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Two alternating voltage generators produce e.m.f. of the same amplitude E₀ but with a phase difference of $π / 3 .$ The resultant e.m.f. is

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$\sqrt{3} E_{0} \sin (ωt + π / 2)$

$E_{0} \sin (ωt + π / 6)$

$E_{0} \sin (ωt + π / 3)$

$\sqrt{3} E_{0} \sin (ωt + π / 6)$

answer is C.

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Detailed Solution

$E_{1} = E_{0} sinωt, E_{2} = E_{0} \sin (ωt + π / 3)$

$E = E_{2} + E_{1}$

$= E_{0} \sin (ωt + π / 3) + E_{0} sinωt$

$= E_{0} [2 \sin (ωt + π / 6) \cos (π / 6)]$

$[∵ sinA + sinB = \sin (\frac{A + B}{2}) \cos (\frac{A - B}{2})]$

$= \sqrt{3} E_{0} \sin (ωt + π / 6)$

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