Question

In a series $LR$ circuit with $X _{ L }= R$, power factor is $P _1$ If a capacitor of capacitance $C$ with $X _{ C }= X _{ L }$ is added to the circuit the power factor becomes $P_2$ The ratio of $P_1$ to $P_2$ will be :

• A. $1: 2$
• B. $1: 3$
• C. $1: 1$
• D. 1 : √2​

Answer 1

The Correct Option is D

To solve the problem of determining the ratio of the original and modified power factors in an $LR$ circuit with the addition of a capacitor, we proceed as follows:

Given:

• Original circuit: Series \(LR\) with resistance, \(R\), and inductive reactance, \(X _{ L } = R\).
• Power factor of original circuit: \(P _1\), which is given by:

\(P_1 = \frac{R}{Z} = \frac{R}{\sqrt{R^2 + X_L^2}} = \frac{R}{\sqrt{R^2 + R^2}} = \frac{R}{\sqrt{2R^2}} = \frac{1}{\sqrt{2}}\).

Next, we consider the modified circuit with a capacitor added:

• Capacitive reactance, \(X _{ C } = X _{ L } = R\).
• Now the total impedance becomes purely resistive, i.e., \(R\) (since inductive and capacitive reactances cancel out: \(X_{L} - X_{C} = 0\)).
• Power factor of the modified circuit: \(P _2 = \frac{R}{R} =\).

The task is to find the ratio of the original power factor to the modified power factor:

\(\text{Ratio } = \frac{P_1}{P_2} = \frac{\frac{1}{\sqrt{2}}}{1} = \frac{1}{\sqrt{2}}\).

Thus, the ratio of the power factors \(P_1 \text{ to } P_2\) is \(1 : \sqrt{2}\).

Conclusion: The correct option is \(1 : \sqrt{2}\), which matches the given answer.