I1 = V/Z1 using polar division: 200/30 = 6.667 A and angle 53.8° - 53.13° = 0.67°

How is the total current found?

Add the phasors I1 + I2 (rectangular sum)

Add magnitudes: 6.667 + 5.963 = 12.63 A

Total current has the same angle as the source voltage. What does this imply?

Unity power factor; reactive currents of the branches cancel

Which gives the real power?

P = I1^2 R1 + I2^2 R2 = 1333.3 W

P = 0 because the branches are reactive

2081 Chaitra (new course) · 6 marks

Parallel RL and RC branch currents, power

Short answer

Z1 = 18+j24 = 30∠53.13° Ω, Z2 = 15-j30 = 33.54∠-63.43° Ω. I1 = 200∠53.8° / Z1 = 6.667∠0.67° A, I2 = 200∠53.8° / Z2 = 5.963∠117.23° A. Total current I = I1+I2 = 6.667∠53.8° A. Real power = I1²×18 + I2²×15 = 1333.3 W.

From the given circuit, find total current (I), branch currents $I_1$ and $I_2$ , and real power. [Figure: source $200\angle 53.8°$ V feeding two parallel branches; branch 1 ( $I_1$ ) has $18\ \Omega$ in series with $j24\ \Omega$ ; branch 2 ( $I_2$ ) has $15\ \Omega$ in series with $-j30\ \Omega$ .]

Given: $V=200\angle53.8^\circ$ V, $Z_1=18+j24=30\angle53.13^\circ\ \Omega$ , $Z_2=15-j30=33.54\angle-63.43^\circ\ \Omega$ .

Branch currents:

I_1=\frac{V}{Z_1}=\frac{200\angle53.8^\circ}{30\angle53.13^\circ}=6.667\angle0.67^\circ\ \text{A}

I_2=\frac{V}{Z_2}=\frac{200\angle53.8^\circ}{33.54\angle-63.43^\circ}=5.963\angle117.23^\circ\ \text{A}

Total current (phasor sum):

I=I_1+I_2=(6.667+j0.078)+(-2.728+j5.302)=3.939+j5.380

In polar form, $I=6.667\angle53.8^\circ$ A. The total current is in phase with the source voltage: the reactive parts of the two branch admittances cancel ( $\frac{1}{Z_1}+\frac{1}{Z_2}=\frac{1}{30}$ S), so the combination behaves as a pure 30 Ω resistance.

Real power:

P=I_1^2R_1+I_2^2R_2=(6.667)^2\times18+(5.963)^2\times15=800+533.3=1333.3\ \text{W}

Check with $P=VI\cos\phi=200\times6.667\times\cos0^\circ=1333.3$ W, where $\phi=0$ since $I$ and $V$ have the same angle.

Answer: $I=6.667\angle53.8^\circ$ A, $I_1=6.667\angle0.67^\circ$ A, $I_2=5.963\angle117.23^\circ$ A, $P=1333.3$ W (unity power factor).

✓ Answer checked by Claude Opus 5.5 (high effort)

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