Engineering: 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

If the phase shift is π/2 rads and T is the period,

hellicuh

10.02.2022

d. - peak

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:

$\omega=\frac{2\pi}{T}$

So the equation will now lok like this:

$V=V_{max}sin(\frac{2\pi}{T} t+\phi)$

we know that $\phi=\frac{\pi}{2}$ and that $t=\frac{T}{2}$ so the equation will now look like this:

$V=V_{max}sin(\frac{2\pi}{T} (\frac{T}{2})+\frac{\pi}{2})$

which can be simplified to:

$V=V_{max}sin(\pi+\frac{\pi}{2})$

$V=V_{max}sin(\frac{3\pi}{2})$

Which solves to:

$V=-V_{max}$

so the answer is d. -peak

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