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A circuit has a section Ab shown in fig....

A circuit has a section `Ab` shown in fig. The emf of the source equals `E = 10V`, the capacitances are equal to `C_(1) = 1.0 muF` and `C_(2) = 2.0 muF`, and the potential difference `varphi_(A) - varphi_(B) = 5.0 V`. Find the voltage across each capacitor.

Text Solution

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The correct Answer is:
10 V,5 V
`10 V,5V`
Let, us make the charge distribution, as shown in the figure.
Now, `varphi_(A)-varphi_(B)=(q)/(C_(1))-xi +(q)/(C_(2))`
or, `q=((varphi_(A)-varphi_(B))+xi)/(C_(1)+C_(2))C_(1)C_(2)`
Hence, voltage across the capacitor `C_(1)`
`= (q)/(C_(1))=((varphi_(A)+varphi_(B))+xi)/(C_(1)+C_(2))C_(2)=10 V`
and voltahe across the capacitor, `C_(2)`
`=(q)/(C_(2))=((varphi_(A)-varphi_(B))+xi)/(C_(1)+C_(2))C_(1)=5V`.
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