---
title: "Classical: inductor current, 10 V with 2 H"
subject: "Electrical Circuit Theory"
chapter: 3
key: "semester_3/electrical_circuit_theory:classical-inductor-current-10v-10ohm-20ohm"
source: https://www.bctengineer.com/q/semester_3/electrical_circuit_theory/classical-inductor-current-10v-10ohm-20ohm
site: BCT Engineering
---

# Classical: inductor current, 10 V with 2 H

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

- Asked in: 2068 Chaitra, 8 marks
- Video: [RL series transient with dc excitaion using classical method](https://www.youtube.com/watch?v=iXo6IuXV9Ds) — 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-10v-10ohm-20ohm
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_circuit_theory/chapter-3

## Question

For the circuit shown in the following figure, use classical method to find the current in the inductor for t > 0.
[Figure: 10 V source. A 10 Ω resistor runs along the top wire from the source positive node to the top-right node (bridging). A second row has a 10 Ω resistor from the source positive node to a middle node, then a 20 Ω resistor from the middle node to the right node. A 2 H inductor connects the right node to the bottom wire. From the middle node, a switch (t = 0) in series with a 10 Ω resistor connects to the bottom wire.]

## Short answer

Initial inductor current (switch open) is 1.333 A, final current (switch closed) is 1.2 A. The Thevenin resistance seen by the inductor is 7.143 Ohms, time constant 0.28 s. i_L(t) = 1.2 + 0.1333 e^(-3.571 t) A.

## Answer

**Assumption (figure not clear):** the switch is open for t < 0 and closes at t = 0. (The result for the opposite case, switch opening, is given at the end.)

#### Initial condition (t = 0⁻, switch open)
The inductor is a short circuit in steady state, so the 10 Ω (top) is in parallel with 10 Ω + 20 Ω = 30 Ω across the 10 V source.

i_L(0⁻) = 10/10 + 10/30 = 1.333 A = i_L(0⁺)

#### Final value (t → ∞, switch closed, L shorted)
Let V_m be the middle-node voltage; the right node is at 0 V.

```text
(10 − Vm)/10 = Vm/10 + Vm/20   →  Vm = 4 V
i_L(∞) = 10/10 + 4/20 = 1 + 0.2 = 1.2 A
```

#### Time constant
Remove the source and look from the inductor terminals: R_th = 10 ∥ (20 + 10 ∥ 10) = 10 ∥ 25 = 7.143 Ω.

τ = L/R_th = 2/7.143 = 0.28 s

#### Classical solution
The differential equation is 2 di/dt + 7.143 i = 8.571 (Thevenin equivalent V_th = 8.571 V). Complete solution = forced + natural:

i_L(t) = 1.2 + A e^(−t/0.28)

i_L(0) = 1.333 → A = 0.1333

**Answer: i_L(t) = 1.2 + 0.1333 e^(−3.571t) A, t > 0** (check: i_L(0) = 1.333 A, i_L(∞) = 1.2 A)

If instead the switch is closed for t < 0 and opens at t = 0: i_L(0) = 1.2 A, i_L(∞) = 10/10 + 10/30 = 1.333 A, R_th = 10 ∥ 30 = 7.5 Ω, τ = 0.2667 s, so i_L(t) = 1.333 − 0.1333 e^(−3.75t) A.

## 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 open?
   A. 1.333 A: 10/10 + 10/30
   B. 1 A
   C. 0.333 A
   D. 1.2 A
   Answer: A. The 10 ohm and the 10+20 path are in parallel with L shorted. (Key point: Initial current)

2. What is i_L(infinity) after closing?
   A. 1.333 A
   B. 1.2 A, as V_m = 4 V so 1 + 4/20
   C. 1 A
   D. 0.2 A
   Answer: B. With L shorted the right node is at 0 V. (Key point: Final current)

3. How is tau obtained?
   A. L x R_th
   B. L/10
   C. L/R_th with R_th = 10 || (20 + 10||10) = 7.143 ohm, so 0.28 s
   D. 2/30 s
   Answer: C. R_th is the resistance seen from the inductor terminals with the source removed. (Key point: Time constant)

4. What is the result?
   A. 1.333 + 0.1333 e^(-3.75t) A
   B. 1.2 - 0.1333 e^(-3.571t) A
   C. 1.333 - 1.2 e^(-3.571t) A
   D. 1.2 + 0.1333 e^(-3.571t) A
   Answer: D. A = 1.333 - 1.2 = 0.1333. (Key point: Result)

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