---
title: "Coil inductance and resistance from two frequencies"
subject: "Basic Electrical Engineering"
chapter: 6
key: "semester_1/basic_electrical_engineering:coil-resistance-inductance-127hz-456hz"
source: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/coil-resistance-inductance-127hz-456hz
site: BCT Engineering
---

# Coil inductance and resistance from two frequencies

Basic Electrical Engineering · Chapter 6: Single-phase AC Circuits

- Asked in: 2067 Asoj, 6 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/coil-resistance-inductance-127hz-456hz
- Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-6

## Question

At 127 Hz frequency, an emf source of 6V sends a current of 100mA through an inductive coil. If the frequency changes to 456Hz, the current drawn by the circuit becomes 50mA. Calculate the inductance and resistance of coil.

## Short answer

From Z = V/I, at 127 Hz Z=60Ω, at 456 Hz Z=120Ω. Solving R²+(2πfL)² gives L≈37.8 mH, R≈51.9 Ω.

## Answer

Let the coil have resistance R and inductance L. Its impedance is Z = √(R² + (2πfL)²).

#### Data
- At f₁ = 127 Hz: Z₁ = V/I = 6/0.1 = 60 Ω
- At f₂ = 456 Hz: Z₂ = 6/0.05 = 120 Ω

#### Equations
R² + (2π × 127 × L)² = 60² = 3600 …(1)
R² + (2π × 456 × L)² = 120² = 14400 …(2)

Subtract (1) from (2):

L² [(2π × 456)² − (2π × 127)²] = 10800

(2π)² (456² − 127²) L² = 10800
39.478 × (207 936 − 16 129) L² = 10800
39.478 × 191 807 L² = 10800
7.572 × 10⁶ L² = 10800 ⟹ L² = 1.426 × 10⁻³

**L = 0.0378 H ≈ 37.8 mH**

#### Resistance
X_L at 127 Hz = 2π × 127 × 0.03777 = 30.14 Ω

R = √(3600 − 30.14²) = √(3600 − 908.4) = **51.9 Ω**

Check at 456 Hz: X_L = 108.2 Ω, √(51.9² + 108.2²) = 120 Ω.

**Answer: L ≈ 37.8 mH, R ≈ 51.9 Ω.**

## Check yourself

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

1. How is impedance found at each frequency?
   A. Same at both
   B. Z = V/I: 60 Ω at 127 Hz and 120 Ω at 456 Hz
   C. Z = I/V
   D. Z = V × I
   Answer: B. 6/0.1 = 60 Ω and 6/0.05 = 120 Ω. (Key point: Impedance values)

2. Why are two equations needed?
   A. To find frequency
   B. R is known
   C. R and L both unknown; Z^2 = R^2 + (2πfL)^2 at each frequency
   D. To find V
   Answer: C. Subtracting the equations eliminates R^2 and gives L. (Key point: Why two equations)

3. What is L?
   A. 37.8 mH
   B. 51.9 mH
   C. 3.78 H
   D. 130 mH
   Answer: A. L^2 = 10800 / [(2π)^2 (456^2 − 127^2)]. (Key point: Inductance)

4. After finding L, R is found by:
   A. R = Z1
   B. R = Z2 − Z1
   C. R = X_L
   D. R = √(Z1^2 − X_L1^2) = 51.9 Ω
   Answer: D. At 127 Hz X_L = 30.14 Ω, so R = √(3600 − 908.4). (Key point: Resistance)

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