An `AC` voltage of `V=220sqrt2 sin (100 pit-pi/2)` is applied across a `DC` voltameter its reading will be:

Options

A. `220sqrt2 V`
B. `sqrt2 V`
C. `220 V`
D. zero

Answer

To solve the problem, we need to analyze the given AC voltage and understand how a DC voltmeter works.

### Step-by-Step Solution:

1. **Identify the AC Voltage Expression**:
The given AC voltage is:
\[ V = 220\sqrt{2} \sin(100\pi t - \frac{\pi}{2}) \]

2. **Understand the Nature of AC Voltage**:
The expression represents an alternating current (AC) voltage, which varies with time. The sine function oscillates between -1 and 1, meaning the voltage will also oscillate between positive and negative values.

3. **Determine the Average Value of AC Voltage**:
The average value of an AC voltage over one complete cycle is calculated. For a sine wave, the average value over one complete cycle is zero because the positive half of the cycle cancels out the negative half. Mathematically, the average value \( V_{avg} \) of a sine wave is given by:
\[ V_{avg} = \frac{1}{T} \int_0^T V(t) dt \]
where \( T \) is the period of the sine wave. Since the integral of the sine function over a complete cycle is zero, we conclude that:
\[ V_{avg} = 0 \]

4. **Understand the Function of a DC Voltmeter**:
A DC voltmeter measures the average value of the voltage applied across it. Since we established that the average value of the AC voltage is zero, the DC voltmeter will read:
\[ V_{DC} = 0 \text{ volts} \]

5. **Conclusion**:
Therefore, when the AC voltage is applied to a DC voltmeter, the reading will be:
\[ \text{Reading of the DC voltmeter} = 0 \text{ volts} \]

### Final Answer:
The reading of the DC voltmeter will be **0 volts**.

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