Calculate the peak value of voltage and mean value of voltage during a positive half cycle in an AC circuit with root mean square value of voltage `50sqrt(2)` V.

Answer

To solve the problem of calculating the peak value of voltage and the mean value of voltage during a positive half cycle in an AC circuit with a given root mean square (RMS) value of voltage \( V_{\text{rms}} = 50\sqrt{2} \) V, we can follow these steps:

### Step 1: Calculate the Peak Voltage
The relationship between the peak voltage \( V_0 \) and the RMS voltage \( V_{\text{rms}} \) for a sinusoidal waveform is given by the formula:

\[ V_{\text{rms}} = \frac{V_0}{\sqrt{2}} \]

To find the peak voltage \( V_0 \), we rearrange the formula:

\[ V_0 = V_{\text{rms}} \times \sqrt{2} \]

Substituting the given RMS value:

\[ V_0 = (50\sqrt{2}) \times \sqrt{2} = 50 \times 2 = 100 \text{ V} \]

### Step 2: Calculate the Mean Value of Voltage during the Positive Half Cycle
The mean value \( V_{\text{avg}} \) for a sinusoidal waveform over a positive half cycle can be calculated using the formula:

\[ V_{\text{avg}} = \frac{1}{T/2} \int_0^{T/2} V(t) \, dt \]

Where \( T \) is the period of the waveform. For a sinusoidal voltage \( V(t) = V_0 \sin(\omega t) \), we can substitute \( V_0 = 100 \) V:

\[ V(t) = 100 \sin(\omega t) \]

The period \( T \) of the waveform is given by \( T = \frac{2\pi}{\omega} \), so \( T/2 = \frac{\pi}{\omega} \).

Now, we can compute the mean value:

\[ V_{\text{avg}} = \frac{1}{\frac{\pi}{\omega}} \int_0^{\frac{\pi}{\omega}} 100 \sin(\omega t) \, dt \]

This simplifies to:

\[ V_{\text{avg}} = \frac{100 \omega}{\pi} \int_0^{\frac{\pi}{\omega}} \sin(\omega t) \, dt \]

### Step 3: Evaluate the Integral
The integral of \( \sin(\omega t) \) is:

\[ \int \sin(\omega t) \, dt = -\frac{1}{\omega} \cos(\omega t) \]

Evaluating the definite integral from \( 0 \) to \( \frac{\pi}{\omega} \):

\[ \int_0^{\frac{\pi}{\omega}} \sin(\omega t) \, dt = -\frac{1}{\omega} \left[ \cos(\omega t) \right]_0^{\frac{\pi}{\omega}} = -\frac{1}{\omega} \left( \cos(\pi) - \cos(0) \right) = -\frac{1}{\omega} \left( -1 - 1 \right) = \frac{2}{\omega} \]

### Step 4: Substitute Back to Find \( V_{\text{avg}} \)
Now substituting back into the mean value formula:

\[ V_{\text{avg}} = \frac{100 \omega}{\pi} \cdot \frac{2}{\omega} = \frac{200}{\pi} \text{ V} \]

### Final Results
- The peak voltage \( V_0 = 100 \text{ V} \)
- The mean voltage during the positive half cycle \( V_{\text{avg}} = \frac{200}{\pi} \text{ V} \)