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In a circuit shown in fig find the pote...

In a circuit shown in fig find the potentail difference between the left and right plates of each capacitor.

Text Solution

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Suppose `E_(2) gt E_(1)`
Therefore, right place of `C_(1)` has charge `+q` and left place of `C_(1)` has charge `-q`.
On the same basics, left plate of `C_(2)` has charge `+q` and right place of `C_(2)` has charge `-q`.
On the same basics, left plate of `C_(2)` has charge `+q` and right plate of `C_(2)` has charge `-q`.
In a closed circuit, pot diff. `dV = 0`.
`(q)/(C_(1)) + E_(1) + (q)/(C_(2)) = 0 or q ((1)/(C_(1)) + (1)/(C_(2))) = E_(2) - E_(1)`
or `q((C_(2) + C_(1)))/(C_(1) C_(2)) = E_(2) - E_(1)`
or `q = ((E_(2) - E_(1)) C_(1) C_(2))/(C_(1) + C_(2))`
Hence, pot, diff across left and right plate of `C_(1)` is
`V_(1) = (q)/(C_(1)) = ((E_(2) - E_(1)) C_(2))/(C_(1) + C_(2))`
Similarly, pot. diff. across left and right plate of `C_(2)` is
`V_(2) = (q)/(C_(2)) = ((E_(2) - E_(1)) C_(1))/((C_(1) + C_(2)))`
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