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Answer 1 · 2016-10-04T16:36:03

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

$V (t) = 60 sin (2 t + 0.8), where t \geq 0$ $V(t)=60sin(2t+0.8)," where "t>=0$

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

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

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

So minimum value $V (t) = - 60 V$ $V(t)=-60V$

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

So putting $V (t) = 30 V$ $V(t)=30V$ we get

$30 = 60 sin (2 t + 0.8)$ $30=60sin(2t+0.8)$

$\Rightarrow sin (2 t + 0.8) = \frac{30}{60} = \frac{1}{2} = sin (\frac{π}{6})$ $=>sin(2t+0.8)=30/60=1/2=sin(pi/6)$

$\Rightarrow 2 t = \frac{π}{6} - 0.8$ $=>2t=pi/6-0.8$

$\Rightarrow 2 t = \frac{3.14}{6} = - 0.8 < 0 \to not possible$ $=>2t=3.14/6=-0.8<0->"not possible"$

So taking

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

$\Rightarrow 2 t = \frac{5 π}{6} = - 0.8 = \frac{5 \times 3.14}{6} - 0.8$ $=>2t=(5pi)/6=-0.8=(5xx3.14)/6-0.8$

$\Rightarrow 2 t = 1.82$ $=>2t=1.82$

$t = 0.91 s$ $t=0.91s$

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