Question

The power factor of an R-L circuit is $\dfrac{1}{\sqrt{2}}$. If the frequency is doubled, what is the power factor?
A)$\dfrac{1}{\sqrt{3}}$
B)$\dfrac{1}{\sqrt{5}}$
C)$\dfrac{1}{\sqrt{7}}$
D)$\dfrac{1}{\sqrt{11}}$

Answer 1

Hint: Write down the power factor of an R-L circuit in terms of frequency. Power factor is used to measure how efficiently a machine is working. It tells us the amount of power that does no useful work in the machine. The apparent power in this formula is also called demand.

Formula used:
$P.F=\dfrac{1}{\sqrt{1+{{({{X}_{L}})}^{2}}}}$

Complete step by step solution:
Power factor is the ratio between true power and apparent power. ${{X}_{L}}$is the inductive reactance of the given R-L circuit. We can write inductive reactance in terms of frequency as
$\begin{align}
  & {{X}_{L}}=L\omega \\
 & {{X}_{L}}=2\pi fL \\
\end{align}$
When frequency is doubled,
$\begin{align}
  & P.F=\dfrac{1}{\sqrt{1+{{(\dfrac{{{X}_{L}}}{R})}^{2}}}} \\
 & \dfrac{1}{\sqrt{2}}=\dfrac{1}{\sqrt{1+{{(\dfrac{{{X}_{L}}}{R})}^{2}}}} \\
 & 2=1+(\dfrac{{{X}_{L}}}{R}) \\
 & 1=\dfrac{{{X}_{L}}}{R} \\
 & \\
\end{align}$
If the frequency is doubled, the above term will also get doubled, so it becomes,
$\begin{align}
  & P.F=\dfrac{1}{\sqrt{1+{{(2)}^{2}}}} \\
 & P.F=\dfrac{1}{\sqrt{5}} \\ +
\end{align}$
Therefore the answer is option B.

       

Additional information:
An R-L circuit is an electrical circuit which consists of a resistor, inductor and a voltage source or current source. As a resistor is present in this circuit, it consumes more energy compared to RC circuit or RLC circuit. Usually, these circuits also consume a small amount of energy as the components and wires are non-zero resistance ones. The impedance of series RL circuit opposes the flow of alternating current. The impedance of series RL Circuit is nothing but the combined effect of resistance (R) and inductive reactance (XL) of the circuit as a whole.

Note: The power factor is used to find out useless current in the machine. Improving the power factor will maximize current carrying capacity, improve voltage to equipment and reduce power losses. In the formula, take care while writing inductive reactance in terms of frequency and amplitude.