The power dissipated across resistance R which is connected across a battery of potential V is P. If resistance is doubled, then the power becomes

Options

A. `1//2`
B. `2`
C. `1//4`
D. `1`

Explanation

To solve the problem, we need to analyze how the power dissipated across a resistor changes when the resistance is doubled.

### Step-by-Step Solution:

1. **Understand the Power Formula**:
The power \( P \) dissipated across a resistor \( R \) connected to a voltage \( V \) is given by the formula:
\[ P = \frac{V^2}{R} \]

2. **Initial Power Calculation**:
Let's denote the initial resistance as \( R \). The initial power dissipated across this resistance is:
\[ P = \frac{V^2}{R} \]

3. **Doubling the Resistance**:
If the resistance is doubled, the new resistance becomes \( 2R \).

4. **New Power Calculation**:
The new power \( P' \) dissipated across the new resistance \( 2R \) is:
\[ P' = \frac{V^2}{2R} \]

5. **Relate New Power to Initial Power**:
To find the relationship between the new power \( P' \) and the initial power \( P \), we can express \( P' \) in terms of \( P \):
\[ P' = \frac{V^2}{2R} = \frac{1}{2} \cdot \frac{V^2}{R} = \frac{P}{2} \]

6. **Conclusion**:
Therefore, when the resistance is doubled, the new power \( P' \) becomes half of the initial power \( P \):
\[ P' = \frac{P}{2} \]

### Final Answer:
If the resistance is doubled, the power becomes \( \frac{P}{2} \).

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