An AC source is 120 V-60 Hz. The value of voltage after 1/720 s from start will be

Options

A. 20.2 V
B. 42.4 V
C. 84.8 V
D. 106.8 V

Explanation

To solve the problem of finding the voltage after \( \frac{1}{720} \) seconds from the start of an AC source with a voltage of 120 V and a frequency of 60 Hz, we can follow these steps:

### Step 1: Calculate the angular frequency (\( \omega \))
The angular frequency is given by the formula:
\[ \omega = 2\pi f \]
where \( f \) is the frequency in hertz. Given \( f = 60 \) Hz, we can calculate:
\[ \omega = 2\pi \times 60 = 120\pi \, \text{rad/s} \]

### Step 2: Determine the peak voltage (\( V_0 \))
The peak voltage can be calculated from the RMS voltage using the formula:
\[ V_0 = \sqrt{2} \times V_{\text{RMS}} \]
Given \( V_{\text{RMS}} = 120 \, \text{V} \):
\[ V_0 = \sqrt{2} \times 120 \approx 169.71 \, \text{V} \]

### Step 3: Calculate the voltage at time \( t = \frac{1}{720} \) seconds
The voltage \( V(t) \) at any time \( t \) can be expressed as:
\[ V(t) = V_0 \sin(\omega t) \]
Substituting \( \omega \) and \( t \):
\[ V\left(\frac{1}{720}\right) = V_0 \sin\left(120\pi \times \frac{1}{720}\right) \]
Calculating the argument of the sine function:
\[ 120\pi \times \frac{1}{720} = \frac{120\pi}{720} = \frac{\pi}{6} \]
Now substituting back:
\[ V\left(\frac{1}{720}\right) = 169.71 \sin\left(\frac{\pi}{6}\right) \]
Since \( \sin\left(\frac{\pi}{6}\right) = \frac{1}{2} \):
\[ V\left(\frac{1}{720}\right) = 169.71 \times \frac{1}{2} = 84.855 \, \text{V} \]

### Step 4: Round the answer
The final voltage after rounding is approximately:
\[ V \approx 84.9 \, \text{V} \]

### Final Answer
The value of the voltage after \( \frac{1}{720} \) seconds is approximately **84.9 V**.

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