---
title: "Classical: inductor current and voltage at 10 ms"
subject: "Electrical Circuit Theory"
chapter: 3
key: "semester_3/electrical_circuit_theory:classical-inductor-current-20v-10v-30ohm"
source: https://www.bctengineer.com/q/semester_3/electrical_circuit_theory/classical-inductor-current-20v-10v-30ohm
site: BCT Engineering
---

# Classical: inductor current and voltage at 10 ms

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

- Asked in: 2069 Chaitra, 8 marks
- Video: [Numericals on RL series Transient ( Classical Method)](https://www.youtube.com/watch?v=eeY8nH1Zyjs) — RSL Lecture Series
- 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-inductor-current-20v-10v-30ohm
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_circuit_theory/chapter-3

## Question

In the circuit shown, the switch is opened for a long time and then it is suddenly closed at t = 0. Obtain the expression for current through inductor for t > 0. Also calculate the voltage across inductor after 10 mSec. [Use classical method]
[Figure: 20 V source (+ at top) in series with a 30 Ω resistor; a 10 V source (+ at top) below it, connected in series with the 20 V source (the junction of the two sources is connected by the switch (t = 0) to the node between the 30 Ω and 20 Ω resistors); then a 20 Ω resistor and a 0.5 H inductor carrying current i(t), returning to the negative terminal of the 10 V source]

## Short answer

Initially iL=0.6 A. After closing, the 20 Ω+L branch is across 10 V, yielding i(t)=0.5+0.1e^{-40t} A for t>0; vL(10 ms)=−1.34 V.

## Answer

**Assumptions from the figure:** the 20 V and 10 V sources are in series aiding (total 30 V, with the 10 V negative terminal at the reference). The + terminal of the 20 V source feeds a 30 Ω resistor to node N. From N, a 20 Ω resistor and the 0.5 H inductor go to the reference. The switch connects the junction of the two sources (10 V above the reference) to N. The switch is open for a long time, and closes at t = 0.

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

```text
i(0⁻) = 30/(30 + 20) = 0.6 A = i(0⁺)
```

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

The switch connects N directly to the 10 V point, so the 20 Ω + L branch is across 10 V (the 30 Ω and 20 V source no longer matter to it).

```text
L di/dt + 20 i = 10
0.5 di/dt + 20 i = 10  →  di/dt + 40 i = 20
```

- Particular (final) value: i(∞) = 10/20 = 0.5 A.
- Complementary: s = −40, τ = L/R = 0.5/20 = 25 ms.

```text
i(t) = 0.5 + K e^(−40t),   i(0⁺) = 0.6  →  K = 0.1
i(t) = 0.5 + 0.1 e^(−40t) A,   t > 0
```

#### Voltage across the inductor at t = 10 ms

```text
v_L = L di/dt = 0.5 × (−40)(0.1) e^(−40t) = −2 e^(−40t) V
v_L(10 ms) = −2 e^(−0.4) = −2 × 0.6703 = −1.341 V
```

**Answer: i(t) = 0.5 + 0.1e^(−40t) A; v_L(10 ms) = −1.34 V.**

## Check yourself

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

1. What is i(0-)?
   A. 30/(30+20) = 0.6 A
   B. 10/20 = 0.5 A
   C. 30/20 = 1.5 A
   D. 0 A
   Answer: A. With the switch open the 30 V total drives 30 + 20 ohm in series with L shorted. (Key point: Initial current)

2. Why does the 30 ohm and 20 V source not matter after closing?
   A. The 30 ohm becomes zero
   B. The switch puts the 20 ohm + L branch directly across 10 V
   C. The inductor becomes open
   D. The 20 V source cancels the 10 V
   Answer: B. L di/dt + 20 i = 10. (Key point: After closing)

3. What is i(t)?
   A. 0.5 - 0.1 e^(-40t) A
   B. 0.6 + 0.5 e^(-40t) A
   C. 0.5 + 0.1 e^(-40t) A
   D. 0.5 + 0.1 e^(-20t) A
   Answer: C. i(infinity) = 0.5 A, tau = 25 ms, K = 0.6 - 0.5. (Key point: Result)

4. How is v_L at 10 ms found?
   A. 20 x i = 10 V
   B. 10 - 20 i at t = 0
   C. +2 e^(-0.4) = +1.34 V
   D. L di/dt = -2 e^(-40t) = -1.34 V
   Answer: D. Differentiating i gives a negative derivative, so v_L is negative. (Key point: Inductor voltage)

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