Question

Two alternating voltage generators produce emfs of the same amplitude $E_0$ but with a phase difference of $\pi /3$. The resultant emf is:

• A. $E_0\sin [\omega t+(\pi /3\left)\right]$
• B. $E_0\sin [\omega t+(\pi /6\left)\right]$
• C. $\sqrt{3}{E}_0\sin [\omega t+(\pi /6\left)\right]$
• D. $\sqrt{3}{E}_0\sin [\omega t+(\pi /2\left)\right]$

Answer 1

The correct option is C $\sqrt{3}{E}_0\sin [\omega t+(\pi /6\left)\right]$

In phasor ,
resultant voltage = $E_0\angle 0+E_0\angle 60=E_0+\frac{E_0}{2}+j\frac{E_0\sqrt{3}}{2}=\frac{3E_0}{2}+j\frac{E_0\sqrt{3}}{2}=\sqrt{3}{E}_0(\frac{\sqrt{3}}{2}+j\frac{1}{2})=\sqrt{3}{E}_0\angle 30$
therefore,
in time domain, resultant voltage = $\sqrt{3}{E}_{0}sin(\omega t+\pi /6)$