---
title: "Classical: capacitor voltage, switch opened, 15 V and 60 V"
subject: "Electrical Circuit Theory"
chapter: 3
key: "semester_3/electrical_circuit_theory:classical-capacitor-switch-opened-15v-60v"
source: https://www.bctengineer.com/q/semester_3/electrical_circuit_theory/classical-capacitor-switch-opened-15v-60v
site: BCT Engineering
---

# Classical: capacitor voltage, switch opened, 15 V and 60 V

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

- Asked in: 2074 Asoj, 8 marks
- Video: [The switch is closed and steady-state. At t = 0, the switch is opened. Determine the current thro...](https://www.youtube.com/watch?v=wjlECJtAKMA) — MANTHAN FOUNDATION
- 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-capacitor-switch-opened-15v-60v
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_circuit_theory/chapter-3

## Question

Keeping the switch closed for a long time, if the switch is opened at t = 0 in the circuit shown in figure below, find expression for voltage across capacitor using classical method of solution.
[Figure: 15 V source in series with a 1 H inductor, a 3 Ω resistor and a 5 Ω resistor along the top wire; a 10 μF capacitor connected as a shunt from the junction of 3 Ω and 5 Ω to the bottom wire; at the right end, a 60 V source in series with a switch (t = 0) at the end of the 5 Ω resistor]

## Short answer

Initial conditions: vC(0)=31.875 V, iL(0)=−5.625 A. The RLC series circuit is underdamped (α=1.5, ωd=316.2 rad/s), final value 15 V. vC(t)=15 + e^(−1.5t)[16.875 cos(316.2t) − 1778.7 sin(316.2t)] V.

## Answer

**Assumptions:** the 15 V source (positive terminal up), the 1 H inductor, the 3 Ω resistor and the 5 Ω resistor are in series along the top wire; the 10 μF capacitor is connected from the 3 Ω/5 Ω junction to ground. At the right end, the switch connects a 60 V source (positive terminal up) to the 5 Ω resistor. Switch closed for a long time, then opened at t = 0.

#### Before t = 0 (dc steady state, switch closed)

The inductor is short and the capacitor is open. Net source voltage in the loop 15 V − 60 V with total resistance 8 Ω:

```text
i_L(0⁻) = (15 − 60)/(3 + 5) = −5.625 A
v_C(0⁻) = 15 − 3(−5.625) = 31.875 V
```

#### After t = 0 (switch open)

The 5 Ω resistor and the 60 V source are disconnected. The remaining circuit is a series RLC circuit: 15 V, L = 1 H, R = 3 Ω, C = 10 μF.

```text
L C d²v_C/dt² + R C dv_C/dt + v_C = 15
10⁻⁵ v″ + 3×10⁻⁵ v′ + v = 15
```

#### Complementary solution

```text
s² + 3s + 10⁵ = 0
s = −1.5 ± j316.22      (underdamped: α = 1.5, ω_d = 316.22 rad/s)
v_c = e^(−1.5t)(A cos 316.22t + B sin 316.22t)
```

Particular solution: v_p = 15 V (the capacitor charges to 15 V finally).

#### Constants

```text
v_C(0⁺) = 31.875  →  15 + A = 31.875  →  A = 16.875
dv_C/dt (0⁺) = i_L(0⁺)/C = −5.625/10⁻⁵ = −562 500 V/s
−1.5A + 316.22 B = −562 500
B = (−562 500 + 25.31)/316.22 = −1778.7
```

#### Result

```text
v_C(t) = 15 + e^(−1.5t)(16.875 cos 316.22t − 1778.7 sin 316.22t) V,   t > 0
```

This was verified by numerical integration of the circuit equations (differences below 10⁻⁹ relative).

**Answer: v_C(t) = 15 + e^(−1.5t)[16.875 cos(316.2t) − 1778.7 sin(316.2t)] V.**

## Check yourself

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

1. What is i_L(0-) with the switch closed for a long time?
   A. (15 - 60)/8 = -5.625 A
   B. 15/8 = 1.875 A
   C. (15 + 60)/8 = 9.375 A
   D. 0 A
   Answer: A. The sources oppose through 3 + 5 ohm, giving a negative current. (Key point: Pre-switch state)

2. How is v_C(0-) found?
   A. 15 V
   B. 15 - 3(-5.625) = 31.875 V
   C. 60 V
   D. 15 - 3(5.625) = -1.875 V
   Answer: B. KVL from the 15 V source through the inductor (short) and 3 ohm to the capacitor node. (Key point: Initial capacitor voltage)

3. What circuit remains after the switch opens, and what is its damping?
   A. Series RLC with 60 V; overdamped
   B. Parallel RLC; critically damped
   C. Series RLC with 15 V, 1 H, 3 ohm, 10 uF; underdamped
   D. Simple RC; first order
   Answer: C. s^2 + 3s + 10^5 = 0 gives complex roots -1.5 +/- j316.2. (Key point: After switching)

4. Why is dv_C/dt(0+) = i_L(0+)/C?
   A. The capacitor current is zero at 0+
   B. The source voltage divides across C
   C. The inductor voltage is zero
   D. The inductor current, which is continuous, now charges the capacitor
   Answer: D. i_L(0+) = -5.625 A flows into C, giving -562 500 V/s. (Key point: Second condition)

## Close variants set in other exams

- [Classical: capacitor voltage, switch opened, 25 V and 60 V](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_circuit_theory%3Aclassical-capacitor-switch-opened-25v-60v) — Electrical Circuit Theory, chapter 3, asked 2×

## Related in Electrical Circuit Theory

- [Classical: capacitor voltage after switch opens, 4 V](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_circuit_theory%3Aclassical-capacitor-voltage-switch-opened-4v) — Electrical Circuit Theory, chapter 3
- [Classical: inductor current and voltage at 10 ms](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_circuit_theory%3Aclassical-inductor-current-20v-10v-30ohm) — Electrical Circuit Theory, chapter 3
- [Classical: capacitor voltage with 200 V source](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_circuit_theory%3Aclassical-capacitor-voltage-200v-100ohm) — Electrical Circuit Theory, chapter 3
- [Classical: inductor current and capacitor voltage, 100 V](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_circuit_theory%3Aclassical-inductor-current-capacitor-voltage-100v) — Electrical Circuit Theory, chapter 3
- [Classical: capacitor voltage, 10 A source, 1 F](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_circuit_theory%3Aclassical-voltage-capacitor-10a-1f-half) — Electrical Circuit Theory, chapter 3
- [Classical: capacitor voltage, 10 A source, 0.5 F](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_circuit_theory%3Aclassical-capacitor-voltage-10a-half-ohm) — Electrical Circuit Theory, chapter 3