Find charge on capacitors in the following cases.
(a)
(b)
(c )
(d)

Explanation

In steady state, no current flows through the capicitor. If there is no information about time. Assume steady state. In steady state, no current in the branch containing capacitance hence resistaces in series with capacitor are redundant and ignore these resistance assume a wire in place of resistance.
(a)
`V_1 = 10/(10+20) xx 2 = 2/3V`
Charge on capacitor
`Q = CV_(1) = 6 xx 2/3 = 4muC`
(b)
`V_1 = 1/(1+2)xx6=2V,V_(2)=6-V_(1)=4V`
`V_(3) = (3)/(3+3)xx6=3V,V_(4)=6-V_(3)=3V`
`Q_(1) = C_1V_1=1xx2=2muC`
`Q_(2) = C_2V_2=2xx4=8muC`
`Q_(3) = C_3V_3=3xx3=9muC`
`Q_(4) = C_4V_4=4xx3=12muC`
(c )
`i = E/(r+R_1)`
`V_1 = iR_(1) = (ER_1)/(r+R_1)`
OR
`V_1 = (R_1 E)/(r+R_1)`
The resistance `R_(2)` is useless, no current in it, repalce it by a wire.

Charge on capacitor
`Q = CV_(1) = (CR_(1)E)/(R_(1)+r)`
(d)
`V_(1) = (15)/((12_15))xx54=30V`
In the resistance `5Omega` no current. repalce it by a wire.

Charge on capacitor
`Q = CV_(1) =2xx30`
`=60muC`.