Question

Two alternating voltage generators produce e.m.f. of the same amplitude $E_{0}$ but with a phase difference of $\pi / 3$. The resultant e.m.f. is

• A. $E_{0} \sin (\omega t+\pi / 3)$
• B. $E_{0} \sin (\omega t+\pi / 6)$
• C. $\sqrt{3} E_{0} \sin (\omega t+\pi / 6)$
• D. $\sqrt{3} E_{0} \sin (\omega t+\pi / 2)$

Answer 1

Answer: (C)

$E_{1}=E_{0} \sin \omega t, E_{2}=E_{0} \sin (\omega t+\pi / 3)$
$E=E_{2}+E_{1}$
$=E_{0} \sin (\omega t+\pi / 3)+E_{0} \sin \omega t$
$=E_{0}[2 \sin (\omega t+\pi / 6) \cos (\pi / 6)]$
$\left[\because \sin A+\sin B=\sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)\right]$
$=\sqrt{3} E_{0} \sin (\omega t+\pi / 6)$