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
Expert's answer
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=VIP = V * I


5000=(100+j200)I5000 = (100 + j200) * I


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


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


Z1=6.25+j12.5ΩZ1 = 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+j1.83ΩZ2 = -10.54 + j1.83 Ω




Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS