Question

In the ac circuit, the current is expressed as i = 100 sin 200 pi t. In this circuit the current rises from zero to peal value in time.

Answer 1

Answer: The time taken = $\dfrac{1}{400}s$

Explanation:

Given that,

$i = 100 sin(200\pi t)$

The general equation,

$i = i_{0}sin(\omega t)$

$\omega = 200\pi$

We know that,

$\dfrac{2\pi}{T}= \omega$

$\dfrac{2\pi}{T} = 200\pi$

$T = \dfrac{1}{100}s$

it is a time period of the wave.

The current reaches to peak value then,

The time taken = $\dfrac{T}{4}$

The time taken = $\dfrac{1}{400}s$

Hence, this is the required solution.

Answer 2

Answer:

The time taken = $\frac{1}{400} s$

Explanation:

we  know that peak value means

$i_{0}$

Then compare the equation with

i = $i_{0}$ sinωt

then we get

$i_{0}$ =100

it is given in the question that

$i_{rms} =i_{0}$

so put 100 in place of i so eq will be

100=100sin 200πt

100/100 =1

then we get

1 =sin 200πt     (we know that sin 90°= 1 =π/2 means sinπ/2)

so sin π/2 =sin 200πt

we get

$\frac{1}{2}$ =200t

t =$\frac{1}{400} s$