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 supply is applied to a resistor, we can follow these steps:

### Step 1: Identify the RMS Voltage
The given RMS voltage (VRMS) is 220 V.

### Step 2: Calculate the Peak Voltage (V0)
The peak voltage (V0) can be calculated using the relationship between RMS voltage and peak voltage:
\[ V_0 = V_{RMS} \times \sqrt{2} \]
Substituting the given value:
\[ V_0 = 220 \, V \times \sqrt{2} \approx 220 \times 1.414 \approx 311.13 \, V \]

### Step 3: Determine the Angular Frequency (ω)
The angular frequency (ω) is related to the frequency (f) by the equation:
\[ \omega = 2\pi f \]
Given that the frequency (f) is 50 Hz:
\[ \omega = 2\pi \times 50 \, Hz = 100\pi \, rad/s \]

### Step 4: Write the Equation for Instantaneous Voltage
The instantaneous voltage (V) as a function of time (t) can be expressed as:
\[ V(t) = V_0 \sin(\omega t) \]
Substituting the values we calculated:
\[ V(t) = 311.13 \sin(100\pi t) \]

### Conclusion
The instantaneous value of voltage as a function of time is:
\[ V(t) = 311.13 \sin(100\pi t) \]

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