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

Updated On: Apr 2, 2025
  • $2\Omega$
  • $\sqrt{2}\Omega$
  • $0.5\Omega$
  • $\frac{1}{\sqrt{2}}\Omega$
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The Correct Option is B

Solution and Explanation

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 \( \sqrt{2} \, \Omega \).

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Concepts Used:

LCR Circuit

An LCR circuit, also known as a resonant circuit, or an RLC circuit, is an electrical circuit consist of an inductor (L), capacitor (C) and resistor (R) connected in series or parallel.

Series LCR circuit

When a constant voltage source is connected across a resistor a current is induced in it. This current has a unique direction and flows from the negative to positive terminal. Magnitude of current remains constant.

Alternating current is the current if the direction of current through this resistor changes periodically. An AC generator or AC dynamo can be used as AC voltage source.