Question

If the current in a primary coil of transformer decreases from $0.8$ A to $0.2$ A in $4$ millisecond, then calculate induced emf in the secondary coil. [Coefficient of mutual induction is $1.76$H].

Answer 1

Rate of change of current in primary coil $\frac{\Delta i}{\Delta t}=\frac{0.2-0.8}{4\times {10}^{-3}}=-150$ $A/s$
Given : $L=1.76H$
Induced emf in the secondary coil is given by $E=-L\frac{\Delta i}{\Delta t}$
$\therefore$ $E=-1.76\times (-150)=264$ Volts