- 2081 Baishakh (new course) · 6 marks
Series-parallel AC circuit currents and power
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).
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.]
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
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
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
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
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) |
✓ Answer checked by Claude Opus 5.5 (high effort)
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