The mutual inductance between a primary and secondary circuit is `0.5H`. The resistance of the primary and the secondary circuits are `20 ohms` and `5ohms` respectvely. To genrate a current of `0.4A` in the secondary,current in the primary must be changed at the rate of

Options

A. `4.0 As^(-1)`
B. `1.6As^(-1)`
C. `16.0As^(-1)`
D. `8.0As^(-1)`

Answer

A

Explanation

emf, `e_(2)=M((di)/(dt))=i_(2)R_(2)`
`therefore"Current"((di_(1))/(dt))=(i_(2)R_(2))/(M)=((0.4)(5))/(0.5)="4 As"^(-1)`