Alternating Current

JEE Main level, single correct, medium

15 of 190

Question 15 of 190

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:

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Answer: (B) 3.3ms3.3ms

Time taken to change current value from half of the peak value to the peak value is,
t=T6t=\dfrac{T}{6}
⇒t=2π6ω=π3ω=π300π=1300=3.33ms\Rightarrow t=\dfrac{2\pi }{6\omega }=\dfrac{\pi }{3\omega }=\dfrac{\pi }{300\pi }=\dfrac{1}{300}=3.33ms

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