An alternating voltage V ( t ) = 220 sin 100 π t 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:

Step-by-step solution

i = V R = 220 50 sin ⁡ 100 π t i = i m a x sin ⁡ 100 π t For i = i m a x 2 :   i m a x 2 = i m a x sin ⁡ 100 πt sin ⁡ 100 π t = 1 2   ⇒ 100 π t = π 6 t = 1 600     s For i = i m a x :   i m a x = i m a x sin ⁡ 100 πt sin ⁡ 100 π t = 1   ⇒ 100 π t = π 2 ⇒ t = 1 200   s so time for   i m a x 2   t o   i m a x is t ( i m a x ) - t i max 2 = 1 200 - 1 600 = 1 300 = 3.