Question

For the circuit shown in the figure, the active power supplied by the source is ________ W (rounded off to one decimal place).

 

Hint:

The active power supplied by an AC source is the real part of the complex power $S = V I^$. Calculate the total current drawn from the source and then use this formula.

Answer 1

Step 1: Calculate the equivalent impedance of the parallel combination of the $5 - j5 \Omega$ branch and the $j5 \Omega$ inductor.
$Z_{eq1} = \frac{(5 - j5)(j5)}{(5 - j5) + j5} = 5 + j5 \Omega$.
Step 2: Calculate the total impedance seen by the source.
$Z_{total} = (10 - j10) || (5 + j5) = \frac{(10 - j10)(5 + j5)}{(10 - j10) + (5 + j5)} = 6 + j2 \Omega$.
Step 3: Calculate the current supplied by the source.
$I = \frac{100\angle 0^\circ}{6 + j2} = 15 - j5 A$.
Step 4: Calculate the active power supplied by the source using $P = Re(V I^)$.
$I^ = 15 + j5 A$.
$P = Re((100\angle 0^\circ)(15 + j5)) = Re(1500 + j500) = 1500 W$.
Step 5: Round off to one decimal place.
$P = 1500.0 W$.