Question

In an AC circuit, current is $3\,A$ and voltage $210\,V$ and power is $63\,W$. The power factor is
A. 0.11
B. 0.09
C. 0.08
D. 0.10

Answer 1

The power factor is the ratio of real power of the electrical component and apparent power. Use the relation between apparent power, voltage and current to determine the apparent current.

Formula used:
$P = {P_A}\left| {\cos \varphi } \right|$

Here, P is the real power, ${P_A}$ is the apparent power and $\left| {\cos \varphi } \right|$ is the power factor.

The apparent power is the product of voltage and current.
${P_A} = IV$

Complete step by step answer: We know power factor is defined as the ratio of real power to the apparent power. The real power is power used to do the work and apparent power is the power supplied to the circuit.

In AC circuits, the power factor is denoted by magnitude of cosine of the apparent power phase angle.

The real power is the product of apparent power and power factor. Therefore,
$P = {P_A}\left| {\cos \varphi } \right|$
$\Rightarrow \left| {\cos \varphi } \right| = \dfrac{P}{{{P_A}}}$

Here, $P$ is the real power and ${P_A}$ is the apparent power. The apparent power is the product of current and voltage given to the circuit.

Therefore, the above equation becomes,
$\Rightarrow \left| {\cos \varphi } \right| = \dfrac{P}{{IV}}$

Substitute 63 W for P, $3\,A$ for I and 210 V for V in the above equation.
$\left| {\cos \varphi } \right| = \dfrac{{63\,W}}{{\left( {3\,A} \right)\left( {210\,V} \right)}}$
$\Rightarrow \left| {\cos \varphi } \right| = \dfrac{{63\,W}}{{630\,W}}$
$\Rightarrow \left| {\cos \varphi } \right| = 0.10$

So, the correct answer is option (D).

Note: We often get confused between real power and apparent power of the electrical component. A electric bulb of power rate 10 W is the real power of the bulb, the supply to this bulb is usually 220 V AC and 5 A current, therefore, the apparent power of the bulb is 1100 W. The angle $\varphi$ is the angle between ${V_{rms}}$ and ${I_{rms}}$ of the supply.