An alternating voltage $\mathrm{V}(\mathrm{t})=220 \sin 100 \ \pi \mathrm{t}$ 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:
An alternating voltage $\mathrm{V}(\mathrm{t})=220 \sin 100 \ \pi \mathrm{t}$ 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:
- [JEE MAIN 2024]
- A
$5 \mathrm{~ms}$
- B
$3.3 \mathrm{~ms}$
- C
$7.2 \mathrm{~ms}$
- D
$2.2 \mathrm{~ms}$
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- [AIPMT 1994]
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