Applying Kirchoff's law to circuit which only has cells

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The discussion revolves around applying Kirchhoff's second law to a circuit with cells that have negligible internal resistance. Initially, it is concluded that the voltage across points A and B (VAB) equals 2E when traversing two different loops. However, when the polarity of one cell is reversed, the calculations yield inconsistent results, suggesting an infinite current, which is deemed impossible. Participants emphasize that all cells inherently possess some internal resistance, challenging the idea of a circuit without it. Ultimately, the conversation highlights the importance of considering resistance and energy conservation when applying Kirchhoff's laws.
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Homework Statement


Find the Voltage across A B in the following circuit. Each cells has negligible internal resistance and an EMF of E volts.
(First diagram in the attached picture)

Homework Equations


I'm trying to use Kirchoff's second law.

The Attempt at a Solution


Now, if I traverse the RSTU loop in the clockwise direction, I find that VAB = 2E.
Similarly, if I traverse the loop RQPU, I also get VAB = 2E.
In this case, traversing either loop gives the same answer for VAB

But what if the polarity of cell TS (the top most cell) was reversed as in the second diagram?
In this case, applying Kirchoff's law to loops RSTU and RQPU yields different results for VAB.. so in such a case, what is the actual value of VAB
 

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The second case has no solution. It would imply an infinite current flowing in the circuit.
It is impossible to determine Vab.
 
hi mahela007! :smile:

sorry, but that makes no sense :redface:

Kirchhoff's rules require current, and you haven't included any

(and there's no such thing as circuits or cells without resistance, you have to include a resistance somewhere on each loop)
 
Do we need to consider the current? Don't the cells maintain a potential difference equal to their EMF because they don't have internal resistance?
 
all cells have internal resistance!

(even you said "negligible" … that's not zero!)
 
lol... ok then.
Let's say for theoretical purposes that a cell had 0 resistance.
Kirchoff's second law says that the algebraic sum of potential differences around a closed loop is 0. The (imaginary) batteries maintain a constant voltage across their terminals. So shouldn't we be able to apply K's law? (regardless of the current that is flowing in this case)

EDIT:
WHOOPS.
I was wondering about this for all of about 5 mins.. then it dawned on me that this model is a violoation of the principle of the conservation of energy. If charge is flowing from one terminal to the other, then some energy must be liberated because of the potential difference. If there is no resistance, no energy can be liberated..

Thanks for all your help.
 
mahela007 said:
lol... ok then.
Let's say for theoretical purposes that a cell had 0 resistance.
Kirchoff's second law says that the algebraic sum of potential differences around a closed loop is 0.

Ok, so try to apply it to the outer loop. You have that Va+Vb=0
Let's say Va=Vb=5V, so we have 10=0 ?
 

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