- 2069 Chaitra · 8 marks
Show i1 = i2 = -V/(R1+R2+R3) at t=0+
Short answer
i1 = i2 = −V/(R1+R2+R3), i3 = 0. vC1 = vC2 = 0, vC3 = V; vL1 = vL2 = −R3V/(R1+R2+R3). Inductor current and capacitor voltage continuity force single-loop behaviour.
In the network shown in the figure below the switch is changed from a to b at t = 0. Show that at t = 0⁺, i1 = i2 = −V/(R1 + R2 + R3) and i3 = 0. Also find the voltage across C1, C2, C3, L1 and L2 at t = 0⁺. [Figure: battery V with a switch whose pole is connected to capacitor C3 in series with R2; position a connects to the battery positive terminal and position b connects through R1 to the battery negative terminal. After R2 there is a node from which L1 goes down to the bottom wire, L2 goes along the top to the right node (to which C2 is connected down to the bottom wire), and R3 in series with C1 runs diagonally from that node to the bottom-right corner. Mesh currents: i1 (left loop with C3, R2, L1), i2 (L1 with R3 and C1), i3 (L2, C2 with R3 and C1).]
Assumptions (figure description): with the switch at a, the battery V is in series with C₃ and R₂ and the circuit has reached dc steady state. Node Y is the junction after R₂. L₁ goes from Y to ground, L₂ from Y to the right node Z, C₂ from Z to ground, and the R₃-C₁ branch from Y to ground. In position b, the pole connects through R₁ to the battery negative terminal (ground).
Conditions at t = 0⁻ (steady state, switch at a)
- C₃ blocks dc, so no current flows: all currents are zero (i_L1 = i_L2 = 0).
- With no current, L₁ makes Y equal to ground, and C₁ and C₂ have no charge: v_C1 = v_C2 = 0, v_C3 = V.
At t = 0⁺ (switch moved to b)
- Inductor currents stay zero: i_L1 = i₁ − i₂ = 0 and i_L2 = i₃ = 0.
- Capacitor voltages stay: v_C3 = V, v_C1 = v_C2 = 0.
So i₃ = 0 and i₁ = i₂ (the inductors act as open circuits). The only loop with current is C₃ (acting as a source V), R₁, R₂, R₃, C₁ (a short, 0 V):
V + (R₁ + R₂ + R₃) i = 0 (loop current i = i₁ = i₂, opposite to the charging direction)
i₁ = i₂ = −V/(R₁ + R₂ + R₃), i₃ = 0 ✓
Voltages at t = 0⁺
| Element | Voltage at 0⁺ |
|---|---|
| C₁ | 0 |
| C₂ | 0 |
| C₃ | V |
| L₁ | Y is at −R₃V/(R₁+R₂+R₃) with respect to ground, so v_L1 = −R₃V/(R₁+R₂+R₃) |
| L₂ | Z = ground (C₂ uncharged), so v_L2 = v_Y − v_Z = −R₃V/(R₁+R₂+R₃) |
Derivation of v_L1: the R₃-C₁ branch carries |i| = V/(R₁+R₂+R₃) and C₁ has no voltage, so the drop across R₃ is R₃|i|. This drop is also the voltage across L₁ (and across L₂ in series with the uncharged C₂), with Y negative with respect to ground.
Answer: i₁ = i₂ = −V/(R₁+R₂+R₃), i₃ = 0; v_C1 = v_C2 = 0, v_C3 = V, v_L1 = v_L2 = −R₃V/(R₁+R₂+R₃).
✓ Answer checked by Claude Opus 5.5 (high effort)
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