---
title: "Coil in series with non-inductive resistance"
subject: "Basic Electrical Engineering"
chapter: 6
key: "semester_1/basic_electrical_engineering:coil-series-30ohm-voltmeter-180v"
source: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/coil-series-30ohm-voltmeter-180v
site: BCT Engineering
---

# Coil in series with non-inductive resistance

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

- Asked in: 2073 Shrawan, 8 marks
- Video: [4.07 How can solve Non Inductive Resistance Coil based Problem | Basic Electrical Engineering](https://www.youtube.com/watch?v=c3855qhNUes) — Mritunjay Lectures
- 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-series-30ohm-voltmeter-180v
- Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-6

## Question

A coil is connected in series with a non-inductive resistance of 30Ω across 240V, 50Hz, 1-φ supply. The reading of voltmeters across the coil is 180 V and across the resistance is 130 V. Calculate, i) Inductance of coil ii) Resistance of coil iii) Power absorbed by coil iv) Power absorbed by whole circuit

## Short answer

Current I=4.333 A from resistor voltage. Coil resistance R_c=7.37 Ω. Inductive reactance 40.88 Ω yields L=0.130 H. Power absorbed by coil 138.3 W, total circuit power 701.7 W.

## Answer

Let the coil have resistance R_c and reactance X_L. The 30 Ω resistor is non-inductive. Supply V = 240 V, f = 50 Hz.

#### Current
I = V_R/R = 130/30 = 4.333 A (same current in the series circuit)

#### Impedances
- Coil: Z_c = 180/4.333 = 41.54 Ω, so R_c² + X_L² = 41.54² = 1725.6 …(1)
- Whole circuit: Z = 240/4.333 = 55.38 Ω, so (R_c + 30)² + X_L² = 55.38² = 3067.1 …(2)

(2) − (1): 60R_c + 900 = 1341.5, so R_c = 7.37 Ω

#### ii) Resistance of coil
**R_c = 7.37 Ω**

#### i) Inductance of coil
X_L = √(1725.6 − 54.3) = √1671.3 = 40.88 Ω

L = X_L/(2πf) = 40.88/314.16 = **0.130 H**

#### iii) Power absorbed by coil
P_c = I²R_c = 4.333² × 7.37 = **138.3 W**

#### iv) Power absorbed by whole circuit
P = I²(R_c + 30) = 18.78 × 37.37 = **701.7 W**

(Power factor of circuit = 37.37/55.38 = 0.675 lagging; P = 240 × 4.333 × 0.675 = 702 W.)

## Check yourself

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

1. How is the circuit current obtained?
   A. I = 180/130
   B. I = 130/30 = 4.333 A
   C. I = 180/30
   D. I = 240/30
   Answer: B. The 30 Ω resistor is non-inductive, so I = V_R/R = 130/30 = 4.333 A (same series current). (Key point: Find the current)

2. Which two equations determine R_c and X_L?
   A. R_c^2 + X_L^2 = (180/I)^2 and (R_c + 30)^2 + X_L^2 = (240/I)^2
   B. R_c = 180/I
   C. R_c + X_L = 180/I
   D. X_L = 240/I
   Answer: A. Coil impedance is 41.54 Ω and total impedance is 55.38 Ω; subtracting gives R_c. (Key point: Equations)

3. Which results follow?
   A. R_c = 30 Ω, L = 0.13 H
   B. R_c = 41.5 Ω, L = 0.13 H
   C. R_c = 7.37 Ω, L = 40.88 H
   D. R_c = 7.37 Ω, L = 0.130 H
   Answer: D. R_c = 7.37 Ω, X_L = 40.88 Ω and L = X_L/(2πf) = 0.130 H. (Key point: Coil resistance and inductance)

4. How is power absorbed by the coil found?
   A. I^2 R_c = 138.3 W
   B. I^2 X_L
   C. 180 × 4.333 × 1
   D. V × I = 780 W
   Answer: A. Only the resistance consumes real power: P_c = 4.333^2 × 7.37; the whole circuit uses I^2(R_c+30) = 701.7 W. (Key point: Power)

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