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Charges q(1) , q(2) and q(3) are placed ...

Charges `q_(1)` , `q_(2)` and `q_(3)` are placed on capacitors of capacitance `C_(1)`, `C_(2)` and `C_(3)`, respectively, arranged in series as shown. Switch `S` is there closed. What are the final charges `q'_(1)`, `q'_(2)` and `q'_(3)` on the capacitors?
Given `q_(1)=30muC`, `q_(2)=muC`, `q_(3)=10muC`, `C_(1)=10muF`, `C_(2)=20muF`, `C_(3)=30muF` and `epsilon=12"volt"`

Text Solution

Verified by Experts

Applying Kirchoff's law on the loop,
`(q_(1)')/(C_(1))+(q_(2)')/(C_(2))+(q_(3)')/(C_(3))=epsilon`.........`(1)`
Net charge on plates `2` and `3` will remain conserved.
`:. -q_(1)'+q_(2)'=-q_(1)+q_(2)` ...........`(2)`
Also, net charge on plates `4` and `5` will remain conserved.
`:. -q_(2)'+q_(3)'=-q_(2)+q_(3)` ..........`(3)`
Using Eqs. `(1)`, `(2)` and `(3)` and putting the values, we get
`q_(1)'=(790)/(11)muC`, `q_(2)'=(680)/(11)muC`, `q_(3)'=(570)/(11)muC`.
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