An alternating voltage V ( t ) = 220 sin 100 π t volt is applied to a purely resistive load of 50 Ω . The…

An alternating voltage V(t)=220sin100πt volt is applied to a purely resistive load of 50 Ω. The time taken for the current to rise from half of the peak value to the peak value is:
  1. 5 ms
  2. 3.3 ms
  3. 7.2 ms
  4. 2.2 ms

Solution

Time taken to change current value from half of the peak value to the peak value is, 

t=T6

⇒t=2π6ω=π3ω=π300π=1300=3.33 ms

Asked in: JEE Main 2024 (30 Jan Shift 2)

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