Answer to Question #167085 in Electrical Engineering for RohithReddy

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

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

$Z1 = 6.25 + j12.5 Ω$

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)

$Z2 = -10.54 + j1.83 Ω$