1.

If φ = 0.02 cos 100xt weber/turns and number of turns is 50 in the coil, the maximum induced emf is.(1) 314 volt(2) 100 volt(3) 31.4 volt(4) 6.28 voltnitude of induced em

Answer»

Given ∅ = 0.02cos(100π)t

and we know that induced emf E = -d∅/dt

so, here d∅/dt = -0.02*100π*(-sin(100π)t) = 2πsin(100π)t

maximum value is = 2π when sin(100π)t = 1.

and for 50 turns it will be E =nd∅/dt = 50*2π = 314v

option 1

thanks a lot



Discussion

No Comment Found

Related InterviewSolutions