Question

The average power in a single-phase ac circuit with a purely capacitive load, for sinusoidal-steady-state excitation, is

(a) Zero

(b) $\raisebox{1ex}{${\mathbf{V}}_{\mathbf{max}}^{\mathbf{2}}$}\!\left/ \!\raisebox{-1ex}{${\mathbf{X}}_{\mathbf{c}}$}\right.$

(c) ${\mathbf{I}}_{\mathbf{rms}}^{\mathbf{2}}\mathbf{}{\mathbf{X}}_{\mathbf{c}}$

[Note: ${\mathbf{X}}_{\mathbf{c}}\mathbf{}\mathbf{=}\raisebox{1ex}{$\mathbf{1}$}\!\left/ \!\raisebox{-1ex}{$\left(\mathrm{\omega C}\right)$}\right.$ is the capacitive reactance]

Short answer

Answer

Therefore, the correct option is (a) zero

Step-by-step solution

Define the average power

The average power absorbed by a load is the product of the rms load voltage $V_{rms}$ , rms load current $I_{rms}$ , and the power factor$\cos \left(\delta -\beta \right)$ .

Use the theory of average power for pure capacitive load

The power factor for pure inductive load is 0 because the voltage lag current in phase$90°$.Thus, average power absorbed by a capacitive load is zero in a cycle, defined as,

$$\begin{gathered} \mathbf{P}\mathbf{=}{\mathbf{V}}_{\mathbf{rms}}\mathbf{}{\mathbf{I}}_{\mathbf{rms}}\mathbf{}\mathbf{cos}\left(\delta ‐\beta \right) \\ \mathbf{}\mathbf{}\mathbf{}={\mathrm{V}}_{\mathrm{rms}}{\mathrm{I}}_{\mathrm{rms}}\cos \left(90°\right) \\ ={\mathrm{V}}_{\mathrm{rms}}{\mathrm{I}}_{\mathrm{rms}}\left(0\right) \\ =0 \end{gathered}$$

…… (1)

From equation (1), the average power absorbed by capacitive load is zero. Thus option (a) is correct.

Explanation for option (b)

From equation (1),the average power absorbed by inductive load is zero.

Since, the average power, in option (b), is a function of maximum voltage, $V_{max}$and not the zero. Thus the option (b) is incorrect.

Explanation for option (b)

From equation (1),the average power absorbed by inductive load is zero.

Since, the average power, in option (c), is a function of rms current,$I_{rms}$and not the zero. Thus the option (c) is incorrect.