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 find the value of the voltage after \( \frac{1}{720} \) seconds from the start for an AC source of 120 V and 60 Hz, we can follow these steps:

### Step 1: Understand the AC Voltage Formula
The voltage in an AC circuit can be expressed as:
\[ V(t) = V_0 \sin(\omega t) \]
where:
- \( V(t) \) is the instantaneous voltage at time \( t \),
- \( V_0 \) is the peak voltage,
- \( \omega \) is the angular frequency,
- \( t \) is the time in seconds.

### Step 2: Calculate the Peak Voltage \( V_0 \)
Given that the root mean square (RMS) voltage \( V_{rms} \) is 120 V, we can find the peak voltage \( V_0 \) using the relationship:
\[ V_0 = V_{rms} \times \sqrt{2} \]
Substituting the given value:
\[ V_0 = 120 \times \sqrt{2} \approx 120 \times 1.414 \approx 169.7 \, V \]

### Step 3: Calculate the Angular Frequency \( \omega \)
The angular frequency \( \omega \) is calculated using the formula:
\[ \omega = 2\pi f \]
where \( f \) is the frequency in Hz. Given \( f = 60 \, Hz \):
\[ \omega = 2\pi \times 60 = 120\pi \, \text{rad/s} \]

### Step 4: Calculate the Time \( t \)
We need to find the voltage at \( t = \frac{1}{720} \) seconds.

### Step 5: Substitute Values into the Voltage Formula
Now we can substitute \( V_0 \), \( \omega \), and \( t \) into the voltage formula:
\[ V\left(\frac{1}{720}\right) = V_0 \sin\left(\omega \cdot \frac{1}{720}\right) \]
Substituting the values:
\[ V\left(\frac{1}{720}\right) = 169.7 \sin\left(120\pi \cdot \frac{1}{720}\right) \]

### Step 6: Simplify the Argument of the Sine Function
Now simplify the argument of the sine function:
\[ 120\pi \cdot \frac{1}{720} = \frac{120\pi}{720} = \frac{\pi}{6} \]

### Step 7: Calculate the Sine Value
Now calculate the sine:
\[ \sin\left(\frac{\pi}{6}\right) = \frac{1}{2} \]

### Step 8: Calculate the Voltage
Now substitute this back into the voltage equation:
\[ V\left(\frac{1}{720}\right) = 169.7 \cdot \frac{1}{2} = 84.85 \, V \]

### Final Answer
Thus, the value of the voltage after \( \frac{1}{720} \) seconds from the start is approximately:
\[ \boxed{84.85 \, V} \]