In the circuit shown, `r = 4 Omega, C = 2mu F`
(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 reducedas
`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)(1)/(2)=Ir=(3)(4)=12V`
Equivalent capacitance between 5 and 6 is
`C_(56)=(C)/(2)=1muF`
`thereforeq1_(56)=C_(56)V_(56)=12muC`
NOW,`V_(97)=V_(17)-V_(19)`
`=-r(1)/(2)+2(rl)/(2)=(rl)/(2)=((4)(3))/(2)=6V`
therefore`q_(97)=(2)(6)=12muC`
Similarly, `q_(89) = 12muC`