---
title: "Series RLC circuit analysis"
subject: "Basic Electrical Engineering"
chapter: 6
key: "semester_1/basic_electrical_engineering:series-rlc-100-0-12h-28-27uf"
source: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/series-rlc-100-0-12h-28-27uf
site: BCT Engineering
---

# Series RLC circuit analysis

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

- Asked in: 2071 Chaitra, 6 marks
- Video: [AC Circuits : Series RLC Circuit  | Electrical Engineering | TheElectricalGuy](https://www.youtube.com/watch?v=Tc1RsqZCq3o) — Gaurav J - TheElectricalGuy
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/series-rlc-100-0-12h-28-27uf
- Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-6

## Question

A series R-L-C circuit having R = 100 Ω, L = 0.12 H and c = 28.27 μF is fed from a 100 V, 50 Hz supply. Find the current flowing active power, reactive power, power factor, and rms values of voltage across each elements. Also draw phasor diagram.

## Short answer

Current is 0.80 A (leading). Active power 64.1 W, reactive power 48.0 VAR capacitive, power factor 0.80 leading. RMS voltages: V_R=80.0 V, V_L=30.2 V, V_C=90.1 V.

## Answer

Given R = 100 Ω, L = 0.12 H, C = 28.27 μF, V = 100 V, f = 50 Hz. ω = 314.16 rad/s.

#### Reactances
- X_L = ωL = 37.70 Ω
- X_C = 1/(ωC) = 1/(314.16 × 28.27 × 10⁻⁶) = 112.60 Ω
- X = X_L − X_C = −74.90 Ω (capacitive)

#### Impedance
Z = 100 − j74.90 = 124.94∠−36.83° Ω

#### Current
I = V/Z = 100/124.94 = **0.80 A** (leading V by 36.83°)

#### Power factor
cos φ = R/Z = 100/124.94 = **0.80 leading**

#### Powers
- Active: P = I²R = 0.80² × 100 = **64.1 W**
- Reactive: Q = I²X = 0.80² × 74.9 = **48.0 VAR** (capacitive)
- Apparent: S = VI = 100 × 0.80 = 80 VA

#### Voltages across elements (rms)
- V_R = I R = **80.0 V**
- V_L = I X_L = 0.80 × 37.70 = **30.2 V**
- V_C = I X_C = 0.80 × 112.60 = **90.1 V**

Check: √(80² + (90.1 − 30.2)²) = √(6400 + 3588) = 100 V.

#### Phasor diagram (I reference)
```text
          ↑ V_L = 30.2 V
          │
          ●──────────────►  I (reference); V_R = 80 V along I
          │  ╲
          │    ╲ 36.8°
          │      ↘ V = 100 V (lags I)
          ↓
          V_C = 90.1 V
```
V_R is along I, V_L is 90° ahead of I and V_C is 90° behind I. Since V_C > V_L, the resultant V lags I by 36.8°, so the circuit is capacitive.

## Check yourself

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

1. Which is correct for X = X_L − X_C?
   A. −74.9 Ω, capacitive
   B. −150 Ω
   C. +74.9 Ω, inductive
   D. 0 Ω
   Answer: A. X_L = 37.70 Ω, X_C = 112.6 Ω, so the circuit is capacitive. (Key point: Net reactance)

2. What is the current and pf?
   A. 0.80 A and 0.80 lagging
   B. 1 A and 0.6 leading
   C. 0.5 A and 0.8 leading
   D. 0.80 A and 0.80 leading
   Answer: D. |Z| = 124.94 Ω, I = 0.80 A, cos φ = 100/124.94 = 0.80 leading. (Key point: Current and power factor)

3. V_R = 80 V, V_L = 30.2 V, V_C = 90.1 V with supply 100 V. How can V_C exceed the supply?
   A. V_L and V_C are opposite in phase so supply = √(V_R^2 + (V_C − V_L)^2)
   B. Resistor drop is subtracted
   C. Error in measurement
   D. Rms values add arithmetically
   Answer: A. V_C and V_L are 180° apart; individual drops can exceed supply, but phasor sum = 100 V. (Key point: Voltages across elements)

4. Which is correct?
   A. Q = 0
   B. P = 80 W
   C. P = 48 W
   D. P = I^2 R = 64.1 W and Q = I^2 X = 48 VAR capacitive
   Answer: D. Only R dissipates power; S = 80 VA. (Key point: Active and reactive power)

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