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My attempt:As the galvanometer shows zero current and the portion encircled contains no such device which can store charge therefore current entered must be zero as well. Moving on these lines I can solve this question easily.

Doubt:By the above logic I get it that the current entered is zero but how does current know beforehand that it should not enter that particular branch as both potential difference and resistance are zero across AP so there is a possibility for current to flow. I understand how current decides to not enter into a loose end wire (because electrons just drift a bit from their place and if electrons before them are not moving further then there would be no current or simply applying KCL) but I can't understand this one by a similar logic.

I hope what I said makes sense, I am not able to phrase myself correctly. Feel free to edit the question for better.

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  • $\begingroup$ You said "both potential difference and resistance are zero". The first part is correct, that the potential difference is zero. The second part is not, as resistance is not zero. The resistance is actually 1//5 = 5/6 Ohms. $\endgroup$ – verdelite May 24 at 2:40
  • $\begingroup$ I was talking about resistance across AP I have edited the question to avoid confusion $\endgroup$ – satyam kumar jha May 24 at 2:56
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I think you are confused why the current entering the branch from the main circuit $=0$

Well what exactly is a cell?

A cell consists of a $+$ve terminal and a $-$ve terminal. The number of electrons that are being supplied by the negatively charged electrode must complete the cycle and reach the positively charged electrode. Now assume some charges flow into the section AB. Now since no charges can escape through the segment QP (otherwise current would have not been equal to zero) can you figure out any such route which the charges can take to reach the other electrode?

Answer is no. Now you can argue that the charges can take the closed loop ABQBA but that isn't possible.

Why? Assume $-Q$ amount of charge comes per unit time out of the negative terminal of the battery. Out of this $-q$ amount of charge sneaks into the section AB. Now some current is there in the loop BQB due to the two cells present here. Hence it will combine with with the already circulating electrons (say $-q2$). Now what is the charge circulating in the circuit. It is $-(q2+q)$. Now here is a trick. Just forget the two $2V$ and $4V$ cells and just replace the two cells by an equivalent cell of suppose emf $V$. This enables you to think that as if there is a single cell whose negative terminal supplies $-q2$ amount of change which subsequently combines with $-q$ amount of charge and reaches the other terminal and here is the problem the amount of charge departing the negative terminal does not equal the charge entering the other terminal. This is the problem. Now you may wonder that what problem will be produced if this does happen.

The answer lies in the electro-chemistry of the cell. The emf value of a cell is fixed and it depends upon the materials that you are using as the electrodes. EMF value is measured as follows :

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Now as you can see since there is no charge storing component in the above circuit the amount of charge entering the $Cu$ electrode must equal to the amount of charge departing the $Zn$ electrode and the "potential difference" between these two electrodes in the isolated set up is the emf of the cell and is always constant (at a given temperature and concentration of the solutions used).

Now if the charge entering one terminal does not equal charge entering another terminal then the "potential difference" between the two electrodes will no longer be same as the emf measured. Hence such a situation is impossible .

Hope I am able to address your doubt.

Note: In the above diagram you have labelled $P$ twice so instead I have replaced one $P$ with $B$ in my answer.

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