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.