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.

Answer

To solve the problem, we need to find two things: the peak voltage (V_peak) and the least possible time (t_min) in which the voltage changes from the RMS value (V_rms) to zero.

### Step-by-Step Solution:

**Step 1: Calculate the Peak Voltage (V_peak)**

The relationship between the RMS voltage (V_rms) and the peak voltage (V_peak) is given by the formula:

\[ V_{rms} = \frac{V_{peak}}{\sqrt{2}} \]

Given that \(V_{rms} = 220 \, V\), we can rearrange the formula to find \(V_{peak}\):

\[ V_{peak} = V_{rms} \times \sqrt{2} \]

Substituting the value of \(V_{rms}\):

\[ V_{peak} = 220 \times \sqrt{2} \]

Calculating \(V_{peak}\):

\[ V_{peak} \approx 220 \times 1.414 \approx 311.08 \, V \]

**Step 2: Calculate the Time Period (T)**

The frequency (f) is given as \(50 \, Hz\). The time period (T) is the reciprocal of the frequency:

\[ T = \frac{1}{f} = \frac{1}{50} \, seconds \]

Calculating T:

\[ T = 0.02 \, seconds \]

**Step 3: Calculate the Minimum Time (t_min)**

The voltage waveform changes from the RMS value to zero in half of the time period, since the waveform is sinusoidal. Therefore, the minimum time (t_min) can be calculated as:

\[ t_{min} = \frac{T}{2} = \frac{0.02}{2} \, seconds \]

Calculating \(t_{min}\):

\[ t_{min} = 0.01 \, seconds \]

### Final Answers:

- Peak Voltage (V_peak) ≈ 311.08 V
- Least Possible Time (t_min) = 0.01 seconds