A circuit containing a two position switch `S` is shown in figure

a. The switch `S` is in position 1. Find the potential differecne `V_A-V_B` and the rte of production of joule heat in `R_1`
b. If now the switch `S` is put in position 2 at `t=0`. Find ltbRrgt i steady current in `R_4` and ii hte time when current in `R_4` is half the steady value. Also calculate the energy stored in the inductor `L` at that time.

Explanation

a. In steady state no current will flow through capacitor.

Applying kirchoff's second law in loop 1.
`-2i_2+_2(i_1-i_2)+12=0`
`:. 2i_1-4i_2=-12`
or `i_1-2i_2=-6`……i
Applying Kirchoff's second law in loop 2,
`-12-2(i_1-i_2)+3-2i_2=0`
`:. 4i_1-2i_2=-9` ..(ii)
Solving Eqn i and ii we get
`i_2=2.5A` and `i_1=-1A`
Now `V_A+3-2i_1=V_B` or `V_A-V_B=2i_1-3`
`=2(-1)-3=-5V`
`P_(R_1)=(i_1-i_2)^2R_1=(-1-2.5)^2(2)`
`=24.5W`
b. In position 2 Circuit is as below

Steady current in `R_4`
`i_0=3/(3+2)=0.6a`
Time when current in `R_4` is half the steady value,
`i =i_(0)(1-e^(-t//tau)L)`
`i=i_0/2at t=t_(1/2)`, where
`t_(1//2) =tau_L(In2) =L/R In(2)=((10xx10^-3))/5 In (2)`
`=1.386x1/210x10^-3 s`
`U =(1)/(2)Li^(2) = (1)/(2) (10 xx 10^(-3)) (0.3)^(2)`
`= 4.5 xx 10^(-4)J`