---
title: "Series-parallel AC circuit currents and power"
subject: "Fundamental of Electrical and Electronics Engineering"
chapter: 3
key: "semester_1/fundamental_of_electrical_and_electronics_engineering:series-parallel-ac-circuit-power"
source: https://www.bctengineer.com/q/semester_1/fundamental_of_electrical_and_electronics_engineering/series-parallel-ac-circuit-power
site: BCT Engineering
---

# Series-parallel AC circuit currents and power

Fundamental of Electrical and Electronics Engineering · Chapter 3: AC Circuit Analysis

- Asked in: 2081 Baishakh (new course), 6 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/fundamental_of_electrical_and_electronics_engineering/series-parallel-ac-circuit-power
- Whole chapter: https://www.bctengineer.com/md/semester_1/fundamental_of_electrical_and_electronics_engineering/chapter-3

## Question

From the given circuit, find the branch current, total current, overall power factor, real power and reactive power. [Figure: a 200 V, 50 Hz AC source supplies impedance ZC = (3 + j2.5) Ω (carrying current IC, from terminal C to node A) in series with a parallel combination between nodes A and B of ZA = (2 + j1.5) Ω (branch current IA) and ZB = (5 − j3.5) Ω (branch current IB); node B returns to the source.]

## Short answer

Branch currents: IA = 28.41∠−51.21° A, IB = 11.64∠20.65° A; total current IC = 33.89∠−32.17° A. Overall power factor is 0.8465 lagging; real power P ≈ 5738 W, reactive power Q ≈ 3609 VAR (inductive).

## Answer

Take the supply voltage as reference: V = 200∠0° V, f = 50 Hz.

ZC = 3 + j2.5 Ω, ZA = 2 + j1.5 Ω, ZB = 5 − j3.5 Ω.

#### Step 1: Parallel impedance ZAB

```text
ZA × ZB = (2 + j1.5)(5 − j3.5) = 15.25 + j0.5
ZA + ZB = 7 − j2
ZAB = (15.25 + j0.5)/(7 − j2)
    = 1.995 + j0.642 Ω  = 2.096∠17.82° Ω
```

#### Step 2: Total impedance and total current

```text
Z  = ZC + ZAB = (3 + j2.5) + (1.995 + j0.642)
   = 4.995 + j3.142 Ω = 5.901∠32.17° Ω
IC = V/Z = 200∠0° / 5.901∠32.17°
   = 33.89∠−32.17° A  (= 28.69 − j18.04 A)
```

#### Step 3: Branch currents

```text
VAB = IC × ZAB = 33.89∠−32.17° × 2.096∠17.82°
    = 71.03∠−14.34° V
IA  = VAB/ZA = 71.03∠−14.34° / 2.5∠36.87°
    = 28.41∠−51.21° A
IB  = VAB/ZB = 71.03∠−14.34° / 6.103∠−34.99°
    = 11.64∠20.65° A
Check: IA + IB = (17.80 − j22.15) + (10.89 + j4.10)
               = 28.69 − j18.04 A = IC  ✓
```

#### Step 4: Power factor and powers

```text
pf = cos 32.17° = 0.8465 (lagging)
S  = V × I = 200 × 33.89    = 6778.5 VA
P  = V I cos φ              = 5738.1 W
Q  = V I sin φ              = 3608.7 VAR (inductive)
Check: P = I² R = 33.89² × 4.995 = 5738 W ✓
```

**Answer:**

| Quantity | Value |
|---|---|
| IA | 28.41∠−51.21° A |
| IB | 11.64∠20.65° A |
| Total current IC | 33.89∠−32.17° A |
| Power factor | 0.8465 lagging |
| Real power P | ≈ 5738 W |
| Reactive power Q | ≈ 3609 VAR (lagging) |

## Check yourself

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

1. How is ZAB of the parallel branches ZA and ZB found?
   A. ZA + ZB
   B. (ZA + ZB)/(ZA x ZB)
   C. (ZA x ZB)/(ZA + ZB)
   D. ZA x ZB
   Answer: C. For two impedances in parallel, Z = product/sum, with complex arithmetic. (Key point: Parallel impedance method)

2. Total impedance is Z = ZC + ZAB = 4.995 + j3.142 ohm. What is the total current from 200 V?
   A. 33.89 A at +32.17 degrees
   B. 5.90 A at -32.17 degrees
   C. 200 A at 0 degrees
   D. 33.89 A at -32.17 degrees
   Answer: D. I = V/Z = 200/5.901 = 33.89 A, lagging the voltage by the impedance angle 32.17 degrees. (Key point: Total impedance and current)

3. How are the branch currents IA and IB obtained?
   A. Find VAB = IC x ZAB, then divide by each branch impedance
   B. Divide 200 V by ZA and ZB
   C. Add IC to ZA and ZB
   D. Multiply IC by each branch impedance
   Answer: A. Both branches share VAB across them, so IA = VAB/ZA and IB = VAB/ZB; IA + IB = IC checks. (Key point: Branch currents)

4. Which results are correct for pf, P and Q?
   A. pf = 0.8465 leading, P about 3609 W, Q about 5738 W
   B. pf = 0.8465 lagging, P about 5738 W, Q about 3609 VAR
   C. pf = 0.5 lagging, P about 6778 W
   D. pf = 1, P = 6778 W, Q = 0
   Answer: B. pf = cos 32.17 degrees; S = 200 x 33.89 = 6778 VA; P = S cos phi, Q = S sin phi. (Key point: Power factor and powers)

## Related in Fundamental of Electrical and Electronics Engineering

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