Question

Given v = 520 sin (30t - 5π/4), what is the peak voltage?Question 7 options:5201040367.6735.3

Answer 1

Given:

`v=520\sin (30t-(5\pi)/(4))`

We know that:

`-1\le\sin \theta\le1`

So the peak voltage can be calculated using the maximum value of the sine function (1 in this case). Then:

`v_{\text{peak}}=520\cdot1=520`