Question

A coil of inductance $L=2.0\mu H$ and resistance $R=1.0\Omega$ is connected to a source of constant emf $ϵ=3.0V$
resistance $R_0=2.0\Omega$ is connected in parallel with the coil.The amount of heat generated in the coil after the switch $S_w$ is disconected (in microjoule) is (The internal resistance of the source is negligible.)

Answer 1

Initially, after a steady current is set up, the current is flowing as shown.
In steady condition, $i_2=\frac{\xi }{R},i_1=\frac{E}{R_0}.$


When the switch is disconnected, the current
through $R_0$ changes from $i_1$ to the right, to
$i_2$ to the left. (The current in the inductance
cannot change suddenly.). We then have the
equation,
$L\frac{d{i}_2}{dt}+(R+R_0)i_2=0$
This equation has the solution $i_2=i_{20}{e}^{\frac{-(R+R_0)t}{L}}$
The heat dissipated in the coil is,
$Q={\int }_0^{\infty }{i}_2^{2}Rdt=i_{20}^{2}R{\int }_0^{\infty }{e}^{\frac{-2(R+R_0)t}{L}}dt$

$=R{i}_{20}^2\times \frac{L}{2(R+R_0)}=\frac{L{\xi }^2}{2R(R+R_0)}=3\mu J$