Question

2) An alternating voltage V(t)= 220sin 100tt volt is applied to
a purely resistive load of 50 S2. The time taken for the current
to rise from half of the peak value to the peak value is


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Answer 1

The time taken for the current to rise from half of the peak value of the peak value is 3.3 ms.

An alternative voltage is given as V(t) = 220sin(100π)t ,

let peak value becomes half at time t.

as we know voltage leads current by π/2.

so, equation of current is in form of cosine. like , I = icos(ωt)

and we know, it will be half at t = π/3 [ because cos(π/3) = 1/2 ]

so, π/3 = 100πt

⇒t = π/300π = 1/300 = 3.3 ms

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Answer 2

$\huge\bold\purple{Answer:-}$

An alternative voltage is given as V(t) = 220sin(100π)t ,

let peak value becomes half at time t.

as we know voltage leads current by π/2.

so, equation of current is in form of cosine. like , I = icos(ωt)

and we know, it will be half at t = π/3 [ because cos(π/3) = 1/2 ]

so, π/3 = 100πt

⇒t = π/300π = 1/300 = 3.3 ms