The maximum voltage that can be applied to 5 K`Omega ` and 8W resistor without exceeding its heat dissipating capacity is

Options

A. A) 100 v
B. B) 400V
C. C) 160V
D. D) 200 V

Answer

To find the maximum voltage that can be applied to a 5 kΩ resistor without exceeding its heat dissipating capacity of 8 W, we can use the formula relating power, voltage, and resistance.

### Step-by-Step Solution:

1. **Identify the given values:**
- Resistance \( R = 5 \, k\Omega = 5000 \, \Omega \)
- Power \( P = 8 \, W \)

2. **Use the power formula:**
The power dissipated in a resistor can be expressed using the formula:
\[ P = \frac{V^2}{R} \]
where \( P \) is the power, \( V \) is the voltage across the resistor, and \( R \) is the resistance.

3. **Rearrange the formula to solve for voltage \( V \):**
\[ V^2 = P \times R \]
\[ V = \sqrt{P \times R} \]

4. **Substitute the known values into the equation:**
\[ V = \sqrt{8 \, W \times 5000 \, \Omega} \]

5. **Calculate the product:**
\[ 8 \times 5000 = 40000 \]

6. **Take the square root:**
\[ V = \sqrt{40000} = 200 \, V \]

7. **Conclusion:**
The maximum voltage that can be applied to the resistor without exceeding its heat dissipating capacity is \( V = 200 \, V \).

### Final Answer:
The maximum voltage is **200 volts**.