Verified Solution • 2025–26 Session moderate

QJEEM24Phe981: JEE Main 2024 Physics - Class 12 JEEM24Phe981 Solution: An alternating voltage volt is applied to a purely resistive…

An alternating voltage V(t)=220sin⁡100πtV(t)=220 \sin 100 \pi t volt is applied to a purely resistive load of 50Ω50 \Omega . The time taken for the current to rise from half of the peak value to the peak value is:

Multiple Choice Options

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

Step-by-Step Verified Solution

Rising half to peak

amp;t=T/6amp;t=2π6ω=π3ω=π300π=1300=3.33 ms\begin{aligned}& \mathrm{t}=\mathrm{T} / 6 \\& \mathrm{t}=\frac{2 \pi}{6 \omega}=\frac{\pi}{3 \omega}=\frac{\pi}{300 \pi}=\frac{1}{300}=3.33 \mathrm{~ms}\end{aligned}

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