The capacitors in are initially uncharged and are cannected as in the diagram with switch S open. The applied potential differnce is `V_(ab)= +360 V`.
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a. What is the potential differnce `V_(cd)`
b What is the potential deffernce across each capacitor afrer switch S is closed?
c. How much charge will flow through the awitch after it is closed?

Answer

a. `120 V` b. 18 V` c. `540 muC` from `C` to `D`.

Explanation

(a) Refer to

`V_(a)V_(d)=6/(6+3)xx360=240 V`
and `V_(a)-V_(c)=3/(6+3)xx360=120 V`
(b) Refer to
.
`V_(a)-V_(c)=V_(a)-V_(d)=V_(d)-V_(b)=V_(c)-V_(b)`
`=(360)/2=180 V`
That is, potential across each capacitor is `180 V`
c. Charge flowing after the switch is closed:
`Deltaq_(1)=1080-720=360 muC`
`Deltaq_(2)=720-540=180 muC`
Now,
`Deltaq=Deltaq_(1) + Deltaq_(2) = 360 + 180 = 540 mu C`
Hence, charge of `540 muC` flows from `c` and `d ` after closing the switch.
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