Question

A single-phase source has a terminal voltage $\mathbf{V}\mathbf{=}\mathbf{120}\mathbf{\angle }{\mathbf{0}}^{\mathbf{°}}$volts and a current localid="1655182868274" $\mathbf{I}\mathbf{=}\mathbf{15}\mathbf{\angle }{\mathbf{30}}^{\mathbf{°}}$A, which leaves the positive terminal of the source. Determine the real and reactive power, and state whether the source is delivering or absorbing each.

Short answer

The real power is 1039.2 W(delivered) and the reactive power is 600 VAR (absorbed).

Step-by-step solution

Obtain the apparent power

The apparent power is obtained by multiplying the line voltage and current as

$\mathbf{S}\mathbf{=}\mathbf{VI}$

Substitute $120\angle 0^{°}$for$V$, and $10\angle -{30}^{°}$for$I$into $S=VI$

$$\begin{gathered} \mathrm{S}=\left(120\angle 0^{°}\right)\left(10\angle -{30}^{°}\right) \\ \mathrm{S}=1200\angle -{30}^{°} \\ \mathrm{S}=1039.2-\mathrm{j}600 \end{gathered}$$

Thus, the apparent power is obtained as$s=1039.2-j600$.

Obtain the real power 

The real power is real part of apparent power $S=1039.2-J600$.thus the real power is 1039.2 W. Since the sign of the power is positive. Thus it is delivered.

Obtain the reactive power

The reactive power is imaginary part of apparent power $\mathrm{S}=1039.2-\mathrm{j}600.$Thus the reactive power is 600 VAR. Since the sign of the power is negative. Thus, it is absorbed.