Question

An alternating voltage \( V(t) = 220 \sin 100\pi t \) volt is applied to a purely resistive load of \( 50\Omega \). The time taken for the current to rise from half of the peak value to the peak value is:

• A. 5 ms
• B. 3.3 ms
• C. 7.2 ms
• D. 2.2 ms

Hint:

The time taken for voltage in an AC waveform to rise from half to peak is always \( \frac{T}{6} \), where \( T \) is the time period.

Answer 1

The Correct Option is B

Step 1: {Understanding the given equation}
The given alternating voltage is: \[ V(t) = 220 \sin 100\pi t \] The angular frequency \( \omega \) is extracted as: \[ \omega = 100\pi { rad/s} \] Step 2: {Time interval calculation for half to peak transition}
The peak value of the voltage is \( V_{{max}} = 220 \) V. The time interval for the voltage to rise from half of the peak value to the peak value in a sinusoidal waveform follows the relation: \[ t = \frac{T}{6} \] where \( T \) is the time period: \[ T = \frac{2\pi}{\omega} = \frac{2\pi}{100\pi} = \frac{2}{100} = 0.02 { s} = 20 { ms} \] Thus, the required time interval: \[ t = \frac{T}{6} = \frac{20}{6} = 3.3 { ms} \] Thus, the correct answer is \( 3.3 \) ms.