Find the charge on the four capacirtors of capacitances `1(mu)F,2(mu)F,3(mu)F,and 4(mu)F, shown in figure.

Explanation

At steady state no current flows through
the capacitors .
` R_eq = (3xx6)/(3+6) = 2 Omega `
` i= 6/2 = 3 `
Since current is divided in the inverse ratio
of resistance in each branch, 2` Omega ` will pass
through ` 1, 2 Omega branch and one through 3,
`3Omega` branch .
` V_AB = 2 xx 1 `
` = 2V `
` Q on 1 = mu f capacitors `
` = 2 xx 1muc `
` =2 muc `
` V_BC = 2 xx 2 = 4V. `
` Q on 2 = 2 mu f capacitor = CV `
` = 2 xx 4 mu c `
` = 8 mu c `
` V_DE = 3 xx 1 = 3V `
` Q on 4 = 4 mu f capacitor `
` = 3 xx 4 mu c `
` =12 mu c `
` V_FE = 3 xx 1 = 3V `
` Q across 3 = 3 mu f capacitor `
` 3 xx 3 mu c `
` = 9 mu c ` .