Question

For the circuit, shown in the figure, the total real power delivered by the source to the load is kW.

• A. 1.866

Answer 1

The correct option is A 1.866

$\overline{I}=\overline{I_1}+\overline{I_2}=5\angle 0^{\circ }+5\angle {30}^{\circ }$

$I=9.659\angle {14.94}^{\circ }$

$P=VIcos\theta$

= 200 $\times$ 9.659 $\times$ cos(14.94)

= 1.866 kW