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:

We are given the alternating (sinusoidal) voltage $$V(t)=220\sin(100\pi t)\ \text{volt}$$ which is applied across a purely resistive load of resistance $$R=50\ \Omega.$$

For a purely resistive circuit, current and voltage remain in the same phase, and we can use Ohm’s law in its instantaneous form

$$i(t)=\frac{v(t)}{R}.$$

Substituting the given voltage expression, we obtain

$$i(t)=\frac{220\sin(100\pi t)}{50}.$$

Simplifying the constant factor term-by-term,

$$i(t)=\left(\frac{220}{50}\right)\sin(100\pi t)=4.4\sin(100\pi t)\ \text{ampere}.$$

Here, the maximum (peak) value of the sine function is 1, so the peak current is

$$I_{\text{peak}}=4.4\ \text{A}.$$

The problem asks for the time required for the current to rise from one-half of this peak value to the full peak value, that is from

$$I_1=\frac{I_{\text{peak}}}{2}=\frac{4.4}{2}=2.2\ \text{A}$$

up to

$$I_2=I_{\text{peak}}=4.4\ \text{A}.$$

First, we locate the instant when the current equals the half-peak value. Setting $$i(t)=2.2\ \text{A}$$ in our current expression, we have

$$4.4\sin(100\pi t_1)=2.2.$$

Dividing both sides by 4.4 keeps every step explicit:

$$\sin(100\pi t_1)=\frac{2.2}{4.4}=\frac12.$$

A sine value of $$\frac12$$ corresponds to an argument of $$\frac{\pi}{6},\; 5\frac{\pi}{6}, \ldots$$ but for the first rising half-cycle we select the principal value

$$100\pi t_1=\frac{\pi}{6}.$$

Now solve for $$t_1$$ by dividing both sides by $$100\pi$$:

$$t_1=\frac{\pi/6}{100\pi}=\frac1{600}\ \text{s}.$$

Next, we find the instant of the peak current. A peak occurs when the sine function reaches 1, hence

$$\sin(100\pi t_2)=1.$$

The first positive solution for this equation is

$$100\pi t_2=\frac{\pi}{2}.$$

Again dividing by $$100\pi$$ gives

$$t_2=\frac{\pi/2}{100\pi}=\frac1{200}\ \text{s}.$$

The desired time interval is the difference $$t_2-t_1$$:

$$t_2-t_1=\frac1{200}-\frac1{600}.$$

Bringing the fractions to a common denominator of 600, we write

$$t_2-t_1=\frac{3}{600}-\frac{1}{600}=\frac{2}{600}=\frac1{300}\ \text{s}.$$

Converting seconds to milliseconds (recall that $$1\ \text{s}=1000\ \text{ms}$$), we have

$$t_2-t_1=\frac1{300}\times1000\ \text{ms}=3.33\ \text{ms}\ (\text{to two decimal places}).$$

Hence, the correct answer is Option D.