Kirchoff's laws, emf and internal resistance

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Homework Help Overview

The problem involves two identical cells with emf E and internal resistance r connected in series and then in parallel with a 7 ohm resistor. The current drawn in each configuration is provided, and participants are tasked with calculating the emf and internal resistance of the cells.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the equations derived from the series and parallel configurations, questioning the validity of the book's answers. Some express suspicion about the book's correctness based on their calculations, while others explore the implications of the current values given in the problem.

Discussion Status

The discussion is active, with participants sharing their equations and results. There is a recognition that the book's answers may not align with their findings, prompting further exploration of the problem setup and assumptions. Some participants suggest that the book may contain errors, while others seek to clarify the reasoning behind the current values.

Contextual Notes

Participants note potential discrepancies in the book's answers and express uncertainty about the assumptions made regarding the current in the parallel circuit. There is a consideration of whether the book's provided values are accurate or if they have misinterpreted the problem.

alexburns1991
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Homework Statement



two identical cells of emf E and internal resistance r are connected in series.
a 7 ohm resistor is connected across the combination and draws a current of 0.333A.
the two cells are now connected in parallel; the 7 ohm resistor now draws a current of 0.375A from the combination.

calculate the emf and internal resistance of the cells.

Homework Equations



E - Ir = IR

Sum of potential differences in a closed loop = 0
Current entering a junction = current leaving a junction

The Attempt at a Solution



i managed to get one equation from the series circuit:
2(E - 0.333r) = 0.333 x 7

and an equation for the parallel circuit:
E - 0.1875r = 0.375 x 7

but this didn't give the correct answers from the book, which says that E = 1.5 and r = 1
 
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alexburns1991 said:

Homework Statement



two identical cells of emf E and internal resistance r are connected in series.
a 7 ohm resistor is connected across the combination and draws a current of 0.333A.
the two cells are now connected in parallel; the 7 ohm resistor now draws a current of 0.375A from the combination.

calculate the emf and internal resistance of the cells.

Homework Equations



E - Ir = IR

Sum of potential differences in a closed loop = 0
Current entering a junction = current leaving a junction

The Attempt at a Solution



i managed to get one equation from the series circuit:
2(E - 0.333r) = 0.333 x 7

and an equation for the parallel circuit:
E - 0.1875r = 0.375 x 7

but this didn't give the correct answers from the book, which says that E = 1.5 and r = 1

Welcome to the PF. Good job showing your work -- that makes it a lot easier.

I don't see how the solution of E = 1.5V can be correct. The 0.333A through the 7 Ohm resistor gives a voltage of 2.33V, so connecting two 1.5V batteries in parallel can't get you up there... Any chance of a typo or misunderstanding in the problem? Your equations look correct to me.
 
im a bit suspicious of the answers in the book, theyve been wrong before!
but i don't want to assume that its always the book that's wrong, so maybe I've missed something?

the equations i got give E = 4.5V and r = 10 ohms, which i think seems too high.
 
alexburns1991 said:
im a bit suspicious of the answers in the book, theyve been wrong before!
but i don't want to assume that its always the book that's wrong, so maybe I've missed something?

the equations i got give E = 4.5V and r = 10 ohms, which i think seems too high.

That's the same numbers that I got with your two equations (which I think are correct equations for the described problem). Either set of numbers (book's or ours) work for the first equation, but only ours fit the 2nd equation. What would be the current for the 2nd case (parallel batteries) to give the book's answers?
 
alexburns1991 said:

The Attempt at a Solution



i managed to get one equation from the series circuit:
2(E - 0.333r) = 0.333 x 7

and an equation for the parallel circuit:
E - 0.1875r = 0.375 x 7

but this didn't give the correct answers from the book, which says that E = 1.5 and r = 1

I agree with the two equations you set up. In fact, the result of the parallel circuit implies an EMF of more than 0.375*7=2.625V, so the 1.5V answer has to be wrong.

EDIT: berkeman beat me.
 
you can get the answers in the book by taking the currents in the parallel equation as being the same, both 0.1875- which gives r = 1, but i don't see why this would be the case at all. surely taking one complete loop and applying kirchhoff's rule, the current through each of the cells would be half that through the resistor?
 
alexburns1991 said:
you can get the answers in the book by taking the currents in the parallel equation as being the same, both 0.1875- which gives r = 1, but i don't see why this would be the case at all. surely taking one complete loop and applying kirchhoff's rule, the current through each of the cells would be half that through the resistor?
Yes, each cell generates 1/2 of the total resistor current.

Berkeman was saying that the parallel current of 0.375A (0.1875A through each cell) is inconsistant with the book answers of 1.5V, 1Ω per cell. So, what would the total current in the parallel circuit have to be if each cell has 1.5V and r=1Ω, and they are connected in parallel to a 7Ω resistor?
 
i worked it out to be 0.2A.

ive just posted another question on this, where my answer disagrees with the answer in the book. i think the solution should be to get a new book!
 
Looks right. So the mistake in the book is either with the final answer, or that they gave the wrong current for the parallel circuit case.

Getting a different book is an option if you are doing self-study.
 
  • #10
alexburns1991 said:
i think the solution should be to get a new book!

:smile: Classic!
 

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