A resistor of resistance R is connected to an ideal battery. If the value of R is decreased,the power dissipated in the resistor will

Options

A. increase
B. decrease
C. remain unchanged
D. None of these

Answer

To solve the problem, we need to analyze how the power dissipated in a resistor changes when its resistance is decreased while connected to an ideal battery.

### Step-by-Step Solution:

1. **Understanding the Circuit**:
- We have a resistor \( R \) connected to an ideal battery. An ideal battery provides a constant voltage (or EMF) \( E \) with no internal resistance.

2. **Power Formula**:
- The power \( P \) dissipated in a resistor can be calculated using the formula:
\[ P = \frac{V^2}{R} \]
- Here, \( V \) is the voltage across the resistor, which is equal to the EMF \( E \) of the battery.

3. **Initial Power Calculation**:
- Initially, the power dissipated in the resistor \( R \) is:
\[ P = \frac{E^2}{R} \]

4. **Changing the Resistance**:
- Now, if we decrease the resistance \( R \) to a new value \( R' \) (where \( R' < R \)), the voltage across the resistor remains the same (still equal to \( E \)).

5. **New Power Calculation**:
- The new power \( P' \) dissipated in the resistor \( R' \) is:
\[ P' = \frac{E^2}{R'} \]

6. **Comparing Powers**:
- Since \( R' < R \), it follows that:
\[ \frac{E^2}{R'} > \frac{E^2}{R} \]
- This implies that:
\[ P' > P \]
- Therefore, the power dissipated in the resistor increases when the resistance is decreased.

7. **Conclusion**:
- Thus, when the resistance \( R \) is decreased, the power dissipated in the resistor increases.

### Final Answer:
The power dissipated in the resistor will **increase** when the resistance is decreased.

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