---
title: "Classical: mesh currents and capacitor voltage, 50 V"
subject: "Electrical Circuit Theory"
chapter: 3
key: "semester_3/electrical_circuit_theory:classical-mesh-currents-50v-2uf"
source: https://www.bctengineer.com/q/semester_3/electrical_circuit_theory/classical-mesh-currents-50v-2uf
site: BCT Engineering
---

# Classical: mesh currents and capacitor voltage, 50 V

Electrical Circuit Theory · Chapter 3: Transient analysis in RLC circuit by direct solution

- Asked in: 2068 Shrawan, 8 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electrical_circuit_theory/classical-mesh-currents-50v-2uf
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_circuit_theory/chapter-3

## Question

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)]

## 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.

## Answer

**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)
```text
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 ✓.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. After eliminating i1, the capacitor voltage obeys which first-order equation?
   A. 5C dvC/dt + vC = 25
   B. 10C dvC/dt + vC = 50
   C. 5C dvC/dt + vC = 50
   D. 20C dvC/dt + vC = 25
   Answer: A. Mesh 1 gives i1 = 2.5 + 0.5 i2; substituting in mesh 2 gives vC + 5 i2 = 25 with i2 = C dvC/dt. (Key point: Reduced equation)

2. Why is the time constant 10 us?
   A. C sees 10 + 10 = 20 ohm, so tau = 20 x 2 uF = 40 us
   B. C sees 10 ohm in parallel with 10 ohm = 5 ohm, so tau = 5 x 2 uF
   C. Only one 10 ohm is seen, so tau = 10 x 2 uF = 20 us
   D. tau = 5 ohm x 2 uF but the result is doubled by the source
   Answer: B. The source is a short for the natural response, so the two 10 ohm resistors are in parallel, giving 5 ohm; tau = 5 x 2e-6 = 10 us. (Key point: Time constant)

3. What are i1 and i2 at t = 0+, and vC at infinity?
   A. i1 = 2.5 A, i2 = 0, vC(inf) = 50 V
   B. i1 = 5 A, i2 = 0, vC(inf) = 25 V
   C. i1 = 5 A, i2 = 5 A, vC(inf) = 25 V
   D. i1 = 2.5 A, i2 = 2.5 A, vC(inf) = 0
   Answer: C. Uncharged C is a short at 0+, so i1 = 50/10 = 5 A and i2 = 5 A; at infinity C is open, i2 = 0 and vC = 25 V. (Key point: Initial and final values)

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