2 A | same as i_L |
| v₃Ω | 6 V | 3 × 2 |
| v_L | 4 V | 10 − 6 |
| i₁Ω | 4 A | (10 − 6)/1 |
| i_C | 4 A | same as i₁Ω |
| v₁Ω | 4 V | 1 × 4 |
| v₂Ω | 0 V | shorted by the switch |
| i₂Ω | 0 A | no current in the resistor (6 A flows through the switch) |
Check: source current = 2 + 4 = 6 A, all through the switch.
Answer: i_L = 2 A, v_C = 6 V, q_C = 12 C, i₃Ω = 2 A, v₃Ω = 6 V, i₁Ω = i_C = 4 A, v₁Ω = 4 V, v_L = 4 V, i₂Ω = 0, v₂Ω = 0.
✓ Answer checked by Claude Opus 5.5 (high effort)
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