An experiment is performed to determine the 1-V characteristics of `R = 100 Omega` and a maximum power of dissipation rating of 1W. The minimum voltage range of the DC source in the circuit is :

Options

A. 0-5V
B. 0-24V
C. 0-12V
D. 0-8V

Explanation

To determine the minimum voltage range of the DC source in the circuit, we can use the relationship between power, voltage, and resistance. The formula we will use is:

\[ P = \frac{V^2}{R} \]

Where:
- \( P \) is the power (in watts),
- \( V \) is the voltage (in volts),
- \( R \) is the resistance (in ohms).

### Step-by-Step Solution:

1. **Identify the given values:**
- Resistance \( R = 100 \, \Omega \)
- Maximum Power \( P = 1 \, W \)

2. **Use the power formula to find voltage:**
Rearranging the formula \( P = \frac{V^2}{R} \) to solve for \( V \):
\[ V^2 = P \cdot R \]

3. **Substitute the known values into the equation:**
\[ V^2 = 1 \, W \cdot 100 \, \Omega \]
\[ V^2 = 100 \, V^2 \]

4. **Calculate \( V \):**
Taking the square root of both sides:
\[ V = \sqrt{100} = 10 \, V \]

5. **Conclusion:**
The minimum voltage range of the DC source in the circuit is \( 10 \, V \).

### Final Answer:
The minimum voltage range of the DC source is **10 V**.

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