1. Given voltage and resistance

   $V(t)=220\sin(100\pi t)$ $R=50\,\Omega$

   Since the load is purely resistive, current is in phase with voltage:

   $i(t)=\frac{V(t)}{R}=\frac{220}{50}\sin(100\pi t)=4.4\sin(100\pi t)$

2. Peak current

   The peak value of current is:

   $I_0=4.4\text{ A}$

   We need the time for current to rise from half of peak to peak, i.e. from

   $i=\frac{I_0}{2} \quad \text{to} \quad i=I_0$

3. Find time when current is half of peak

   $I_0\sin(100\pi t_1)=\frac{I_0}{2}$

   $\sin(100\pi t_1)=\frac{1}{2}$

   In the rising part of the cycle,

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

   Hence,

   $t_1=\frac{\pi/6}{100\pi}=\frac{1}{600}\text{ s}$

4. Find time when current is peak

   $I_0\sin(100\pi t_2)=I_0$

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

   Therefore,

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

   $t_2=\frac{\pi/2}{100\pi}=\frac{1}{200}\text{ s}$

5. Time taken to rise from half peak to peak

   $\Delta t=t_2-t_1=\frac{1}{200}-\frac{1}{600}$

   $\Delta t=\frac{3-1}{600}=\frac{2}{600}=\frac{1}{300}\text{ s}$

   $\Delta t=3.33\times 10^{-3}\text{ s}=3.3\text{ ms}$

6. Option check

   • A: $7.2\text{ ms}$ ❌
   • B: $3.3\text{ ms}$ ✅
   • C: $5\text{ ms}$ ❌
   • D: $2.2\text{ ms}$ ❌

Therefore, the correct answer is Option B.