Calculate the charge on each capacitor and the potential difference across it in the circuits shown in figure for the cases :


(i) switch S is closed and
(ii) switch `S` is open.
(iii) In. figure (b), what is the potential of point `A` when `S` is open?

Answer

A

Explanation

a. Fig a. `V_6=90/3=30V, q_6=6xx30=180muC`
`V_3=90/3=30V`
`q_3=30xx3=90muC`
Fig b. Capacitor `1muF` is short circuited.
Therefore `q_1=0`
`V_26=(20/(20+20+10))xx100=40V`
This `40V` will distribute in inverse ratio of capacity.
`:. V_6=2/8xx40=10V`
`V_2=6/8xx40=30V`
`:. q_6=60muC, q_2=60muC`
b. Fig a When `S` is open `6muF` is short-circuited
or `V_6=0, q_6=0`
and `V_3=90V, q_3=270muC`
Fig b. When `S` is open `V_1=100V`
`q_1=100muC`
`V_26=100V`
`:.V_6=2/8xx100=25V`
and `V_2=6/8xx100=75V`
`:. q_6=150muC, q_2=150muC`
Further, `V_A-0=V_2`
`:. V_A=V_2=75V`