Answer to Question #165525 in Quantum Mechanics for Anurag

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, Current in the first branch be $I_1 = 16A$

Resistance in the first branch be $R_1 = 5\Omega$

Applied voltage be $V = 100+j200.$

Total Power $P = 5 kW$

and, let the total current be I

We know,

$P=VI$

$5000 = (100+j200)I$

$I = \dfrac{5000}{100+j200}$

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

$I = 10-j20 A$

Impedance$Z_1$ in the first branch can be calculated as $\dfrac{100+j200}{16}$

$Z_1$ = $6.25+j12.5 \Omega$

Current $I_2$ in the second branch can be calculated as,

$I_2 = I - I_1$

$I_2 = 10-j20-16$

$I_2 = -6-j20 A$

Impedance $Z_2$ in second branch can be calculated as $\dfrac{100+j200}{-6-j20}$

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