220 V, 50 Hz , AC is applied to a resistor . The instantaneous value of voltage is

Options

A. 220`sqrt 2`sin100`pi`t
B. 220sin100`pi`t
C. 220`sqrt 2`sin50`pi`t
D. 220sin50`pi`t

Explanation

To find the instantaneous value of voltage when a 220 V, 50 Hz AC is applied to a resistor, we can use the formula for the instantaneous voltage in an AC circuit. The formula is given by:

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

Where:
- \( V(t) \) is the instantaneous voltage at time \( t \).
- \( V_m \) is the maximum (peak) voltage.
- \( \omega \) is the angular frequency in radians per second.

### Step 1: Determine the RMS Voltage
The given RMS voltage \( V_{rms} \) is 220 V. The peak voltage \( V_m \) can be calculated using the relationship:

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

Substituting the given value:

\[ V_m = 220 \times \sqrt{2} \]

### Step 2: Calculate the Angular Frequency
The angular frequency \( \omega \) can be calculated using the formula:

\[ \omega = 2\pi f \]

Where \( f \) is the frequency in hertz. Given \( f = 50 \) Hz:

\[ \omega = 2 \times \pi \times 50 \]

Calculating this gives:

\[ \omega = 100\pi \, \text{radians/second} \]

### Step 3: Write the Instantaneous Voltage Equation
Now, substituting \( V_m \) and \( \omega \) into the instantaneous voltage formula:

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

Substituting the values we calculated:

\[ V(t) = (220\sqrt{2}) \sin(100\pi t) \]

### Conclusion
Thus, the instantaneous value of voltage is:

\[ V(t) = 220\sqrt{2} \sin(100\pi t) \]

### Final Answer
The correct option is **A: \( 220\sqrt{2} \sin(100\pi t) \)**.

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