The coefficient of mutual inductance of two circuits `A` and `B` is `3 mH` and their respective resistances are `10 Omega` and `4 Omega`. How much current should change in `0.02 s` in circuit `A`, so that the induced current in `B` should be `0.0060 A`?

Options

A. 0.24 A
B. 1.6 A
C. 0.18 A
D. 0.16 A

Answer

D

Explanation

Induced current in `B=0.006A =6xx10^(-3)A`
Induced emf in `B=6xx10^(-3)xx4V=24 xx10^(-3)V`
Now, `M(dI)/(dt)=24xx10^(-3)`
or `dI=(24xx10^(-3)xx0.02)/(3xx10^(-3))A=0.16 A`