An electric bulb operates 10 V d.c. If this bulb is connected to an a.c. source and gives normal brightness, then peak value of the source is …….

Options

A. 141.4 V
B. 14.14 V
C. 1.414 V
D. 0.1414 V

Answer

To find the peak value of the AC source that allows the electric bulb, which operates at 10 V DC, to give normal brightness, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Relationship Between RMS and Peak Voltage**:
The relationship between the root mean square (RMS) voltage (Vrms) and the peak voltage (V0) for an AC source is given by the formula:
\[ V_{rms} = \frac{V_0}{\sqrt{2}} \]

2. **Identify the Given Value**:
In this problem, the bulb operates at 10 V DC, which means the RMS voltage of the AC source must also be 10 V for the bulb to give normal brightness:
\[ V_{rms} = 10 \, \text{V} \]

3. **Rearrange the Formula to Find Peak Voltage**:
To find the peak voltage (V0), we can rearrange the formula:
\[ V_0 = V_{rms} \times \sqrt{2} \]

4. **Substitute the Known Value**:
Substitute the value of Vrms into the equation:
\[ V_0 = 10 \, \text{V} \times \sqrt{2} \]

5. **Calculate the Value**:
Using the approximate value of \(\sqrt{2} \approx 1.414\):
\[ V_0 = 10 \, \text{V} \times 1.414 \approx 14.14 \, \text{V} \]

6. **Conclusion**:
Therefore, the peak value of the AC source is approximately:
\[ V_0 \approx 14.14 \, \text{V} \]

### Final Answer:
The peak value of the source is **14.14 V**.