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

Question

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 (A) E0 sin (ω t+π / 3) (B) E0 sin (ω t+π / 6) (C) √3 E0 sin (ω t+π / 6) (D) √3 E0 sin (ω t+π / 2)

Tardigrade Admin · Accepted Answer

E1=E0 sin ω t, E2=E0 sin (ω t+π / 3) E=E2+E1 =E0 sin (ω t+π / 3)+E0 sin ω t =E0[2 sin (ω t+π / 6) cos (π / 6)] [∵ sin A+ sin B= sin ((A+B/2)) cos ((A-B/2))] =√3 E0 sin (ω t+π / 6)

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

$E_{0} sin (ω t + π /3)$

$E_{0} sin (ω t + π /6)$

$3 E_{0} sin (ω t + π /6)$

$3 E_{0} sin (ω t + π /2)$

$E_{1} = E_{0} sin ω t, E_{2} = E_{0} sin (ω t + π /3)$
$E = E_{2} + E_{1}$
$= E_{0} sin (ω t + π /3) + E_{0} sin ω t$
$= E_{0} [2 sin (ω t + π /6) cos (π /6)]$
$[∵ sin A + sin B = sin (\frac{A + B}{2}) cos (\frac{A - B}{2})]$
$= 3 E_{0} sin (ω t + π /6)$

Q. 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

E0​sin(ωt+π/3)

E0​sin(ωt+π/6)

3​E0​sin(ωt+π/6)

3​E0​sin(ωt+π/2)