Question

A series \( L, R \) circuit connected with an AC source \( E = (25 \sin 1000 \, t) \, \text{V} \) has a power factor of \( \frac{1}{\sqrt{2}} \). If the source of emf is changed to \( E = (20 \sin 2000 \, t) \, \text{V} \), the new power factor of the circuit will be:

• A. \( \frac{1}{\sqrt{2}} \)
• B. \( \frac{1}{\sqrt{3}} \)
• C. \( \frac{1}{\sqrt{5}} \)
• D. \( \frac{1}{\sqrt{7}} \)

Answer 1

The Correct Option is C

Step 1: Determine Initial Reactance \( X_L \): Since the initial power factor \( \cos \theta = \frac{1}{\sqrt{2}} \), we have \( \tan \theta = 1 \), meaning \( X_L = R \). With the initial angular frequency \( \omega_1 = 1000 \, \text{rad/s} \), we can write \( X_L = \omega_1 L = R \).

Step 2: Calculate Reactance at New Frequency: For the new frequency, \( \omega_2 = 2000 \, \text{rad/s} \), the new inductive reactance becomes:

\[ X'_L = \omega_2 L = 2 \omega_1 L = 2R \]

Step 3: Determine New Power Factor: With the new reactance \( X'_L = 2R \), we find \( \tan \theta' = \frac{X'_L}{R} = 2 \), which gives:

\[ \cos \theta' = \frac{1}{\sqrt{1 + (2)^2}} = \frac{1}{\sqrt{5}} \]

Conclusion: The new power factor is therefore:

\[ \frac{1}{\sqrt{5}} \]