Question

The power factor of an R-L circuit is $\frac{1}{\sqrt{2}}$. If the frequency of AC is doubled the power factor will now be

• A. 0
• B. 1
• C. $\frac{1}{\sqrt{2$
• D. Cannot be determined

Hint:

Power factor in an R-L circuit is $R/Z$. Doubling frequency increases inductive reactance ($X_L$), thus increasing impedance ($Z$) and decreasing power factor.

Answer 1

The Correct Option is A

Step 1: Formula of power factor
For a series R-L circuit: \[ \text{Power Factor} = \cos\phi = \frac{R}{Z} \] where \[ Z = \sqrt{R^2 + X_L^2}, \quad X_L = 2\pi f L \]
Step 2: Use given power factor
Given: \[ \cos\phi = \frac{1}{\sqrt{2}} \] \[ \frac{R}{Z} = \frac{1}{\sqrt{2}} \Rightarrow Z = R\sqrt{2} \]
Step 3: Relation between $R$ and $X_L$
\[ \sqrt{R^2 + X_L^2} = R\sqrt{2} \] Squaring: \[ R^2 + X_L^2 = 2R^2 \] \[ X_L^2 = R^2 \Rightarrow X_L = R \]
Step 4: Frequency is doubled
\[ X_L' = 2X_L = 2R \]
Step 5: New impedance
\[ Z' = \sqrt{R^2 + (2R)^2} = \sqrt{R^2 + 4R^2} = \sqrt{5R^2} = R\sqrt{5} \]
Step 6: New power factor
\[ \text{PF}' = \frac{R}{Z'} = \frac{R}{R\sqrt{5}} = \frac{1}{\sqrt{5}} \]
Final Answer:
\[ \boxed{\frac{1}{\sqrt{5}}} \]
Note: If this option is not present, then the given options likely contain an error.