---
title: "Classical: inductor current, 5 V with 2 H"
subject: "Electrical Circuit Theory"
chapter: 3
key: "semester_3/electrical_circuit_theory:classical-inductor-current-5v-2h-switch"
source: https://www.bctengineer.com/q/semester_3/electrical_circuit_theory/classical-inductor-current-5v-2h-switch
site: BCT Engineering
---

# Classical: inductor current, 5 V with 2 H

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

- Asked in: 2072 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-5v-2h-switch
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_circuit_theory/chapter-3

## Question

In the circuit of the figure below, the switch K is open and the circuit reaches a steady state. At t = 0, K is closed. Find the current in the inductor for t > 0. Use classical method.
[Figure: 5 V source. A 10 Ω resistor runs from the source positive terminal to a middle node, then a 20 Ω resistor to the top-right node; a 2 H inductor is connected in parallel across the series 10 Ω + 20 Ω (from the source positive node to the top-right node). From the middle node, switch K in series with a 10 Ω resistor connects to the bottom wire. A 10 Ω resistor connects the top-right node to the bottom wire.]

## Short answer

Initially iL=0.5 A. After closing, the Thevenin equivalent seen by the inductor has Vth=4.286 V, Rth=7.143 Ω, giving iL(∞)=0.6 A, τ=0.28 s, so iL(t)=0.6−0.1e^{-3.571t} A for t>0.

## Answer

**Assumptions from the figure:** 5 V source (node A at its + terminal). 10 Ω from A to node M, 20 Ω from M to node R; the 2 H inductor is connected from A to R (across the 10 Ω + 20 Ω series pair). From M, switch K and a 10 Ω resistor go to ground; a 10 Ω resistor connects R to ground. K is open until t = 0.

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

The inductor is a short, so V_R = 5 V and there is no voltage across the 10 Ω + 20 Ω path, hence no current in it. The inductor current flows only to the 10 Ω load at R:

```text
i_L(0⁻) = 5/10 = 0.5 A = i_L(0⁺)
```

#### After t = 0 (K closed): Thevenin equivalent seen by the inductor

**Open-circuit voltage** (inductor removed), unknowns V_M, V_R with A = 5 V:

```text
M:  (5 − V_M)/10 = V_M/10 + (V_M − V_R)/20
R:  (V_M − V_R)/20 = V_R/10   →  V_M = 3V_R
→  2(5 − 3V_R) = 2(3V_R) + 2V_R  →  10 = 14 V_R,  V_R = 0.714 V
V_oc = 5 − 0.714 = 4.286 V
```

**Thevenin resistance** (source shorted, so A = ground): looking between A and R,

```text
R_th = 10 ∥ [20 + (10 ∥ 10)] = 10 ∥ 25 = 7.143 Ω
```

**Final current and time constant:**

```text
i_L(∞) = V_oc/R_th = 4.286/7.143 = 0.6 A
τ = L/R_th = 2/7.143 = 0.28 s
```

#### Solution

```text
i_L(t) = 0.6 + (0.5 − 0.6) e^(−t/0.28)
i_L(t) = 0.6 − 0.1 e^(−3.571t) A,   t > 0
```

Check: at t = 0⁺ the node equations with i_L = 0.5 A give a voltage 0.714 V across L, and L di/dt = 2 × 0.1 × 3.571 = 0.714 V ✓ (verified numerically). At t → ∞ the voltage across L is zero ✓.

**Answer: i_L(t) = 0.6 − 0.1e^(−3.571t) A.**

## Check yourself

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

1. Why is i_L(0-) = 0.5 A?
   A. The inductor is a short, so no current flows in the 10 + 20 ohm path and only the 10 ohm load carries 5/10
   B. It equals 5/30 A
   C. It equals 5/(10+20+10) A
   D. It is zero since the inductor is open
   Answer: A. The inductor shorts the 10+20 path and puts 5 V across the load resistor. (Key point: Initial current)

2. How is the final current found after K closes?
   A. 5 V divided by 10 ohm
   B. Thevenin equivalent seen by the inductor: V_oc/R_th = 4.286/7.143 = 0.6 A
   C. Use i_L(0+) as the final value
   D. Take the average of 0.5 A and 0.6 A
   Answer: B. With L shorted at dc, i_L(infinity) = V_th/R_th. (Key point: Method)

3. What is R_th?
   A. 10 + 20 + 10 = 40 ohm
   B. 10 || 30 = 7.5 ohm
   C. 10 || [20 + (10 || 10)] = 7.143 ohm
   D. 20 ohm
   Answer: C. With the source shorted the 10 ohm from A to R is in parallel with 20 plus 5 ohm. (Key point: Thevenin resistance)

4. Which is the current?
   A. 0.6 + 0.1 e^(-3.571t) A
   B. 0.5 + 0.1 e^(-3.571t) A
   C. 0.6 - 0.1 e^(-0.28t) A
   D. 0.6 - 0.1 e^(-3.571t) A
   Answer: D. tau = 2/7.143 = 0.28 s and K = 0.5 - 0.6 = -0.1. (Key point: Result)

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