Question

In an AC circuit, supply voltage \( V_{\text{rms}} = 1000\,\text{V} \), \( R = 80\,\Omega \), \( X_L = 80\,\Omega \), and source frequency \( f = 50\,\text{Hz} \). Find the power factor.

• A. \( \dfrac{1}{\sqrt{2}} \)
• B. \( \dfrac{1}{\sqrt{3}} \)
• C. \( \dfrac{1}{\sqrt{5}} \)
• D. \( \dfrac{1}{\sqrt{6}} \)

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The power factor in an AC circuit is defined as the cosine of the phase angle (\(\phi\)) between the voltage and the current. It represents the ratio of true power to apparent power.
Step 2: Key Formula or Approach:
The formula for the power factor is: \[ \cos \phi = \frac{R}{Z} \] where the total impedance \(Z\) for a series RL circuit is given by: \[ Z = \sqrt{R^2 + X_L^2} \]
Step 3: Detailed Explanation:
We are given \(R = 80 \, \Omega\) and \(X_L = 80 \, \Omega\). Calculate the impedance \(Z\): \[ Z = \sqrt{80^2 + 80^2} = \sqrt{2 \times 80^2} = 80\sqrt{2} \, \Omega \] Now compute the power factor: \[ \cos \phi = \frac{R}{Z} = \frac{80}{80\sqrt{2}} = \frac{1}{\sqrt{2}} \]
Step 4: Final Answer:
The power factor of the circuit is \(\frac{1}{\sqrt{2}}\).