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?

Explanation

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

### Step-by-Step Solution:

1. **Understand the relationship between AC and DC voltages**:
- The brightness of the bulb is related to the power it dissipates, which depends on the voltage across it. For an AC source, we use the root mean square (RMS) voltage to compare it with a DC voltage.
- The RMS voltage (V_rms) of an AC source is the effective voltage that produces the same power in a resistive load as a DC voltage.

2. **Set the RMS voltage equal to the DC voltage**:
- Since the bulb operates at 12 volts DC, we set the RMS voltage of the AC source equal to 12 volts:
\[ V_{rms} = 12 \text{ volts} \]

3. **Relate RMS voltage to peak voltage**:
- The relationship between the peak voltage (V_peak) and the RMS voltage for a sinusoidal AC source is given by:
\[ V_{rms} = \frac{V_{peak}}{\sqrt{2}} \]
- Rearranging this equation to find the peak voltage gives:
\[ V_{peak} = V_{rms} \times \sqrt{2} \]

4. **Substitute the known RMS voltage**:
- Now, substitute \( V_{rms} = 12 \) volts into the equation:
\[ V_{peak} = 12 \times \sqrt{2} \]

5. **Calculate the peak voltage**:
- The value of \( \sqrt{2} \) is approximately 1.414. Thus:
\[ V_{peak} = 12 \times 1.414 \approx 16.97 \text{ volts} \]

6. **Final answer**:
- Therefore, the peak voltage of the AC source that would make the bulb emit the same brightness as when connected to 12 volts DC is approximately:
\[ V_{peak} \approx 16.97 \text{ volts} \]