two impedances Z1 and Z2 are connected in parallel across applied voltage of (100+j 200) volts the total power supplied to the circuit is 5 kW .the first branch takes a leading current of 16 A and has a resistance of 5ohms while the second branch takes a lagging current at 0.8 power factor .
Answer (1)
15th Jul, 2021
Its given that the current in the first branch is 1A, R1=5 ohms and voltage is 100+j200 volts.Power is 5kw.
so we know that P=V*I so we get I as 10 - j20 Amp by substituting values.
Hence Z1 = 6.25 + j12.5 Ω
I 2 = - 6 - j20 Amp so then finally z2 will be −10.54+j1.83Ω
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