Question

Given v = 520 sin (30t - 5π/4), what is the phase angle?Question 4 options:-225 degrees90 degrees-135 degrees-90 degrees

Answer 1

Given data:

The voltage can be expressed as,

v = 520 sin (30t - 5π/4) ...... (1)

Now, the general equation of the sine wave can be given as,

`v=V_m\text{ sin}(\omega t+\phi)\ldots\ldots\text{ }(2)`

Here,

`\phi`

is the phase angle,

`V_m`

is the maximum voltage, and

`\omega`

is the angular frequency.

Compare equations (1) and (2), we get:

`\begin{gathered} \phi=-\frac{5\pi\text{ rad}}{4}((180\degree)/(\pi)) \\ =-225\degree \end{gathered}`

Thus, the phase angle is

`-225\degree`

and the first option (-225 degrees) is correct.