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Answer to Question #167085 in Electrical Engineering for RohithReddy

Question #167085

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.

1
2021-02-26T05:40:23-0500

Let the current in the first branch to be I1 = 16 A

Resistance in the first branch to be R1 = 5 Ω

Applied Voltage to be V = 100 + j 200

Total Power P = 5 kW

Let the total current to be I, then as we know

"P = V * I"

"5000 = (100 + j200) * I"

"I = 5000\/(100+j200)" I = 10 - j20 Amp

Impedance Z1 in the first branch can be calculated as (100 + j200)/16

"Z1 = 6.25 + j12.5 \u03a9"

Current I2 = I - I1

I2 = 10 - j20 - 16

I2 = - 6 - j20 Amp

Impedance Z1 in the first branch can be calculated as (100 + j200)/(-6 -j20)

"Z2 = -10.54 + j1.83 \u03a9"

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