Question

If the phase shift is π/2 rads and T is the period, then the voltage at (T/2) is _____

a. zero
b. +peak/2
c. +peak
d. -peak

Answer 1

d. - peak

Step-by-step explanation:

In alternating current, the voltage is represented by the following formula:

`V=V_(max)sin(\omega t+\phi)`

where,

`V_(max)`=Maximum voltage

`\omega`=Angular frequency

`\phi`=phase shift

t=time

The angular frequency can be written in terms of the period (T), so:

`\omega=(2\pi)/(T)`

So the equation will now lok like this:

`V=V_(max)sin((2\pi)/(T) t+\phi)`

we know that
`\phi=(\pi)/(2)` and that
`t=(T)/(2)` so the equation will now look like this:

`V=V_(max)sin((2\pi)/(T) ((T)/(2))+(\pi)/(2))`

which can be simplified to:

`V=V_(max)sin(\pi+(\pi)/(2))`

`V=V_(max)sin((3\pi)/(2))`

Which solves to:

`V=-V_(max)`

so the answer is d. -peak