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Classical: mesh currents and capacitor voltage, 50 V

Short answer

After switch closure, the capacitor sees a Thevenin resistance of 5 Ohms, giving a time constant of 10 microseconds. Final capacitor voltage is 25 V. v_C(t) = 25(1 - e^(-100000 t)) V, i2(t) = 5 e^(-100000 t) A, i1(t) = 2.5(1 + e^(-100000 t)) A.

In the two mesh network shown in the figure below, the switch is closed at t = 0. Find the mesh currents i1(t) and i2(t) as shown, and the capacitor voltage vc(t). [Use classical approach]. [Figure: 50 V source (+ at top) with a switch (t = 0) and a 10 Ω resistor in the top wire of the left mesh (i1); a 10 Ω resistor in the middle shared branch; a 2 μF capacitor on the right (right mesh current i2)]

Assumption: the capacitor is initially uncharged (switch open before t = 0). Both mesh currents are clockwise, the left mesh has 10 Ω (source branch) and the shared 10 Ω; the right mesh has the shared 10 Ω and the 2 μF capacitor.

Mesh equations (t > 0)

Mesh 1:  50 = 10 i1 + 10 (i1 − i2)   →  20 i1 − 10 i2 = 50
Mesh 2:  10 (i2 − i1) + vC = 0,   i2 = C dvC/dt

From mesh 1: i1 = 2.5 + 0.5 i2. Put in mesh 2:

vC + 10 (i2 − 2.5 − 0.5 i2) = 0 → vC + 5 i2 = 25 → 5C dvC/dt + vC = 25

Solution

Time constant τ = 5C = 5 × 2×10⁻⁶ = 10 μs (equivalent resistance seen by C is 10 Ω ∥ 10 Ω = 5 Ω).

Forced response: vC(∞) = 25 V. Natural: A e^(−t/τ). With vC(0) = 0, A = −25.

vC(t) = 25 (1 − e^(−10⁵ t)) V

i2 = C dvC/dt = (25 − vC)/5 = 5 e^(−10⁵ t) A

i1 = 2.5 + 0.5 i2 = 2.5 + 2.5 e^(−10⁵ t) A

Answer:

  • v_C(t) = 25 (1 − e^(−10⁵t)) V
  • i2(t) = 5 e^(−10⁵t) A
  • i1(t) = 2.5 (1 + e^(−10⁵t)) A

Check: at t = 0 the capacitor is a short, so i1 = 50/10 = 5 A and i2 = 5 A (i1 = 2.5 + 2.5 = 5 ✓). At t = ∞, i1 = 50/20 = 2.5 A, i2 = 0, vC = 25 V ✓.

✓ Answer checked by Claude Opus 5.5 (high effort)

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