Question

A coil of 10-2 h inductance carries a current i =2sin (100t)

Answer 1

Explanation:

Given that,

I=2sin(100t)I=2sin(100t)… (I)

Now,

  imax=2Aimax=2A

 imax2=1Aimax2=1A

We know that,

  1=2sin(100t)1=2sin⁡(100t)

 sin(100t)=12sin⁡(100t)=12

 100t=π6100t=π6

Now,

 Differentiate of equation (I)

  I=2sin(100t)I=2sin⁡(100t)

 dIdt=2×100cos(100t)dIdt=2×100cos⁡(100t)

 dIdt=200×cos(π6)dIdt=200×cos⁡(π6)

 dIdt=200×√32dIdt=200×32

 dIdt=100√3dIdt=1003

Now, the induced e m f in coil

  eind=L×dIdteind=L×dIdt

 eind=10−2×100√3eind=10−2×1003

 eind=√3eind=3

Hence, the induced e m f in coil is √3 volt