A wire when connected to 220 V mains supply has power dissipation `P_1`. Now the wire is cut into two equal pieces which are connected in parallel to the same supply. Power dissipation in this case is `P_2`. Then `P_2: P_1` is

Options

A. 1
B. 2
C. 3
D. 4

Answer

To solve the problem, we need to analyze the power dissipation in two scenarios: when the wire is intact and when it is cut into two equal pieces and connected in parallel.

### Step 1: Determine the power dissipation \( P_1 \) when the wire is intact.
- Let the resistance of the wire be \( R \).
- The voltage across the wire is \( V = 220 \, \text{V} \).
- The power dissipation \( P_1 \) can be calculated using the formula:
\[ P_1 = \frac{V^2}{R} \]

### Step 2: Cut the wire into two equal pieces.
- When the wire is cut into two equal pieces, the resistance of each piece becomes \( \frac{R}{2} \).

### Step 3: Calculate the equivalent resistance when the two pieces are connected in parallel.
- The equivalent resistance \( R_{eq} \) of two resistors \( R_1 \) and \( R_2 \) in parallel is given by:
\[ \frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} \]
- In our case, both resistors are \( \frac{R}{2} \):
\[ \frac{1}{R_{eq}} = \frac{1}{\frac{R}{2}} + \frac{1}{\frac{R}{2}} = \frac{2}{\frac{R}{2}} = \frac{4}{R} \]
- Therefore, the equivalent resistance is:
\[ R_{eq} = \frac{R}{4} \]

### Step 4: Calculate the power dissipation \( P_2 \) when the two pieces are connected in parallel.
- The power dissipation \( P_2 \) can be calculated using the equivalent resistance:
\[ P_2 = \frac{V^2}{R_{eq}} = \frac{V^2}{\frac{R}{4}} = 4 \cdot \frac{V^2}{R} \]
- Since \( P_1 = \frac{V^2}{R} \), we can express \( P_2 \) as:
\[ P_2 = 4P_1 \]

### Step 5: Calculate the ratio \( \frac{P_2}{P_1} \).
- The ratio of the power dissipation when the wire is cut and connected in parallel to the power dissipation when the wire is intact is:
\[ \frac{P_2}{P_1} = \frac{4P_1}{P_1} = 4 \]

### Final Answer:
Thus, the ratio \( P_2 : P_1 \) is \( 4 : 1 \).

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