Answer to Question #167097 in Electric Circuits for Vinit Kumawat

Question

. Two impedances Z1 and Z2 are connected in parallel. The first branch takes a leading current of 16 A and has a resistance of 5 Ω, while the second branch takes a lagging current at 0.8pf. The applied voltage is 100+j200 V and the total power is 5 kW. Find branch impedances, total circuit impedance, branch currents and total circuit current.

Answer 1

Current in the first branch: $I_1=16\ A$

Applied Voltage: $U=100+j200\ V$

Total Power: $P=5\ kW$

Then:

$P=UI$

$5000 =I (100 + j200)$

$I=5000/(100+j200)=10-j20\ A$

Impedance in the first branch:

$Z_1=U/I_1=(100 + j200)/16=6.25+j12.5 \Omega$

$I_2=I-I_1=10-j20-16=-6-j20\ A$

$Z_2=U/I_2=−10.54+j1.83\ \Omega$

Total impedance:

$Z=U/I=\frac{100+j200}{10-j20}=\frac{(100+j200)(10-j20)}{500}=\frac{5000}{500}=10\ \Omega$