Question

equation of alternating current is given by I = 10(2)1/2sin(100piet + pie/6). the time taken by current to reach the root mean square value from t=0 is t1 then the value of t1 is




1) 1/1200s




2) 1/250s




3) 1/200s




4) 1/800s

Answer 1

Answer: (a). The value of $t_{1}$ is $t_{1} = \dfrac{1}{1200}s$.

Explanation:

Given equation,

$I = 10(2)^{\dfrac{1}{2}} sin(100\pi t+\dfrac{\pi}{6})$....(I)

At t = 0

The general equation

$I =I_{0}sin(\omega t +\phi)$

Compare to equation (I) from general equation

$I_{0} = 10\sqrt2$

Now, $I_{rms}$ is

$I_{rms} = \dfrac{I_{0}}{\sqrt2}$

$I_{rms} = \dfrac{10\sqrt2}{\sqrt2}$

$I_{rms} = 10$

Now, put the value of $I_{rms}$ in equation (I)

$10 = 10\sqrt2 sin (100\pi t+\dfrac{\pi}{6})$

$\dfrac{1}{\sqrt2}=sin(100\pi t+\dfrac{\pi}{6})$

$sin^{-1}(\dfrac{1}{\sqrt2})= 100\pi t+ \dfrac{\pi}{6}$

$\dfrac{\pi}{4} = 100\pi t + \dfrac{\pi}{6}$

$\dfrac{\pi}{4}-\dfrac{\pi}{6} = 100\pi t$

$\dfrac{\pi}{12} = 100\pi t$

$t = \dfrac{1}{1200}s$

So, we can say $t=t_{1}$

$t_{1} = \dfrac{1}{1200}s$

Hence, the value of $t_{1}$ is $t_{1} = \dfrac{1}{1200}s$.