A bulb is designed to operate at `12` volts constant direct current.If this bulb is connected to an alternating current source and gives same brightness.What would be the peak voltage of the source?

Answer

To solve the problem, we need to find the peak voltage of an alternating current (AC) source that would make a bulb, designed for 12 volts direct current (DC), operate at the same brightness.

### Step-by-Step Solution:

1. **Understanding the Relationship Between RMS and Peak Voltage**:
The brightness of the bulb when connected to AC is determined by the root mean square (RMS) voltage of the AC source. For a bulb rated at 12 volts DC, we can assume that this is equivalent to the RMS voltage of the AC source.

2. **Setting the RMS Voltage**:
Since the bulb operates at 12 volts DC, we can equate this to the RMS voltage of the AC source:
\[ V_{\text{rms}} = 12 \text{ volts} \]

3. **Using the RMS to Peak Voltage Formula**:
The relationship between the peak voltage (\(V_{\text{peak}}\)) and the RMS voltage (\(V_{\text{rms}}\)) for an AC source is given by:
\[ V_{\text{rms}} = \frac{V_{\text{peak}}}{\sqrt{2}} \]
Rearranging this formula to find the peak voltage gives:
\[ V_{\text{peak}} = V_{\text{rms}} \times \sqrt{2} \]

4. **Calculating the Peak Voltage**:
Now, substituting the RMS voltage into the equation:
\[ V_{\text{peak}} = 12 \times \sqrt{2} \]
We can calculate \(\sqrt{2}\) which is approximately \(1.414\):
\[ V_{\text{peak}} = 12 \times 1.414 \approx 16.97 \text{ volts} \]

5. **Final Answer**:
Therefore, the peak voltage of the AC source that would allow the bulb to operate at the same brightness as when connected to 12 volts DC is approximately:
\[ V_{\text{peak}} \approx 17 \text{ volts} \]