A circuit containing a two position switch S is shown in fig.

(a) The switch S is in position '1'. Find the potential difference `V_(A)-V_(B)` and the rate of production of joule heat in `(R_1)`.
(b) If now the switch S is put in position 2 at t=0 find
(i) steady current in `(R_4)` and
(ii) the time when current in `(R_4)` is half the steady value.
Also calculate the energy stroed in the inductor L at that time.

Answer

A, B, C, D

Explanation

(a) (i) In this case S and I are connected


Using Kirchhoff's law in ABCDGA
`+3-(I_2)xx2-12+(I_1)xx2=0`
`2(I_1)-2(I_2)=9`...(i)
Applying Kirchhoff's law in DEFGD
`=2(I_1)+12-(I_(1)+I_(2))2=0`...(ii)
From (i) and (ii) `(I_1)=21/6 amp.
From (ii) `(I_2)=-1amp`.
To find potential difference between A and B
`V_(A)+3-(-1)xx2=V_(B) implies (V_A)-(V_B)=-5V`
The rate of production of heat of heat in `(R_1)`
`=(I_(1)^(2))(R_1)=(21/6)^(2)xx2=24.5w`
(i) when the switch is put in position 2 then the active circuit will be as shown in the figure.

When the steady state current is reached then the inductor plays no role in the circuit.
`(E_2)=I(R_(2)+R_(4))`
`implies I=3/5=0.6amp.`
(ii) the growth of current in L-R circuit is given by the expression
`I=(I_0)[1-e^(-(R)/(L)t)]`
When `I=(I_0)/(2), then (I_0)/(2)=(I_0)[1-e^(-(R)/(L)t)]`
implies 1/2=1-e^(-(R)/(L)t) implies e^(-(R)/(L)t)=1/2`
Taking log on both side
`log_(e) e^(-(R)/(L)t)=log_(e)1/2`
When `R=(R_2)+(R_4)`
`=t=1.386xx10^(-3) sec`.Thus this much time is required for current to reach hald of its steady value.
The energy stored by the inductor at that time is given
bu `E=1/2 LI^(2)=1/2xx10xx10^(-3)xx(0.6/2)^(2)=4.5xx10^(-4)J`.