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2069 Asar · 8 marks

Series-parallel circuit with I1, I2, I3 phasors

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.

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.]

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)

        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.

✓ Answer checked by Claude Opus 5.5 (high effort)

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