---
title: "Series-parallel circuit with I1, I2, I3 phasors"
subject: "Basic Electrical Engineering"
chapter: 6
key: "semester_1/basic_electrical_engineering:series-parallel-ac-v100-z1-z2-z3"
source: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/series-parallel-ac-v100-z1-z2-z3
site: BCT Engineering
---

# Series-parallel circuit with I1, I2, I3 phasors

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

- Asked in: 2069 Asar, 8 marks
- Video: [Lesson 15 - Series And Parallel Circuits In Phasor Domain (AC Circuit Analysis)](https://www.youtube.com/watch?v=goEOoJvf3Yw) — Math and Science
- 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-v100-z1-z2-z3
- Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-6

## Question

Following 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. Draw phasor diagram showing V, I1, I2 and I3. [Figure: Z1 = 2 Ω in series with a 3 Ω inductive reactance; then three parallel branches: Z2 = 1 Ω with 5 Ω capacitive reactance, Z3 = 4 Ω in series with 2 Ω; supply V = 100∠0° V.]

## Short answer

Total impedance 5.866∠15.71° Ω, drawn current 17.05 A ∠-15.71°. Voltage across parallel branches 66.7 V ∠-36.87°. Branch currents: I2=13.07 A ∠41.82°, I3=14.91 A ∠-63.43°. Power factor 0.963 lagging, active 1641 W, reactive 461.5 VAR lagging, apparent 1705 VA.

## Answer

Circuit (from the figure): Z₁ = 2 + j3 Ω (2 Ω with an inductor of 3 Ω reactance) in series with two parallel branches. Z₂ = 1 − j5 Ω (1 Ω with a capacitor of 5 Ω reactance) and Z₃ = 4 + j2 Ω (4 Ω with an inductor of 2 Ω reactance). Supply V = 100∠0° V. I₁ is the source current and I₂, I₃ are the branch currents. The element types are read as inductive and capacitive from the figure's symbols.

#### (i) Total impedance
Z₂Z₃ = (1 − j5)(4 + j2) = 4 + j2 − j20 + 10 = 14 − j18

Z₂ + Z₃ = 5 − j3

Z_p = (14 − j18)/(5 − j3) = (14 − j18)(5 + j3)/34 = (124 − j48)/34 = 3.647 − j1.412 Ω

Z_T = Z₁ + Z_p = 5.647 + j1.588 = **5.866∠15.71° Ω**

#### (ii) Current drawn
I₁ = 100/5.866 = **17.05 A**, ∠−15.71° (= 16.42 − j4.62 A)

#### (iii) Voltage across parallel branches
V_p = I₁ Z_p = 17.05 × 3.911 = **66.7 V**, ∠−36.87°

#### (iv) Branch currents
- I₂ = V_p/Z₂ = 66.67/5.099 = **13.07 A**, ∠+41.82° (leading)
- I₃ = V_p/Z₃ = 66.67/4.472 = **14.91 A**, ∠−63.43° (lagging)

Check: I₂ + I₃ = (9.76 + j8.71) + (6.67 − j13.33) = 16.42 − j4.62 A = I₁.

#### (v) Power factor
cos φ = 5.647/5.866 = **0.963 lagging**

#### (vi) Powers
- Apparent: S = VI = 100 × 17.05 = **1705 VA**
- Active: P = VI cos φ = **1641 W**
- Reactive: Q = VI sin φ = **461.5 VAR** (lagging)

#### Phasor diagram (V reference)
```text
        I2 (+41.8°, 13.07 A)
         ↗
  ──────────────────────────► V (0°)
        ↘ I1 (−15.7°, 17.05 A)
          ↘ I3 (−63.4°, 14.91 A)
```
V_p = 66.7 V lies at −36.87°, and I₁ = I₂ + I₃.

## Check yourself

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

1. Z2 and Z3 are:
   A. 5 Ω and 6 Ω
   B. 1 − j5 and 4 − j2
   C. 1 − j5 Ω and 4 + j2 Ω
   D. 1 + j5 and 4 − j2
   Answer: C. Z2 has a capacitor (negative reactance), Z3 has an inductor. (Key point: Branch impedances)

2. Z_T = Z1 + Z2Z3/(Z2+Z3). Result is:
   A. 3.647 − j1.412 Ω
   B. 6 + j5 Ω
   C. 2 + j3 Ω
   D. 5.647 + j1.588 = 5.866∠15.71° Ω
   Answer: D. Z_p = 3.647 − j1.412; adding Z1 = 2 + j3 gives Z_T. (Key point: Total impedance)

3. How are I1, I2, I3 related?
   A. I1 = I2 − I3 in magnitudes
   B. I1 = I2 = I3
   C. I1 = I2 + I3 (phasor sum)
   D. I2 = I1 + I3
   Answer: C. I1 is the source current; the parallel branches share it (16.42 − j4.62 A). (Key point: Source current relation)

4. With Z_T = 5.647 + j1.588 Ω, the pf is:
   A. 0.963 leading
   B. 0.27 lagging
   C. 1
   D. 0.963 lagging
   Answer: D. cos φ = 5.647/5.866 and reactance is positive. (Key point: Power factor)

## Close variants set in other exams

- [Series-parallel AC circuit analysis](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Aseries-parallel-ac-r1-15-l1-0-05-200v) — Basic Electrical Engineering, chapter 6, asked 2×

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