---
title: "RLC series choke coil and capacitor"
subject: "Basic Electrical Engineering"
chapter: 6
key: "semester_1/basic_electrical_engineering:choke-coil-10ohm-0-05h-100uf"
source: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/choke-coil-10ohm-0-05h-100uf
site: BCT Engineering
---

# RLC series choke coil and capacitor

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

- Asked in: 2073 Chaitra, 6 marks
- Video: [RLC Series Circuit: Impedance, Power Factor, Active Power, & Reactive Power Analysis with AC Supply](https://www.youtube.com/watch?v=QpRtsF6RlBQ) — Electrical and Electronics Engineering
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/choke-coil-10ohm-0-05h-100uf
- Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-6

## Question

A choke coil having a resistance of 10Ω and inductance of 0.05H is connected in series with a condenser of 100 μF. The whole circuit has been connected to 200V, 50 Hz supply. Calculate (i) impedance (ii) current (iii) power factor (iv) power input (v) Apparent and reactive power of circuit.

## Short answer

Net capacitive reactance 16.12 Ω, impedance 18.97 Ω. Current 10.54 A leading 58.2°, power factor 0.527 leading. Active power 1111 W, apparent 2108 VA, reactive 1792 VAR.

## Answer

Given R = 10 Ω, L = 0.05 H, C = 100 μF in series; V = 200 V, 50 Hz. ω = 314.16 rad/s.

- X_L = ωL = 15.71 Ω
- X_C = 1/(ωC) = 31.83 Ω
- Net reactance X = X_L − X_C = −16.12 Ω (capacitive)

#### (i) Impedance
Z = R + jX = 10 − j16.12 Ω

**|Z| = √(10² + 16.12²) = 18.97 Ω**, angle −58.2°

#### (ii) Current
I = V/Z = 200/18.97 = **10.54 A** (leading the voltage by 58.2°)

#### (iii) Power factor
cos φ = R/Z = 10/18.97 = **0.527 leading**

#### (iv) Power input
P = I²R = 10.54² × 10 = **1111 W** (= VI cos φ = 200 × 10.54 × 0.527)

#### (v) Apparent and reactive power
- Apparent power S = VI = 200 × 10.54 = **2108 VA**
- Reactive power Q = VI sin φ = I²|X| = 10.54² × 16.12 = **1792 VAR** (capacitive)

Check: √(1111² + 1792²) = 2108 VA.

## Check yourself

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

1. What is the net reactance of the series circuit?
   A. X_L − X_C = −16.12 Ω (capacitive)
   B. X_L − X_C = +16.12 Ω (inductive)
   C. X_L + X_C = 47.5 Ω inductive
   D. 0 Ω
   Answer: A. X_L = 15.71 Ω, X_C = 31.83 Ω; X_C is larger so the net is capacitive. (Key point: Net reactance)

2. What are the impedance and current?
   A. 18.97 Ω and 10.54 A lagging
   B. 10 Ω and 20 A
   C. 41.8 Ω and 4.8 A
   D. 18.97 Ω and 10.54 A leading
   Answer: D. |Z| = √(10^2 + 16.12^2) = 18.97 Ω; I = 200/18.97 = 10.54 A leading. (Key point: Impedance and current)

3. What is the power factor?
   A. 1
   B. 0.527 leading
   C. 0.527 lagging
   D. 0.85 leading
   Answer: B. cos φ = R/Z = 10/18.97 = 0.527; it is leading as the circuit is capacitive. (Key point: Power factor)

4. Which equation relation holds for the powers?
   A. P = S sin φ
   B. S = √(P^2 + Q^2) = 2108 VA
   C. S = P + Q
   D. Q = I^2 R
   Answer: B. P = 1111 W, Q = 1792 VAR, so S = 2108 VA. (Key point: Powers)

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