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.

Answer 2

Step 1: Understand the form of the voltage
Given alternating voltage:
\( V(t) = 220 \sin(100\pi t) \) volts
This is a sine wave of the form \( V(t) = V_0 \sin(\omega t) \), where:
- \( V_0 = 220 \) V (peak voltage)
- \( \omega = 100\pi \) rad/s (angular frequency)

Step 2: Since load is purely resistive
The current is in phase with the voltage and also sinusoidal:
\( I(t) = \frac{V(t)}{R} = \frac{220}{50} \sin(100\pi t) = 4.4 \sin(100\pi t) \) A
Peak current \( I_0 = 4.4 \) A
We want the time taken for the current to rise from half of its peak value to the peak value:

Step 3: Set up the condition
Let \( I(t_1) = \frac{I_0}{2} = 2.2 \) A
We solve for \( t_1 \):
\[ 4.4 \sin(100\pi t_1) = 2.2 \Rightarrow \sin(100\pi t_1) = \frac{1}{2} \]
This occurs when \( 100\pi t_1 = \frac{\pi}{6} \) → \( t_1 = \frac{1}{600} \) s = 1.667 ms

Next, let \( I(t_2) = I_0 = 4.4 \) A
\[ \sin(100\pi t_2) = 1 \Rightarrow 100\pi t_2 = \frac{\pi}{2} \Rightarrow t_2 = \frac{1}{200} \text{ s} = 5 \text{ ms} \]

Step 4: Time taken for the rise from half to peak
\[ \Delta t = t_2 - t_1 = 5 \text{ ms} - 1.667 \text{ ms} = 3.33 \text{ ms} \]

Step 5: Final Answer
The time taken for the current to rise from half the peak to the peak value is:
3.3 ms