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

Options

A. 200%
B. 100%
C. 60%
D. 40%

Answer

To solve the problem step by step, we need to analyze the power factor changes in an AC series circuit and how it relates to impedance.

### Step 1: Understand the Power Factor
The power factor (PF) in an AC circuit is defined as:
\[ \text{PF} = \cos \theta = \frac{R}{Z} \]
where \( R \) is the resistance and \( Z \) is the impedance.

### Step 2: Set Up Initial Conditions
Initially, we have:
\[ \text{PF}_1 = \frac{1}{2} \]
This implies:
\[ \frac{R}{Z_1} = \frac{1}{2} \]
From this, we can express the initial impedance \( Z_1 \):
\[ Z_1 = 2R \]

### Step 3: Set Up Final Conditions
Now, the power factor changes to:
\[ \text{PF}_2 = \frac{1}{4} \]
This implies:
\[ \frac{R}{Z_2} = \frac{1}{4} \]
From this, we can express the final impedance \( Z_2 \):
\[ Z_2 = 4R \]

### Step 4: Calculate the Change in Impedance
Now, we need to find the change in impedance:
\[ \Delta Z = Z_2 - Z_1 = 4R - 2R = 2R \]

### Step 5: Calculate the Percentage Change in Impedance
The percentage change in impedance is given by:
\[ \text{Percentage Change} = \frac{\Delta Z}{Z_1} \times 100 = \frac{2R}{2R} \times 100 = 100\% \]

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

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