Question

A voltage of $10\mathrm{V}$ and frequency $1000\mathrm{Hz}$ is applied to a $\left(\frac{1}{\mathrm{\pi }}\right)\mathrm{\mu F}$ capacitor. Find the power factor of the circuit and the average power dissipated.

• A. $40.018\mathrm{W}$
• B. $0.1\mathrm{W}$
• C. $10.18\mathrm{W}$
• D. $20.018\mathrm{W}$

Answer 1

The correct option is B

$0.1\mathrm{W}$

Step 1: Given

Voltage, $V=10\mathrm{V}$

Frequency, $\mathrm{f}=1000\mathrm{Hz}$

Resistance, $\mathrm{R}=500\mathrm{\Omega }$

Capacitor, $\mathrm{C}=\left(\frac{1}{\mathrm{\pi }}\right)\mathrm{\mu F}$

Step 2: To find

1. The power factor of the circuit.
2. The average power dissipated.

Step 3: Finding the power factor

We know the formula,

$$\begin{gathered} \mathrm{Power}\mathrm{factor}=\frac{\mathrm{actual}\mathrm{power}}{\mathrm{apparent}\mathrm{power}} \\ =\frac{{\mathrm{V}}_{\mathrm{rms}}\times {\mathrm{I}}_{\mathrm{rms}}\times \mathrm{Cos}\mathrm{\phi }}{{\mathrm{V}}_{\mathrm{rms}}\times {\mathrm{I}}_{\mathrm{rms}}} \\ =\mathrm{Cos}\mathrm{\phi } \end{gathered}$$

Here,

${\mathrm{V}}_{\mathrm{rms}}$ is the rms voltage.

${\mathrm{I}}_{\mathrm{rms}}$ is the rms current.

$\mathrm{rms}$ is the root mean square.

Now,

The reactive power will be:

$$\begin{gathered} \mathrm{Xc}=\frac{1}{\mathrm{\omega C}} \\ =\frac{1}{2\mathrm{\pi f}C} \\ =\frac{1}{(2\times \mathrm{\pi }\times 1000\times \left({\displaystyle \frac{1}{\mathrm{\pi }}}\right)\times {10}^{-6})} \\ =500 \end{gathered}$$

and,

$$\begin{gathered} \tan \mathrm{\phi }=\frac{\mathrm{Xc}}{\mathrm{R}} \\ =\frac{500}{500} \\ =1 \\ \mathrm{\phi }=45° \end{gathered}$$

$$\begin{gathered} \mathrm{Power}\mathrm{factor}=\mathrm{Cos}\mathrm{\phi } \\ =Cos45° \\ =\frac{1}{\sqrt{2}} \end{gathered}$$

Step 4: Finding the average power dissipated

$$\begin{gathered} {\mathrm{I}}_{\mathrm{rms}}=\frac{\mathrm{V}}{\mathrm{Xc}\sqrt{2}} \\ {\mathrm{I}}_{\mathrm{rms}}=\frac{10}{500\sqrt{2}} \\ =\frac{1}{50\sqrt{2}} \end{gathered}$$

$$\begin{gathered} \mathrm{Power}\mathrm{dissipated}={\mathrm{V}}_{\mathrm{rms}}\times {\mathrm{I}}_{\mathrm{rms}}\times \mathrm{Cos}\mathrm{\phi } \\ =10\times \left(\frac{1}{50\sqrt{2}}\right)\times \left(\frac{1}{\sqrt{2}}\right) \\ =0.1\mathrm{W} \end{gathered}$$

Therefore, the power factor of the circuit is $\frac{1}{\sqrt{2}}$ and the average power dissipated is $0.1\mathrm{W}$.