By what percentage the impedance in an `AC` series circuit should be increased so that the power factor changes from `(1//2)` to `(1//4)` (when `R` is constant)?

Options

A. `200%`
B. `100%`
C. `50%`
D. `400%`

Explanation

To solve the problem, we need to determine by what percentage the impedance in an AC series circuit should be increased so that the power factor changes from \( \frac{1}{2} \) to \( \frac{1}{4} \), given that the resistance \( R \) remains constant.

### Step-by-Step Solution:

1. **Understanding Power Factor**:
The power factor \( \cos \phi \) is defined as the ratio of the resistance \( R \) to the impedance \( Z \):
\[ \cos \phi = \frac{R}{Z} \]

2. **Initial Power Factor**:
For the initial power factor of \( \frac{1}{2} \):
\[ \frac{1}{2} = \frac{R}{Z_1} \]
Rearranging gives:
\[ Z_1 = 2R \]

3. **Final Power Factor**:
For the final power factor of \( \frac{1}{4} \):
\[ \frac{1}{4} = \frac{R}{Z_2} \]
Rearranging gives:
\[ Z_2 = 4R \]

4. **Calculating the Change in Impedance**:
The change in impedance is:
\[ \Delta Z = Z_2 - Z_1 = 4R - 2R = 2R \]

5. **Calculating the Percentage Increase**:
The percentage increase in impedance is given by:
\[ \text{Percentage Increase} = \frac{\Delta Z}{Z_1} \times 100\% \]
Substituting the values:
\[ \text{Percentage Increase} = \frac{2R}{2R} \times 100\% = 100\% \]

### Final Answer:
The impedance should be increased by **100%**.