Question

An a.c. source is applied to a series LR circuit with $\text{X}_{\text{L}} = 3\text{R}$ and power factor is $\text{X}_1$. Now a capacitor with $\text{X}_{\text{c}} = \text{R}$ is added in series to LR circuit and the power factor is $\text{X}_2$. The ratio $\text{X}_1$ to $\text{X}_2$ is

• A. $1 : 2$
• B. $2 : 1$
• C. $1 : \sqrt{2}$
• D. $\sqrt{2} : 1$

Hint:

Power factor increases as reactance decreases (moving closer to resonance).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Power factor is defined as the cosine of the phase angle between voltage and current in an AC circuit.
It can be calculated as the ratio of true resistance to total impedance in a series circuit.
Adding a capacitor introduces capacitive reactance which partially cancels the inductive reactance, changing the total impedance and thus the power factor.
Step 2: Key Formula or Approach:
Power factor $\cos\phi = \frac{R}{Z}$.
Impedance for LR circuit: $Z_1 = \sqrt{R^2 + X_L^2}$.
Impedance for LCR circuit: $Z_2 = \sqrt{R^2 + (X_L - X_C)^2}$.
Step 3: Detailed Explanation:
Case 1: Series LR Circuit
Given $X_L = 3R$.
Calculate impedance $Z_1$: \[ Z_1 = \sqrt{R^2 + (3R)^2} = \sqrt{R^2 + 9R^2} = \sqrt{10R^2} = R\sqrt{10} \] Power factor $X_1$: \[ X_1 = \frac{R}{Z_1} = \frac{R}{R\sqrt{10}} = \frac{1}{\sqrt{10}} \] Case 2: Series LCR Circuit
A capacitor is added in series, with $X_C = R$.
The net reactance is $X_{\text{net}} = X_L - X_C = 3R - R = 2R$.
Calculate new impedance $Z_2$: \[ Z_2 = \sqrt{R^2 + X_{\text{net}}^2} = \sqrt{R^2 + (2R)^2} = \sqrt{R^2 + 4R^2} = \sqrt{5R^2} = R\sqrt{5} \] New power factor $X_2$: \[ X_2 = \frac{R}{Z_2} = \frac{R}{R\sqrt{5}} = \frac{1}{\sqrt{5}} \] Find the ratio $\text{X_1 : \text{X}_2$:}
\[ \text{Ratio} = \frac{X_1}{X_2} = \frac{1/\sqrt{10}}{1/\sqrt{5}} = \frac{\sqrt{5}}{\sqrt{10}} \] \[ \text{Ratio} = \sqrt{\frac{5}{10}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}} \] Step 4: Final Answer:
The ratio is $1 : \sqrt{2}$.