Q.5 (c) Two impedances of 12 + j16 Ω and 10 − j20 Ω are connected in parallel across 230 V, 50 Hz, single phase ac supply. Find admittance, current and power factor of each branch.
Best answer
Take V = 230∠0° V.
Branch 1: Z1 = 12 + j16 = 20∠53.13° Ω.
\( Y_1 = \dfrac{1}{Z_1} = \dfrac{12-j16}{12^2+16^2} = \dfrac{12-j16}{400} = 0.03 - j0.04 \) S = 0.05∠−53.13° S.
\( I_1 = VY_1 = 230\times0.05\angle-53.13^\circ = \) 11.5∠−53.13° A (= 6.9 − j9.2 A).
pf1 = cos 53.13° = R1/Z1 = 12/20 = 0.6 lagging.
Branch 2: Z2 = 10 − j20 = 22.36∠−63.43° Ω.
\( Y_2 = \dfrac{10+j20}{10^2+20^2} = \dfrac{10+j20}{500} = 0.02 + j0.04 \) S = 0.04472∠63.43° S.
\( I_2 = 230\times0.04472\angle63.43^\circ = \) 10.29∠63.43° A (= 4.6 + j9.2 A).
pf2 = cos 63.43° = 10/22.36 = 0.447 leading.
Total (for completeness): Y = Y1 + Y2 = 0.05 + j0 S; I = 230 × 0.05 = 11.5∠0° A (= 6.9 − j9.2 + 4.6 + j9.2 = 11.5 A ✓); overall pf = 1 (unity); total power = 230 × 11.5 = 2645 W (check: 11.5² × 12 + 10.29² × 10 = 1587 + 1058 = 2645 W ✓).
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