Answer:
d. - peak
Explanation:
In alternating current, the voltage is represented by the following formula:
[tex]V=V_{max}sin(\omega t+\phi)[/tex]
where,
[tex]V_{max}[/tex]=Maximum voltage
[tex]\omega[/tex]=Angular frequency
[tex]\phi[/tex]=phase shift
t=time
The angular frequency can be written in terms of the period (T), so:
[tex]\omega=\frac{2\pi}{T}[/tex]
So the equation will now lok like this:
[tex]V=V_{max}sin(\frac{2\pi}{T} t+\phi)[/tex]
we know that [tex]\phi=\frac{\pi}{2}[/tex] and that [tex]t=\frac{T}{2}[/tex] so the equation will now look like this:
[tex]V=V_{max}sin(\frac{2\pi}{T} (\frac{T}{2})+\frac{\pi}{2})[/tex]
which can be simplified to:
[tex]V=V_{max}sin(\pi+\frac{\pi}{2})[/tex]
[tex]V=V_{max}sin(\frac{3\pi}{2})[/tex]
Which solves to:
[tex]V=-V_{max}[/tex]
so the answer is d. -peak
