In the circuit shown, `r=4Omega`, `C=2muF`
`(a)` Find the current coming out of the battery just after the switch is closed.
`(b)` Find charge on each capacitor in the steady state condition.

Explanation

`(a)` Due to symmetry the points of equal potential are joined together, and the circuit may be reduced as
`R_(13)=(r )/(2)+(r )/(4)+(r )/(4)+(r )/(2)=(3)/(2)r`
Here, `r=4Omega`, `R_(13)=(3)/(2)(4)=6Omega`
Thus, `I=(24)/(R_(13))=(24)/(6)=4A`

`(b)` In the steady-state condition, the circuit may be reduced as
`R_(13)=2r=2(4)Omega=8Omega`
`I=(24)/(8)=3A`
`V_(56)=(2r)(I)/(2)=Ir=(3)(4)=12V`
Equivalent capacitance between `5` and `6` is
`C_(56)=(C )/(2)=1muF`
`:. q_(56)=C_(56)V_(56)=12muC`
Now, `V_(97)=V_(17)-V_(19)= -r(I)/(2)+2(rI)/(2)=(rI)/(2)=((4)(3))/(2)=6V`
`:. q_(97)=CV_(97)=(2)(6)=12muC`
Similarly, `q_(89)=12muC`