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

$\epsilon _{1}=L_{1}\frac{d \overset{\cdot }{i}_{1}}{d t}+\frac{M d i_{2}}{d t}$
$dW=\epsilon _{1}i_{1}dt$
$dW_{1}=L_{1}\left(d i_{1}\right)i_{1}+M\left(d i_{2}\right)i_{1}$
$dW_{2}=L_{2}\left(d i_{2}\right)i_{2}+M\left(d i_{1}\right)i_{2}$
$\displaystyle \int _{0}^{W}\left(\left(dW\right)_{1} + \left(dW\right)_{2}\right)=\displaystyle \int _{0}^{i_{1}}L_{1}\left( \left(di\right)_{1}\right)i_{1}+\displaystyle \int _{0}^{i_{2}}L_{2}\left( \left(di\right)_{2}\right)i_{2}+M\displaystyle \int _{0}^{i_{1} i_{2}}d\left(i_{1} i_{2}\right)$
$W=\frac{1}{2}L_{1}i_{1}^{2}+\frac{1}{2}L_{2}i_{2}^{2}+M\left(i_{1} i_{2}\right)$
$W=\frac{1}{2}\times 10\times 10^{- 3}\times 1+\frac{1}{2}\times 20\times 10^{- 3}\times 4+5\times 10^{- 3}\times 2$
$W=55\times 10^{- 3}=55mJ$