$A$ transformer is based on the principle of

$X$ and $Y$ are two circuits having a coefficient of mutual inductance $3 \text{ mH}$ and resistances $10 \text{ } \Omega$ and $4 \text{ } \Omega$ respectively. To have an induced current of $60 \times 10^{-4} \text{ A}$ in circuit $Y$,the amount of current to be changed in circuit $X$ in $0.02 \text{ s}$ is: (in $A$)

$A$ pair of adjacent coils has a mutual inductance of $2 \text{H}$. If the current in one coil changes from $0 \text{A}$ to $30 \text{A}$ in $0.15 \text{s}$, what is the change of flux linkage with the other coil (in $\text{ Wb}$)?

If a change in current of $0.01 \,A$ in one coil produces a change in magnetic flux of $2 \times 10^{-2} \,Wb$ in the other coil,then the mutual inductance of the two coils in henry is (in $\,H$)

Two coils have a mutual inductance of $0.005 \ H$. The current changes in the first coil according to the equation $I = I_0 \sin \omega t$,where $I_0 = 10 \ A$ and $\omega = 100 \pi \ rad/s$. The maximum value of the induced emf in the second coil is:

The mutual inductance between two coils is $1.25 \ H$. If the current in the primary coil changes at the rate of $80 \ A/s$, then the induced $e.m.f.$ in the secondary coil is ...... $V$.