Two alternating voltage generators produce emfs of the same amplitude E0 but with a phase difference of π/3. The resultant emf is

E1=E0sinωt; E2=E0sinωt+π/3E = E2 + E1=E0 sinωt+π/3+E0sinωt=E02sinωt+π/6 cosπ/6=3E0sin ωt+π/6

Two alternating voltage generators produce emfs of the same amplitude E0 but with a phase difference of π/3. The resultant emf is

E1=E0sinωt; E2=E0sinωt+π/3E = E2 + E1=E0 sinωt+π/3+E0sinωt=E02sinωt+π/6 cosπ/6=3E0sin ωt+π/6

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

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$E_{0} \sin [ω t + (π / 6)]$

$E_{0} \sin [ω t + (π / 3)]$

$\sqrt{3} E_{0} \sin [ω t + (π / 2)]$

$\sqrt{3} E_{0} \sin [ω t + (π / 6)]$

answer is C.

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Detailed Solution

$E_{1} = E_{0} \sin ω t; E_{2} = E_{0} \sin [ω t + (π / 3)]$

E = E₂ + E₁

$= E_{0} \sin [ω t + (π / 3)] + E_{0} \sin ω t$

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

$= \sqrt{3} E_{0} \sin [ω t + (π / 6)]$

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