---
title: "Series-parallel circuit currents and power"
subject: "Basic Electrical Engineering"
chapter: 6
key: "semester_1/basic_electrical_engineering:series-parallel-z1-6-0-01h-100v"
source: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/series-parallel-z1-6-0-01h-100v
site: BCT Engineering
---

# Series-parallel circuit currents and power

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

- Asked in: 2073 Shrawan, 8 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/series-parallel-z1-6-0-01h-100v
- Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-6

## Question

Determine the value of current I1, I2 and I and overall power factor of the circuit shown in figure below for series and parallel circuit. Also draw the phasor diagram and find the total power consumed by the circuit. [Figure: Z1 = 6 Ω in series with 0.01 H, followed by two parallel branches: Z2 = 4 Ω in series with 0.02 H (current I1) and a branch 2 Ω in series with 200 μF capacitor (current I2); supply 100 V, 50 Hz.]

## Short answer

Total impedance 17.29 Ω, supply current I=5.79 A lagging 30.76°. Parallel voltage 60.9 V gives I₁=8.18 A (lag), I₂=3.80 A (lead). Power factor 0.859 lagging, total power 497 W.

## Answer

Circuit: Z₁ = 6 Ω + 0.01 H in series with two parallel branches. Branch 1 (current I₁): Z₂ = 4 Ω + 0.02 H. Branch 2 (current I₂): 2 Ω + 200 μF. Supply 100 V, 50 Hz as reference. ω = 314.16 rad/s.

#### Reactances
- X_L1 = 3.142 Ω; X_L2 = 6.283 Ω; X_C = 15.92 Ω

#### Impedances
- Z₁ = 6 + j3.142 Ω
- Z_a = 4 + j6.283 = 7.45∠57.5° Ω
- Z_b = 2 − j15.92 = 16.04∠−82.8° Ω

#### Parallel part
Z_p = Z_a Z_b/(Z_a + Z_b) = 8.854 + j5.697 = 10.53∠32.8° Ω

#### Total impedance
Z_T = Z₁ + Z_p = 14.85 + j8.84 = 17.29∠30.76° Ω

#### Currents
- **I = V/Z_T = 100/17.29 = 5.79 A**, lagging V by 30.76°
- Parallel voltage V_p = I Z_p = 5.79 × 10.53 = 60.9 V
- **I₁ = 60.9/7.45 = 8.18 A** (∠−55.5°)
- **I₂ = 60.9/16.04 = 3.80 A** (∠+84.8°)

Check: I₁ + I₂ = (4.63 − j6.74) + (0.35 + j3.78) = 4.97 − j2.96 → 5.79∠−30.8° A.

#### Power factor and power
- **cos φ = 14.85/17.29 = 0.859 lagging**
- P = VI cos φ = 100 × 5.79 × 0.859 = **497 W**
 (check: 5.79² × 6 + 8.18² × 4 + 3.80² × 2 = 201 + 268 + 29 = 498 W)
- Q = VI sin φ = 296 VAR, S = 579 VA

#### Phasor diagram
```text
                I2 (+84.8°, 3.80 A)
               ↑
               │
   ────────────┼──────────────► V (0°)
               │  ↘ I (−30.8°, 5.79 A)
               │     ↘ I1 (−55.5°, 8.18 A)
```
I₁ lags V (inductive), I₂ leads V (capacitive), and I = I₁ + I₂.

## Check yourself

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

1. Which set of reactances is correct at 50 Hz?
   A. X_L1 = 31.4, X_L2 = 62.8, X_C = 15.92 Ω
   B. X_L1 = 3.142, X_L2 = 6.283, X_C = 63.7 Ω
   C. X_L1 = 3.142, X_L2 = 6.283, X_C = 15.92 Ω
   D. X_L1 = 6.283, X_L2 = 3.142, X_C = 15.92 Ω
   Answer: C. X_L = ωL with 0.01 H and 0.02 H, and X_C = 1/(ωC) with 200 μF. (Key point: Reactances)

2. Z_T = Z1 + Z_p. With Z1 = 6 + j3.142 and Z_p = 8.854 + j5.697, Z_T is:
   A. 17.29∠−30.76° Ω
   B. 14.85 + j8.84 = 17.29∠30.76° Ω
   C. 10.53∠32.8° Ω
   D. 6 + 8.854 Ω only
   Answer: B. Add real and imaginary parts: 14.85 + j8.84. (Key point: Total impedance)

3. What is I and the voltage across the parallel part?
   A. I = 8.18 A, V_p = 100 V
   B. I = 5.79 A, V_p = 100 V
   C. I = 3.80 A, V_p = 60.9 V
   D. I = 5.79 A, V_p = 60.9 V
   Answer: D. I = 100/17.29 = 5.79 A and V_p = I Z_p = 5.79 × 10.53 = 60.9 V. (Key point: Supply current)

4. Which statement is correct?
   A. I1 lags V, I2 leads V; P = 497 W (equals sum of I^2R in all resistors)
   B. P = 100 × 8.18 W
   C. Both lag V; P = 579 W
   D. I1 leads V; I2 lags V
   Answer: A. Branch 1 is inductive, branch 2 capacitive; real power equals losses in 6 Ω, 4 Ω and 2 Ω resistors. (Key point: Branch current nature and power)

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