---
title: "Inductor voltage and time for i = i0/2"
subject: "Engineering Physics"
chapter: 8
key: "semester_1/engineering_physics:inductance-voltage-half-current"
source: https://www.bctengineer.com/q/semester_1/engineering_physics/inductance-voltage-half-current
site: BCT Engineering
---

# Inductor voltage and time for i = i0/2

Engineering Physics · Chapter 8: Electromagnetism

- Asked in: 2070 Asar, 5 marks; 2068 Shrawan, 5 marks (asked 2 times)
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/engineering_physics/inductance-voltage-half-current
- Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-8

## Question

An inductance L is connected to a battery of emf E through a resistance. Show that the potential difference across the inductance after time t is V_L = ε e^(−Rt/L). At what time is the potential difference across the inductance equal to that across the resistance such that i = i0/2.

## Short answer

In an LR circuit, V_L = ε e^(-Rt/L). Setting V_L = V_R gives ε e^(-Rt/L) = ε/2, so t = (L/R) ln 2 ≈ 0.693 L/R. At that time, i = (ε/R)(1 - 1/2) = i0/2.

## Answer

#### Part 1: V_L = ε e^(−Rt/L)
Circuit: battery ε, resistance R, inductance L in series, switched on at t = 0.

1. Current growth: i = (ε/R)(1 − e^(−Rt/L)), obtained from ε = iR + L di/dt.
2. Voltage across the resistor: V_R = iR = ε(1 − e^(−Rt/L))
3. Loop rule: V_L = ε − V_R = ε − ε(1 − e^(−Rt/L))

**V_L = ε e^(−Rt/L)**

(Equivalently V_L = L di/dt = L · (ε/R)(R/L) e^(−Rt/L) = ε e^(−Rt/L).) At t = 0, V_L = ε (all of the emf), and it decays to zero as the current settles at ε/R.

#### Part 2: when V_L = V_R
1. Set V_L = V_R: since V_L + V_R = ε, each equals ε/2.
2. ε e^(−Rt/L) = ε/2 → e^(−Rt/L) = ½
3. Rt/L = ln 2

**t = (L/R) ln 2 = 0.693 τ**, where τ = L/R.

#### Current at that time
i = (ε/R)(1 − ½) = ½ (ε/R) = i₀/2, where i₀ = ε/R is the final current. So at t = 0.693 L/R the voltage across L equals that across R, and the current is half its final value.

## Check yourself

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

1. How is V_L obtained?
   A. V_L = ε − V_R
   B. V_L = ε + V_R
   C. V_L = iR
   D. V_L = ε R
   Answer: A. With V_R = ε(1 − e^(−Rt/L)), V_L = ε e^(−Rt/L). (Key point: Loop rule)

2. At t = 0 the inductor voltage is
   A. Zero
   B. ε/2
   C. ε (the full emf)
   D. ε/e
   Answer: C. The current is zero, so V_R = 0 and V_L = ε. (Key point: Initial value)

3. When V_L = V_R, each equals
   A. ε
   B. ε/2
   C. ε/e
   D. 2ε
   Answer: B. Since V_L + V_R = ε. (Key point: Equal voltages)

4. The time when V_L = V_R is
   A. L/R
   B. 0.693 L/R
   C. 2 L/R
   D. L/(R ln 2)
   Answer: B. e^(−Rt/L) = ½ gives t = (L/R) ln 2, where i = i0/2. (Key point: Time)

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