---
title: "Energy stored in coil, time constants"
subject: "Engineering Physics"
chapter: 8
key: "semester_1/engineering_physics:coil-energy-stored-time-constants"
source: https://www.bctengineer.com/q/semester_1/engineering_physics/coil-energy-stored-time-constants
site: BCT Engineering
---

# Energy stored in coil, time constants

Engineering Physics · Chapter 8: Electromagnetism

- Asked in: 2067 Asar, 5 marks
- Video: [RL Circuits (6 of 8) Charging an Inductor, Time Constant, Voltage, Current, An Explanation](https://www.youtube.com/watch?v=OeaSZPTYClA) — Step by Step Science
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/engineering_physics/coil-energy-stored-time-constants
- Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-8

## Question

A coil has an inductance of 53 mH and a resistance of 0.35Ω. If a 12V emf is applied across the coil, how much energy is stored in the magnetic field after the current has built up to its equilibrium value? After how many time constants will half this equilibrium energy be stored?

## Short answer

Maximum current I_max = 12 V / 0.35 Ω = 34.3 A, energy U_max = 0.5 × 0.053 × (34.3)^2 ≈ 31.2 J. Half energy when I = I_max/√2 → 1−e^(−t/τ) = 1/√2 → t/τ ≈ 1.23.

## Answer

**Given:** L = 53 mH, R = 0.35 Ω, ε = 12 V.

#### Energy at equilibrium
At equilibrium the current is steady and limited only by R: I_max = ε/R.

```text
I_max = 12 / 0.35 = 34.29 A
U_max = ½ L I_max² = ½ × 0.053 × (34.29)²
      = ½ × 0.053 × 1175.5
      = 31.2 J
```

#### Time for half this energy
The current grows as I = I_max(1 − e^(−t/τ)), with τ = L/R = 0.053/0.35 = 0.1514 s. Energy U = ½LI² is half of U_max when I² = I_max²/2, that is, I = I_max/√2.

```text
1 − e^(−t/τ) = 1/√2 = 0.7071
e^(−t/τ)     = 0.2929
t/τ          = −ln 0.2929 = 1.228
t = 1.228 × 0.1514 = 0.186 s
```

**Answer: U_max ≈ 31.2 J; half of it is stored after about 1.23 time constants (t ≈ 0.19 s).**

## Check yourself

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

1. At equilibrium the current is limited only by:
   A. the inductance, I = ε/L
   B. the resistance, I_max = ε/R ≈ 34.3 A
   C. the product LR
   D. the time constant
   Answer: B. At steady state dI/dt = 0 and the inductor has no voltage. (Key point: Equilibrium current)

2. The energy stored at equilibrium is:
   A. ½LI_max² ≈ 31.2 J
   B. LI_max² ≈ 62 J
   C. ½ε I_max ≈ 206 J
   D. ½LI_max ≈ 0.9 J
   Answer: A. ½ × 0.053 × 1175.5. (Key point: Energy)

3. Half the energy is stored when the current is:
   A. I_max/2
   B. I_max/4
   C. I_max/√2
   D. 0.37 I_max
   Answer: C. U ∝ I², so I² = I_max²/2. (Key point: Half energy)

4. The time for that is about:
   A. 0.69 time constants
   B. 2.0 time constants
   C. 0.5 time constants
   D. 1.23 time constants (0.19 s)
   Answer: D. t/τ = −ln(0.2929) = 1.228. (Key point: Time constants)

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