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

Options

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

Answer

To find the maximum voltage that can be applied to a 5 kΩ (5000 Ω) and 8 W resistor without exceeding its heat dissipating capacity, 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 as:
\[ P = \frac{V^2}{R} \]
where \( V \) is the voltage across the resistor.

3. **Rearrange the formula to solve for voltage**:
To find the voltage, we can rearrange the formula:
\[ V^2 = P \times R \]
Taking the square root of both sides gives:
\[ V = \sqrt{P \times R} \]

4. **Substitute the known values**:
Now, substitute the values of \( P \) and \( R \):
\[ V = \sqrt{8 \, W \times 5000 \, \Omega} \]

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

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

7. **Final answer**:
The maximum voltage that can be applied to the resistor without exceeding its heat dissipating capacity is:
\[ V = 200 \, V \]