Electromotive Force: E=Vab=IR Explained

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Homework Help Overview

The discussion revolves around the concept of electromotive force (emf) and its relationship with potential difference in electrical circuits, specifically focusing on the equation E = Vab = IR. Participants are exploring the implications of this relationship in the context of circuit behavior.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to understand the meaning of the relationship between potential rise and drop in a circuit, particularly how it relates to an ideal source of emf and the behavior of charge in a closed loop.

Discussion Status

Some participants have expressed confusion regarding the implications of the equations and concepts presented. Others have provided clarifications and examples, such as referencing Kirchhoff's voltage law, which may help in understanding the relationships involved. There appears to be a productive exchange of ideas, with at least one participant indicating improved understanding.

Contextual Notes

There is an emphasis on the ideal conditions of the source of emf and the assumptions made regarding the circuit components. The discussion also highlights the importance of understanding the relationship between potential differences in various parts of the circuit.

AGGENGR
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The potential difference between the ends of a wire is given by Vab = IR. Combing that with E = Vab we have
E = Vab = IR (ideal source of emf).

" That is, when a positive charge flows around the circuit, the potential rise as it passes through the ideal source is numerically equal to the potential drop as it passes through the remainder of the circuit. Once and are known, this relationship determines the current in the circuit."- I can't seem to understand what this means?
 
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Imagine going around a closed loop in the circuit. As charge goes through the resistor it drops down in potential. As it goes through the battery it rises in potential. Those two potentials are equal in magnitude.
 
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...the potential rise as it passes through the ideal source is numerically equal to the potential drop as it passes through the remainder of the circuit.

If you think about it this is because the "ideal source" and the "remainder of the circuit" are connect to the same two nodes...
KVL.png


See also Kirchhoff's voltage law.
 
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Thanks i seem to get it after reading these posts and rereading the chapter! :):):):):):):):):):):):):):):):):):):)
 

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