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Look at the above circuit, I didn't get anything of it. Firstly i tried it using Kirchoff's voltage rule, I failed. I'm confused with those batteries. Can someone please explain me the role of those batteries and the solution to this problem.

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  • $\begingroup$ V1 may be $$\frac{(E2-E1)C2}{(C1+C2)} $$ $\endgroup$ – Nehal Samee Feb 24 '18 at 10:14
  • $\begingroup$ may be? ........ $\endgroup$ – Selena Feb 24 '18 at 10:16
  • $\begingroup$ I'm only confused with those batteries... $\endgroup$ – Selena Feb 24 '18 at 10:17
  • $\begingroup$ The term Cx in the question is unclear ... So I found the individual potential difference of the capacitors ... $\endgroup$ – Nehal Samee Feb 24 '18 at 10:18
  • $\begingroup$ $C_x$ means $C_1$ and $C_2$ $\endgroup$ – Selena Feb 24 '18 at 10:18
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Define $V_1$ and $V_2$ such that if $q>0$, then $V_1>0$ and $V_2>0$ (i.e. $V_1$ is the left plate potential minus right plate potential, and $V_2$ is the right plate potential minus left plate potential). Write KVL, adding voltage drops along the circuit: $$ -E_1 + V_1 + E_2 + V_2 = 0 $$ Since the charge induced in the capacitors are equal, $$ q=C_1 V_1 = C_2 V_2 \Rightarrow V_1 = \frac{C_2}{C_1}V_2, V_2=\frac{C_1}{C_2}V_1$$ To find $V_1$, substitute the expression for $V_2$ in the KVL equation. This yields $$V_1 = \frac{C_2}{C_1+C_2}(E_1 - E_2). $$ Similarly, substituting the expression for $V_1$ gives $$V_2 = \frac{C_1}{C_1+C_2}(E_1 - E_2).$$

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See..this is the answer! U understood? if yes, please clear, me not.

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  • $\begingroup$ Well...It's the direct formula...What's new...? $\endgroup$ – Nehal Samee Feb 24 '18 at 10:33
  • $\begingroup$ I only not got how (1) Came, everything else is alright...the last part $\endgroup$ – Selena Feb 24 '18 at 10:34
  • $\begingroup$ They are trying to say that , $V_2$ is connected at positive plates of both cells and so have a positive potential...Opposite for $V_1$... $\endgroup$ – Nehal Samee Feb 24 '18 at 10:36
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    $\begingroup$ i'm not understanding what you are even saying, can you please elaborate the concept and be specific! $\endgroup$ – Selena Feb 24 '18 at 10:42
  • $\begingroup$ Are you not understanding the loop part ? $\endgroup$ – Nehal Samee Feb 24 '18 at 10:47
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The charge polarity on the capacitor due to the cell $E_1$ is not a problem I think . Excess electrons flow from the negative plate and flow into the positive plate . But your confusion lies in the cell $E_2$ . As you notice that $E_2$ has a higher tendency to flow current in its own direction . From the figure , electricity flows from + of top capacitor to - of bottom capacitor . And this direction matches with the current flow for cell $E_2$ . enter image description here

I think the opposite polarity is justified ...

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