Question

The inductors of two $LR$ circuits are placed next to each other, as shown in the figure. The values of the self-inductance of the inductors, resistances, mutual-inductance and applied voltages are specified in the given circuit. After both the switches are closed simultaneously, then total work done by the batteries against the induced $EMF$ in the inductors by the time the currents reach their steady state values is______________ $mJ$

Answer 1

Answer: 55

${ϵ}_1=L_1\frac{d{\stackrel{\cdot }{i}}_1}{dt}+\frac{Md{i}_2}{dt}$
$dW={ϵ}_1{i}_{1}dt$
$d{W}_1=L_1\left(d{i}_1\right)i_1+M\left(d{i}_2\right)i_1$
$d{W}_2=L_2\left(d{i}_2\right)i_2+M\left(d{i}_1\right)i_2$
${\int }_0^W\left({\left(dW\right)}_1+{\left(dW\right)}_2\right)={\int }_0^{i_1}{L}_1\left({\left(di\right)}_1\right)i_1+{\int }_0^{i_2}{L}_2\left({\left(di\right)}_2\right)i_2+M{\int }_0^{i_1{i}_2}d\left(i_1{i}_2\right)$
$W=\frac{1}{2}{L}_1{i}_1^2+\frac{1}{2}{L}_2{i}_2^2+M\left(i_1{i}_2\right)$
$W=\frac{1}{2}\times 10\times 10^{-3}\times 1+\frac{1}{2}\times 20\times 10^{-3}\times 4+5\times 10^{-3}\times 2$
$W=55\times 10^{-3}=55mJ$