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In the circuit shown each capacitor has a capcitance C. The emf of the celll is E. If the switch S is closed then <br> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/DCP_VOL_4_C25_E01_121_Q01.png" width="80%">

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Answer Text

Positive charge wil flow out of the positive terminal of the cell Positive charge will enter the positive terminal of the cellthe amount of the charge flowing through the cell will be `1/3` CEthe amount of charge flowing through the cell is `(4/3)CE`

Answer :
A::D
Solution :
`C_i=((C)(2C))/(C+2C)=2/3C` <br> `q_i=C_iE=2/3EC` <br> `C_f=2C` <br> `:. q_f=2EC` <br> `q_f=2EC` <br> `/_\q=q_f-q_i` <br> `=4/3CE`

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