Question

In a three phase system, what is the instantaneos voltage of phase 2 when phase 1 is -85 volts and phase 3 is 116 volts?

Answer 1

\[V(t)=V_0\cos(\omega t + \phi)\]
phases are separated by 120 degrees ( 2pi/3)
I can assume that theta is 0 at first phase so:
\[-85=V_0\cos(\omega t )\]
\[x=V_0\cos(\omega t + \frac{2 \pi}{3})\]
\[116=V_0\cos(\omega t + \frac{4 \pi}{3})\]
use also the equation:
\[\cos(a+b)=cosa*cosb-sina*sinb\]
and you should be able to find x :)

Answer 2

or simply sum of instantaneous voltages of each phase must be zero:
-85+x+116=0
x=-31