In an AC circuit, the current is expressed as i = 100 sin 200 π t. In this circuit the current rises from zero to peak value in time

Question

In an AC circuit, the current is expressed as i = 100 sin 200 π t. In this circuit the current rises from zero to peak value in time (A)  ( 1/ 400) s (B)  ( 1/ 300) s (C)  ( 1/ 100) s (D)  ( 1/ 200) s

Tardigrade Admin · Accepted Answer

The simple type of alternating current is one,
which varies with time simple harmonically,
i.e,

i = i_{0} sin ω t

...(i)
where

ω

is angular frequency

(ω = \frac{2 π}{T})

.
Given,

i = 100 sin 200 π t

...(ii)
Comparing Eqs. (i) and (ii), we get

ω = 200 π \Rightarrow \frac{2 π}{T} = 200 π

\Rightarrow T = \frac{2}{200} = \frac{1}{100} s

The current rises from zero to peak value in time

T /4

.

∴

T =

\frac{T}{4} = \frac{1}{400} s

Q. In an AC circuit, the current is expressed as $i = 100 sin 200 π t$ . In this circuit the current rises from zero to peak value in time

$\frac{1}{400}$ s

$\frac{1}{300}$ s

$\frac{1}{100}$ s

$\frac{1}{200}$ s

Q. In an AC circuit, the current is expressed as i=100sin200πt. In this circuit the current rises from zero to peak value in time

4001​ s

3001​ s

1001​ s

2001​ s

Q. In an AC circuit, the current is expressed as $i = 100 sin 200 π t$ . In this circuit the current rises from zero to peak value in time

$\frac{1}{400}$ s

$\frac{1}{300}$ s

$\frac{1}{100}$ s

$\frac{1}{200}$ s