How is the total impedance found?

Z_T = Z_series + Z1Z2/(Z1+Z2), using complex impedances

Add magnitudes of Z1 and Z2 then series

How is the voltage across the parallel part and branch currents obtained?

V_p = I Z_p, then I1 = V_p/Z1 and I2 = V_p/Z2

With Z_T = 25.90 + j18.57 Ω, what is the power factor and nature?

0.813 leading as the capacitor dominates

Asked 2/15
2074 Asoj · 8 marks

Series-parallel AC circuit analysis

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.

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.]

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)

                 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).

✓ Answer checked by Claude Opus 5.5 (high effort)

Series-parallel circuit with I1, I2, I3 phasors · Basic Electrical Engineering, chapter 6 · asked 2×
Parallel RL and RC branches, total current · Basic Electrical Engineering, chapter 6 · asked 2×
Total power of series-parallel circuit · Basic Electrical Engineering, chapter 6
Series-parallel circuit currents and power · Basic Electrical Engineering, chapter 6
Branch currents in series-parallel AC · Basic Electrical Engineering, chapter 6
Series RLC circuit analysis · Basic Electrical Engineering, chapter 6
Parallel RL and RC branches · Basic Electrical Engineering, chapter 6 · asked 2×
Power factor and series-parallel circuit calculation · Fundamental of Electrical and Electronics Engineering, chapter 3
Series-parallel AC circuit currents and power · Fundamental of Electrical and Electronics Engineering, chapter 3
Parallel RL and RC branch currents, power · Basic Electrical and Electronics Engineering, chapter 1
Zener circuit output voltage and current · Basic Electrical and Electronics Engineering, chapter 3

All Basic Electrical Engineering chapter 6 questions →