---
title: "Total power of series-parallel circuit"
subject: "Basic Electrical Engineering"
chapter: 6
key: "semester_1/basic_electrical_engineering:series-parallel-total-power-265uf-19mh"
source: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/series-parallel-total-power-265uf-19mh
site: BCT Engineering
---

# Total power of series-parallel circuit

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

- Asked in: 2067 Asoj, 8 marks
- Video: [Determine Power Consumed in a Series Parallel Circuit](https://www.youtube.com/watch?v=goAyE0K0urA) — Engineer4Free
- 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-parallel-total-power-265uf-19mh
- Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-6

## Question

Determine total power consumed by the following circuit. Also draw the phasor diagram. [Figure: 4 Ω in series with 19.1 mH, followed by two parallel branches: 10 Ω in series with 265.15 μF capacitor; and 6 Ω in series with 31.8 mH; supply 200 V, 50 Hz.]

## Short answer

Reactances: series L=6Ω, C=12Ω, branch B L=10Ω. Parallel impedance Zp=10.85+j3.11Ω, total Z=14.85+j9.11Ω. Total current 11.48A. Power consumed P = 11.48²×14.85 = 1957 W.

## Answer

Circuit: 4 Ω in series with L = 19.1 mH, then two parallel branches: Branch A is 10 Ω in series with C = 265.15 μF, Branch B is 6 Ω in series with L = 31.8 mH. Supply 200 V, 50 Hz (reference). ω = 314.16 rad/s.

#### Reactances
- X_L (series) = 314.16 × 0.0191 = 6.00 Ω
- X_C = 1/(314.16 × 265.15 × 10⁻⁶) = 12.0 Ω
- X_L (branch B) = 314.16 × 0.0318 = 9.99 ≈ 10.0 Ω

#### Impedances
- Z_s = 4 + j6.0 Ω
- Z_A = 10 − j12.0 = 15.62∠−50.2° Ω
- Z_B = 6 + j10.0 = 11.66∠59.0° Ω

#### Parallel combination
Z_p = Z_A Z_B/(Z_A + Z_B) = 10.85 + j3.11 Ω

#### Total impedance
Z_T = Z_s + Z_p = 14.85 + j9.11 = 17.43∠31.5° Ω

#### Total current
I = 200/17.43 = 11.48 A, lagging V by 31.5°

#### Branch currents
V_p = I Z_p = 11.48 × 11.29 = 129.6 V

- I_A = 129.6/15.62 = 8.29 A (leading V by 34.7°)
- I_B = 129.6/11.66 = 11.12 A (lagging V by 74.6°)

#### Total power
P = I²R_T = I²(Re Z_T) = 11.48² × 14.85 = **1957 W** (≈ 1.96 kW)

Check by summing resistor losses:
- 4 Ω: 11.48² × 4 = 527 W
- 10 Ω: 8.29² × 10 = 688 W
- 6 Ω: 11.12² × 6 = 742 W
- Total = 1957 W

Also Q = 1200 VAR (lagging), S = 2296 VA, power factor = 0.852 lagging.

#### Phasor diagram
```text
         I_A (+34.7°, 8.29 A)
        ↗
  ─────────────────────────► V (0°)
        ↘ I (−31.5°, 11.48 A)
          ↘ I_B (−74.6°, 11.12 A)
```
I = I_A + I_B. The drop across the series part (4 Ω, 6 Ω reactance) is I Z_s and V = I Z_s + V_p.

## Check yourself

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

1. Branch A has C = 265.15 μF. Its X_C is:
   A. 12.0 Ω
   B. 6.0 Ω
   C. 10 Ω
   D. 31.8 Ω
   Answer: A. X_C = 1/(314.16 × 265.15×10^-6). (Key point: Reactances)

2. Z_T equals:
   A. Z_A Z_B
   B. Z_A + Z_B
   C. Z_s only
   D. Z_s + Z_AZ_B/(Z_A+Z_B) = 14.85 + j9.11 Ω
   Answer: D. Combine the parallel branches then add the series part. (Key point: Total impedance)

3. Total power is:
   A. V × I = 2296 W
   B. 200^2/17.43
   C. I^2 × Re(Z_T) = 11.48^2 × 14.85 = 1957 W
   D. I^2 × |Z|
   Answer: C. Only resistive part counts; check by adding I^2R in 4, 10 and 6 Ω resistors. (Key point: Power calculation)

4. Which resistors carry which currents?
   A. 4 Ω carries I_A
   B. All carry total
   C. 4 Ω carries total current; 10 Ω carries I_A; 6 Ω carries I_B
   D. 10 Ω carries I
   Answer: C. The series 4 Ω resistor carries the supply current, branches split it. (Key point: Branch current)

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