Applying Kirchoff's law to circuit which only has cells

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SUMMARY

This discussion focuses on applying Kirchhoff's laws to circuits composed solely of ideal cells with negligible internal resistance. The user initially calculates the voltage across points A and B (VAB) as 2E volts using Kirchhoff's second law for two loops, RSTU and RQPU. However, when the polarity of one cell is reversed, the calculations yield conflicting results, leading to the conclusion that VAB cannot be determined in this scenario due to the violation of energy conservation principles. The discussion emphasizes the necessity of including resistance in circuit analysis.

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  • Understanding of Kirchhoff's laws, specifically Kirchhoff's voltage law.
  • Basic knowledge of electrical circuits and components, including cells and resistances.
  • Familiarity with the concept of electromotive force (EMF) in circuits.
  • Awareness of energy conservation principles in electrical systems.
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  • Study the implications of Kirchhoff's voltage law in circuits with varying resistance.
  • Explore the effects of internal resistance in real-world batteries and cells.
  • Learn about the conservation of energy in electrical circuits and its applications.
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mahela007
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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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