An AC source is rated `220 V, 50 Hz`. The average voltage is calculated in a time interval of `0.01 s`. It

Options

A. must be zero
B. may be zero
C. is never zero
D. is `(220/(sqrt 2)) V`

Answer

To solve the problem of calculating the average voltage of an AC source rated at `220 V, 50 Hz` over a time interval of `0.01 s`, we will follow these steps:

### Step 1: Determine the angular frequency (ω)
The angular frequency (ω) is calculated using the formula:
\[ \omega = 2\pi f \]
where \( f \) is the frequency in Hz. Given \( f = 50 \, \text{Hz} \):
\[ \omega = 2\pi \times 50 = 100\pi \, \text{rad/s} \approx 314 \, \text{rad/s} \]

**Hint:** Remember that the angular frequency is related to the frequency by a factor of \( 2\pi \).

### Step 2: Write the equation for the AC voltage
The voltage of an AC source can be expressed as:
\[ V(t) = V_0 \sin(\omega t) \]
where \( V_0 \) is the peak voltage. For a source rated at \( 220 \, \text{V} \), the peak voltage \( V_0 \) can be calculated as:
\[ V_0 = \sqrt{2} \times 220 \approx 311.8 \, \text{V} \]

**Hint:** The peak voltage \( V_0 \) is related to the RMS voltage by the factor \( \sqrt{2} \).

### Step 3: Calculate the average voltage over the time interval
The average voltage over a time interval \( T \) is given by:
\[ V_{\text{avg}} = \frac{1}{T} \int_0^T V(t) \, dt \]
For our case, \( T = 0.01 \, \text{s} \):
\[ V_{\text{avg}} = \frac{1}{0.01} \int_0^{0.01} V_0 \sin(\omega t) \, dt \]

**Hint:** The average voltage is calculated by integrating the voltage function over the specified time interval and dividing by the time interval.

### Step 4: Perform the integration
Substituting \( V(t) \):
\[ V_{\text{avg}} = \frac{V_0}{0.01} \int_0^{0.01} \sin(314 t) \, dt \]
The integral of \( \sin(314 t) \) is:
\[ -\frac{1}{314} \cos(314 t) \bigg|_0^{0.01} \]
Calculating the limits:
\[ = -\frac{1}{314} \left( \cos(3.14) - \cos(0) \right) = -\frac{1}{314} \left( -1 - 1 \right) = \frac{2}{314} \]

**Hint:** When integrating trigonometric functions, remember to evaluate the antiderivative at the limits of integration.

### Step 5: Substitute back to find the average voltage
Now substituting back into the average voltage formula:
\[ V_{\text{avg}} = \frac{V_0}{0.01} \cdot \frac{2}{314} \]
Substituting \( V_0 \approx 311.8 \):
\[ V_{\text{avg}} = \frac{311.8}{0.01} \cdot \frac{2}{314} \approx \frac{31180 \cdot 2}{314} \approx 198.7 \, \text{V} \]

### Conclusion
The average voltage over the time interval of \( 0.01 \, \text{s} \) is approximately \( 198.7 \, \text{V} \).