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:

Question

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:, 5 ms, 3 . 3 ms, 7 . 2 ms, 2 . 2 ms

MARKS App · Accepted Answer

Time taken to change current value from half of the peak value to the peak value is, t = T 6 ⇒ t = 2 π 6 ω = π 3 ω = π 300 π = 1 300 = 3 . 33 ms

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