Question

The frequency of an alternating voltage is 50 Hz. The time taken for instantaneous voltage to increase from zero to half of its peak voltage is

• A. \( \frac{1}{800} \) s
• B. \( \frac{1}{600} \) s
• C. \( \frac{1}{300} \) s
• D. \( \frac{1}{200} \) s

Hint:

For AC sine waves, the time to reach half peak value is \( T/12 \). Time to reach peak value is \( T/4 \). Here \( T = 1/50 \) s, so \( t = (1/50)/12 = 1/600 \) s.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The instantaneous voltage equation for AC starting from zero is \( V = V_0 \sin(\omega t) \). We need to find the time \( t \) when \( V = \frac{V_0}{2} \).
Step 2: Key Formula or Approach:
1. \( V = V_0 \sin(2\pi f t) \) 2. Given condition: \( V = \frac{V_0}{2} \)
Step 3: Detailed Explanation:
Substitute \( V = \frac{V_0}{2} \) into the voltage equation: \[ \frac{V_0}{2} = V_0 \sin(2\pi f t) \] \[ \sin(2\pi f t) = \frac{1}{2} \] The smallest angle for which sine is \( 1/2 \) is \( \frac{\pi}{6} \) (or \( 30^\circ \)). \[ 2\pi f t = \frac{\pi}{6} \] \[ 2 f t = \frac{1}{6} \] \[ t = \frac{1}{12f} \] Given \( f = 50 \) Hz: \[ t = \frac{1}{12 \times 50} = \frac{1}{600} \, \text{s} \]
Step 4: Final Answer:
The time taken is \( \frac{1}{600} \) s.