The power factor of an R-L circuit is 1/`sqrt 2` if the frequency of AC is doubled , what will be the power

Options

A. `1/sqrt 3`
B. `1/sqrt(5)`
C. `1/sqrt 7`
D. `1/sqrt 11`

Explanation

To solve the problem step by step, we will analyze the given information about the R-L circuit, calculate the new power factor after the frequency is doubled, and derive the final result.

### Step 1: Understand the given power factor
The power factor (PF) of the R-L circuit is given as \( \cos \phi = \frac{1}{\sqrt{2}} \). This can be expressed in terms of resistance (R) and impedance (Z):
\[ \cos \phi = \frac{R}{Z} \]
Thus, we can write:
\[ Z = \frac{R}{\cos \phi} = \frac{R}{\frac{1}{\sqrt{2}}} = R \sqrt{2} \]

### Step 2: Relate impedance to resistance and inductive reactance
The impedance \( Z \) of an R-L circuit is also defined as:
\[ Z = \sqrt{R^2 + X_L^2} \]
where \( X_L \) is the inductive reactance given by \( X_L = \omega L = 2 \pi f L \). From the previous step, we have:
\[ R \sqrt{2} = \sqrt{R^2 + X_L^2} \]
Squaring both sides gives:
\[ 2R^2 = R^2 + X_L^2 \]
This simplifies to:
\[ X_L^2 = R^2 \implies X_L = R \]

### Step 3: Determine the new inductive reactance when frequency is doubled
If the frequency \( f \) is doubled, the new inductive reactance \( X_L' \) can be calculated as:
\[ X_L' = \omega' L = 2 \pi (2f) L = 2X_L \]
Since we found that \( X_L = R \), we have:
\[ X_L' = 2R \]

### Step 4: Calculate the new impedance
Now we can find the new impedance \( Z' \):
\[ Z' = \sqrt{R^2 + (X_L')^2} = \sqrt{R^2 + (2R)^2} = \sqrt{R^2 + 4R^2} = \sqrt{5R^2} = R\sqrt{5} \]

### Step 5: Find the new power factor
The new power factor \( \cos \phi' \) can now be calculated:
\[ \cos \phi' = \frac{R}{Z'} = \frac{R}{R\sqrt{5}} = \frac{1}{\sqrt{5}} \]

### Final Answer
Thus, the new power factor when the frequency is doubled is:
\[ \cos \phi' = \frac{1}{\sqrt{5}} \]