Switch `S` is closed at time `t=0`. in the circuit shown in figure.
a. Find the time varying quantities in the circuit.
b. Find the their values at time `t=0`
c. Find their vales at time `t=oo`
Find the time constant of all time varying function
e. Make their exponetial graphs and write their exponential equations.
f. just write the equations to slove them to find different time varying functions.

Explanation

a.
There are four times variable functions `i_1,i_2` and `q`.
b.At `t=0` equivalent resistance of capacitor is zero. So, the simple circuit is as shown below

`R_(n et)=3+(3xx6)/(3=6)=5Omega`
`i_1=15/5=3Aimplies i_2/i_3=6/3=2/1`
`:. i_2=2/(2+1)(3A)=2A`
`i_3=(1/(2+1))(3A)=1A` and `q=0`
c. `t=oo` equilavent resistance of capacitor is infinite.So, equivalent circuit is as shown below `i_1=i_3=15/(3+6)=5/3A`
`i_2=0`
`V_(2F)=V_(6Omega)=iR=(5/3)(6)=10vol t`
`q=CV=(2)(10)=20C`

d.By short circuiting the battery the simplified circuit is as shown below

Net resistance across capacitor or `ab` is
`R_(n et) =3+(3xx6)/(3+6)=5Omega`
`:. tau_C=CR_(net)=(2)(5)=10s`
e. Exponential graphs and their exponential equations are as under.

`i_1=5/3+(3-5/3)e^((-t)/(e^(tauc)))`
`i_2=2e^(-t/(tauC))=2e^(-t/10)`
`i_3=1+(5/3-1)(1-e^(-t(tauC)))=1+2/3(1-e^(-t/10))` ltbr gt `q=20(1-e^(-t/(tauC)))=20(1-e^(-t/10))`
f. Unknowns are four `i_1,i_2,i_3` and `q` . So corresponding to the figure of part a four equations are
`i_1=i_2+i_3`............i
`i_2=(dq)/(dt)`..........ii
Applying loop equation in left hand side loop.
`+15-3i_1-3i_2-q/2=0` ..............iii
Applying loop equation in right hand side loop,
`+q/2+3i_2-6i_3=0` ..............iv.
Soving these equations (with some integration), we can find same time function as we have obtained in part e.