TITLE: Cells in parallel combination
QUESTION [0 upvotes]: This is an excerpt from my book and I am not able to understand how could “V” be equal to Ε1- I1r1 as well as Ε2- I2r2.
 I mean One battery could have emf of 50 volt and another just 5 volt. So, How could 50 be equal to 5? How could voltage across both the batteries be same?
  And I also wonder if both the batteries were of equal emfs, would there be any current? Because that would be like two persons pushing with equal force but in opposite direction.
 I mean electrons coming from two equal cells E1 and E2 in parallel combination (as in above figure), won’t they push against each other with equal force and hence no current should flow?

REPLY [1 votes]: Perhaps a numerical example will help with the assumption that there is no external circuit connected to the batteries?

Applying Kirchhoff's voltage law to loop $ABCDA$,
$-5-v_1-v_8+50 = 0 \Rightarrow -5-i\cdot 1-i\cdot 8+50 = 0$
Hence $ i= 5\,\rm A$, $v_1 = 5\,\rm V$ and $v_2 = 40\,\rm V$.
So in that loop there is a voltage drop of $5+40 = 45 \,\rm V$ which is the difference between the emfs of the two batteries.
How could voltage across both the batteries be same?
The potential of node $A$ relative to node $C$ is $5\color{red}+ 5\cdot 1 = 10\,\rm V = 50 \color{red} - 5\cdot 8$
Note also that the  $50\,\rm V$ battery is being discharged and $5\,\rm V$ battery is being "charged" and can think of the $50\,\rm V$ battery being discharged as wasted energy and over time will mean that it has supplied no useful energy to an external circuit.
I also wonder if both the batteries were of equal emfs, would there be any current?
Without an external circuit present there would be no current on the loop $ABCDA$.
In practice (nearly) identical batteries are put in parallel so that each battery delivers a portion of the current demand of an external load.
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