If the power factor changes from `1/2 ` to `1/4` then what is the increase in impedance in AC?

Options

A. 0.2
B. 0.5
C. 0.25
D. 1

Answer

To solve the problem of finding the increase in impedance when the power factor changes from \( \frac{1}{2} \) to \( \frac{1}{4} \), we can follow these steps:

### Step 1: Understand the Power Factor Formula
The power factor (PF) is given by the formula:
\[ \text{PF} = \cos \phi = \frac{R}{Z} \]
where \( R \) is the resistance and \( Z \) is the impedance.

### Step 2: Calculate Initial Impedance \( Z_1 \)
Given the initial power factor \( \cos \phi_1 = \frac{1}{2} \):
\[ Z_1 = \frac{R}{\cos \phi_1} = \frac{R}{\frac{1}{2}} = 2R \]

### Step 3: Calculate Final Impedance \( Z_2 \)
Now, for the new power factor \( \cos \phi_2 = \frac{1}{4} \):
\[ Z_2 = \frac{R}{\cos \phi_2} = \frac{R}{\frac{1}{4}} = 4R \]

### Step 4: Calculate the Increase in Impedance
The increase in impedance \( \Delta Z \) can be calculated as:
\[ \Delta Z = Z_2 - Z_1 = 4R - 2R = 2R \]

### Step 5: Conclusion
Thus, the increase in impedance when the power factor changes from \( \frac{1}{2} \) to \( \frac{1}{4} \) is:
\[ \Delta Z = 2R \]