The household suply of electricity is at `220 V` (rms value) and `50 Hz`. Find the peak voltage and the least possible time in which the voltage an change from the rms value to zero.

Explanation

To solve the problem, we need to find the peak voltage and the least possible time for the voltage to change from the RMS value to zero.

### Step-by-Step Solution:

**Step 1: Calculate the Time Period (T)**
The time period \( T \) of an AC signal is given by the formula:
\[ T = \frac{1}{f} \]
where \( f \) is the frequency. Given that \( f = 50 \, \text{Hz} \):
\[ T = \frac{1}{50} = 0.02 \, \text{seconds} \]

**Step 2: Calculate the Peak Voltage (V_peak)**
The relationship between the RMS voltage (\( V_{rms} \)) and the peak voltage (\( V_{peak} \)) is given by:
\[ V_{rms} = \frac{V_{peak}}{\sqrt{2}} \]
Rearranging this formula to find \( V_{peak} \):
\[ V_{peak} = V_{rms} \times \sqrt{2} \]
Substituting the given \( V_{rms} = 220 \, \text{V} \):
\[ V_{peak} = 220 \times \sqrt{2} \approx 220 \times 1.414 \approx 311 \, \text{V} \]

**Step 3: Determine the Least Possible Time to Change from RMS to Zero**
In one complete cycle of the AC waveform, the voltage reaches the RMS value twice (once going up and once going down). Therefore, the time taken to go from the RMS value to zero is half of the time period:
\[ \text{Time to change from } V_{rms} \text{ to } 0 = \frac{T}{2} = \frac{0.02}{2} = 0.01 \, \text{seconds} \]

### Final Answers:
- **Peak Voltage**: \( V_{peak} \approx 311 \, \text{V} \)
- **Least Possible Time to Change from RMS to Zero**: \( 0.01 \, \text{seconds} \)