Question

63. The mutual inductance between a primary and a
secondary circuit is 0.5H. The resistance of the
primary and the secondary circuits are 20 ohm and
5ohm respectively. To generate a current of 0.4A in
the secondary, current in the primary must be
changed at the rate of (A/S)​

Answer 1

Answer:

The answer will be  4 A/s

Explanation:

According to the problem there are two circuits , primary and secondary ,

The mutual inductance is given as 0.5 H

Now the general equation of mutual inductance is

NØ = MI

Now differentiating both side with respect to time

=> N dØ/dt = M dI/dt

Now N dØ/dt = induced emf, e

Therefore  e = M dI/dt

Now we know V = IR

  we can write as ,  IR= M dI/dt

Now putting the value I as 0.4 A and resistance of secondary circuit as 5 ohm

0.4 x 5 = 0.5 dI/dt

=> dI/dt = 4 A/s