# Question #ed6ce

Oct 4, 2016

The function representing the variation of voltage with time is given by

$V \left(t\right) = 60 \sin \left(2 t + 0.8\right) , \text{ where } t \ge 0$

(a) As it is a sine function , It will have

(1) a maximum value
when $\sin \left(2 t + 0.8\right)$ is maximum = 1
So maximum value $V \left(t\right) = 60 V$

(2) a minimum value
when $\sin \left(2 t + 0.8\right)$ is maximum =-1

So minimum value $V \left(t\right) = - 60 V$

(b) we are to findout the time when first time $V \left(t\right) = 30 V$

So putting $V \left(t\right) = 30 V$ we get

$30 = 60 \sin \left(2 t + 0.8\right)$

$\implies \sin \left(2 t + 0.8\right) = \frac{30}{60} = \frac{1}{2} = \sin \left(\frac{\pi}{6}\right)$

$\implies 2 t = \frac{\pi}{6} - 0.8$

$\implies 2 t = \frac{3.14}{6} = - 0.8 < 0 \to \text{not possible}$

So taking

$\sin \left(2 t + 0.8\right) = \sin \left(\pi - \frac{\pi}{6}\right)$

$\implies 2 t = \frac{5 \pi}{6} = - 0.8 = \frac{5 \times 3.14}{6} - 0.8$

$\implies 2 t = 1.82$

$t = 0.91 s$

Hence first time at 0.91s $V \left(t\right)$ will reach to 30V value