---
title: "Series-parallel AC circuit analysis"
subject: "Basic Electrical Engineering"
chapter: 6
key: "semester_1/basic_electrical_engineering:series-parallel-ac-r1-15-l1-0-05-200v"
source: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/series-parallel-ac-r1-15-l1-0-05-200v
site: BCT Engineering
---

# Series-parallel AC circuit analysis

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

- Asked in: 2074 Asoj, 8 marks
- Video: [AC Analysis: Series/Parallel RLC Circuit](https://www.youtube.com/watch?v=s3Daf4GC_u4) — ElectronX Lab
- 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-ac-r1-15-l1-0-05-200v
- Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-6

## Question

Below Figure shows a series parallel circuit. Find: (i) total impedance (ii) current drawn from the circuit (iii) voltage across the parallel branches (iv) current flowing through each parallel branch (v) power factor (vi) Active, reactive and apparent power. Also, draw the phasor diagram of the circuit. [Figure: R1 = 15 Ω in series with L1 = 0.05 H, followed by two parallel branches: branch 1 is R2 = 10 Ω in series with L2 = 0.02 H; branch 2 is R3 = 20 Ω in series with C = 100 μF; supply 200 V, 50 Hz.]

## Short answer

Total impedance = 31.87∠35.64° Ω; source current = 6.28 A ∠-35.64°; parallel voltage = 70.7 V; branch currents: I1=5.99 A ∠-53.1°, I2=1.88 A ∠36.9°. Power factor = 0.813 lagging. P=1020 W, Q=731 VAR, S=1255 VA.

## Answer

Circuit: R₁ = 15 Ω and L₁ = 0.05 H in series with two parallel branches: branch 1 is R₂ = 10 Ω with L₂ = 0.02 H, and branch 2 is R₃ = 20 Ω with C = 100 μF. Supply 200 V, 50 Hz, taken as reference (V = 200∠0°). ω = 2π × 50 = 314.16 rad/s.

#### Reactances
- X_L1 = ωL₁ = 15.71 Ω
- X_L2 = ωL₂ = 6.283 Ω
- X_C = 1/(ωC) = 31.83 Ω

#### Impedances
- Z_series = 15 + j15.71 Ω
- Z₁ = 10 + j6.283 = 11.81∠32.1° Ω
- Z₂ = 20 − j31.83 = 37.60∠−57.9° Ω

#### (i) Total impedance
Z_p = Z₁Z₂ / (Z₁ + Z₂) = 10.90 + j2.86 = 11.27∠14.7° Ω

Z_T = Z_series + Z_p = (15 + 10.90) + j(15.71 + 2.86)

**Z_T = 25.90 + j18.57 = 31.87∠35.64° Ω**

#### (ii) Current drawn
I = V/Z_T = 200∠0° / 31.87∠35.64° = **6.28∠−35.64° A**

#### (iii) Voltage across parallel branches
V_p = I Z_p = 6.28 × 11.27 = **70.7 V** (∠−20.9°)

#### (iv) Branch currents
- I₁ = V_p/Z₁ = 70.7/11.81 = **5.99 A** (∠−53.1°)
- I₂ = V_p/Z₂ = 70.7/37.60 = **1.88 A** (∠+36.9°)

Check: I₁ + I₂ = (3.60 − j4.79) + (1.50 + j1.13) = 5.10 − j3.66 → ≈ 6.28∠−35.6° A.

#### (v) Power factor
cos φ = 25.90 / 31.87 = **0.813 lagging**

#### (vi) Powers
- Apparent: S = VI = 200 × 6.276 = **1255 VA**
- Active: P = VI cos φ = **1020 W** (check: 6.276² × 15 + 5.99² × 10 + 1.88² × 20 = 591 + 359 + 71 = 1021 W)
- Reactive: Q = VI sin φ = **731 VAR** (lagging)

#### Phasor diagram (V reference)
```text
                 I2 (+36.9°, 1.88 A)
                ↗
  ─────────────────────► V (0°)
      ↘
        I (−35.6°, 6.28 A), I1 (−53.1°, 5.99 A)
```
I₁ and I are lagging V (inductive branch dominates), I₂ leads V (capacitive branch).

## Check yourself

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

1. At 50 Hz, X_L1 for L1 = 0.05 H is:
   A. 31.83 Ω
   B. 0.05 Ω
   C. 6.28 Ω
   D. 15.71 Ω
   Answer: D. X_L = 2πfL = 314.16 × 0.05 = 15.71 Ω. (Key point: Reactance values)

2. How is the total impedance found?
   A. Z_T = Z_series + Z1Z2/(Z1+Z2), using complex impedances
   B. Add all impedances directly
   C. Add magnitudes of Z1 and Z2 then series
   D. Use the largest impedance
   Answer: A. Combine the two parallel branches as complex numbers, then add the series branch. (Key point: Method for total impedance)

3. How is the voltage across the parallel part and branch currents obtained?
   A. V_p = supply voltage 200 V
   B. V_p = I Z_p, then I1 = V_p/Z1 and I2 = V_p/Z2
   C. V_p = I × Z_series
   D. I1 = I2 = I/2
   Answer: B. The supply voltage is shared between series part and parallel part, so use V_p = I Z_p (70.7 V). (Key point: Branch current method)

4. With Z_T = 25.90 + j18.57 Ω, what is the power factor and nature?
   A. 0.813 leading as the capacitor dominates
   B. 0.813 leading
   C. 0.58 lagging
   D. 0.813 lagging
   Answer: D. cos φ = R/Z = 25.90/31.87 = 0.813; positive reactance means current lags. (Key point: Power factor and power)

## Close variants set in other exams

- [Series-parallel circuit with I1, I2, I3 phasors](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Aseries-parallel-ac-v100-z1-z2-z3) — Basic Electrical Engineering, chapter 6, asked 2×

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