Question

An alternating voltage v (t) 220 sin100πI volt is applied to a purely resistive load of 50Ω . The time taken for the current to rise from half of the peak value of the peak value is:
(A) 2.2 ms (B) 3.3 ms
(C) 5 ms (D) 7.2 ms

Answer 1

Answer:

A Options is Right Answer

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Answer 2

The time taken for the current to rise from half of the peak value of the peak value is:

(B) 3.3 ms

We know that, in a purely resistive circuit,

i(t) = Vo/R sin(wt+Ф)  ...(1)

Given that

v (t) = 220 sin100πt V, so Vo = 220V

and R = 50 ohms

ω = 100П

Ф = 0°

Replacing the values in (1), we get

i(t) = 220/50 sin(100Пt)

So, peak value is 220/50 A  = 22/5 A

For half of peak value , let us calculate 't'.

22/10 =  220/50 sin(100Пt)

⇒ 100Пt = 1/2 = sin(П/6)

⇒ t = 1/(6*100) s = 1/600 s

Again, for peak value,

100Пt' = П/2

⇒ t' = 1/(2*100) s = 1/200 s

So, the time taken to rise from t to t' is

= t' - t

= (1/200 - 1/600) s

= 2/600 s

= 1/300 s

= 3.3 ms                                 [1ms = 10⁻³ s]

Option (B) is correct.