Question #165525

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
2021-02-22T09:46:00-0500

Let, Current in the first branch be I1=16AI_1 = 16A


Resistance in the first branch be R1=5ΩR_1 = 5\Omega


Applied voltage be V=100+j200.V = 100+j200.


Total Power P=5kWP = 5 kW


and, let the total current be I


We know,

P=VIP=VI


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


I=5000100+j200I = \dfrac{5000}{100+j200}


I=(5000)(100j200)50000I = \dfrac{(5000)(100-j200)}{50000}


I=10j20AI = 10-j20 A


ImpedanceZ1Z_1 in the first branch can be calculated as 100+j20016\dfrac{100+j200}{16}


Z1Z_1 = 6.25+j12.5Ω6.25+j12.5 \Omega


Current I2I_2 in the second branch can be calculated as,

I2=II1I_2 = I - I_1


I2=10j2016I_2 = 10-j20-16


I2=6j20AI_2 = -6-j20 A


Impedance Z2Z_2 in second branch can be calculated as 100+j2006j20\dfrac{100+j200}{-6-j20}

Z2=10.54+j1.83ΩZ2 = -10.54+j1.83 \Omega







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