Question

The power factor of R-L circuit is $\frac{1}{\sqrt{3}}$. If the inductive reactance is $2\Omega$. The value of resistance is

• A. $2\Omega$
• B. $\sqrt{2}\Omega$
• C. $0.5\Omega$
• D. $\frac{1}{\sqrt{2}}\Omega$

Answer 1

The Correct Option is B

The power factor of an R-L circuit is given by the formula: \[ \text{Power Factor} = \frac{R}{\sqrt{R^2 + X_L^2}} \] Where:
\( R \) is the resistance,
\( X_L \) is the inductive reactance.
Given that the power factor is \( \frac{1}{\sqrt{3}} \) and \( X_L = 2 \, \Omega \),
we substitute these values into the formula: \[ \frac{1}{\sqrt{3}} = \frac{R}{\sqrt{R^2 + 2^2}} \] \[ \frac{1}{\sqrt{3}} = \frac{R}{\sqrt{R^2 + 4}} \] Squaring both sides: \[ \frac{1}{3} = \frac{R^2}{R^2 + 4} \] Cross-multiplying: \[ R^2 + 4 = 3R^2 \] \[ 4 = 2R^2 \] \[ R^2 = 2 \quad \Rightarrow \quad R = \sqrt{2} \, \Omega \] Thus, the value of resistance is \( \sqrt{2} \, \Omega \).

Answer 2

he power factor \( \text{pf} \) of an R-L circuit is given by: \[ \text{pf} = \frac{R}{\sqrt{R^2 + X_L^2}} \] Where:
\( R \) is the resistance,
\( X_L \) is the inductive reactance.

We are given that the power factor \( \text{pf} = \frac{1}{\sqrt{3}} \) and the inductive reactance \( X_L = 2 \, \Omega \). Substitute these values into the formula: \[ \frac{1}{\sqrt{3}} = \frac{R}{\sqrt{R^2 + 2^2}} \] \[ \frac{1}{\sqrt{3}} = \frac{R}{\sqrt{R^2 + 4}} \] Now, square both sides to eliminate the square roots: \[ \frac{1}{3} = \frac{R^2}{R^2 + 4} \] Cross-multiply: \[ R^2 + 4 = 3R^2 \] \[ 4 = 2R^2 \] \[ R^2 = 2 \] \[ R = \sqrt{2} \, \Omega \]

Thus, the value of the resistance is \( \sqrt{2} \, \Omega \).