Question

In the circuit shown below, the average power supplied by the dependent source is ______ W.

• A. 96

Answer 1

The correct option is A 96
The circuit can be redrawn as shown below,
$I=2\angle 0{}^{\circ }A$
By using source transformation,



Applying KVL in the loop-1:

$-9.60\angle 0{}^{\circ }-(4.8+jL92)I_x=V_c=0$

$V_s=\left(9.6\angle 0{}^{\circ }\right)+(4.8+j1.92)$ ... (i)

Applying KCL at node $V_s,$

$1.6I_x=I_x=\frac{V_x}{8}$

$1.6I_x=I_x+\frac{\left(9.6\right)+(4.8+j1.92)I_x}{8}$
Solving the above equation,

$I_x=5\angle 90{}^{\circ }A$
Substituting the value of $I_x$ in equation (i), we get

$V_x=\left(9.6\angle 0{}^{\circ }\right)+\left(\right(4.8+j1.92)\times 5\angle 90{}^{\circ })$

$=24\angle 90{}^{\circ }V$

$and1.6I_x=1.6\times 5\angle 90{}^{\circ }=8\angle 90{}^{\circ }A$
Power delivered by dependent source,

$P=V_{rms}{I}_{rms}\cos \varphi$

$=\frac{24}{\sqrt{2}}\times \frac{8}{\sqrt{2}}\times 1$

$P=96W$