Why Does Current Direction Affect Voltage Sign in KVL Analysis?

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Homework Help Overview

The discussion revolves around the application of Kirchhoff's Voltage Law (KVL) in circuit analysis, particularly focusing on how current direction affects voltage sign across components like resistors and voltage sources. Participants are exploring the implications of current flow and loop traversal direction in the context of mesh analysis.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants are examining the relationship between current direction and voltage sign, particularly in the context of entering and exiting voltage sources. Questions are raised about the conventions used in different resources regarding voltage polarity when current interacts with voltage sources.

Discussion Status

Some participants have provided insights into the nature of voltage sources and how their polarity affects voltage calculations. There is an ongoing exploration of differing interpretations and conventions regarding voltage signs in relation to current direction, with no explicit consensus reached.

Contextual Notes

There are references to different textbooks and resources that may present varying conventions regarding voltage signs, indicating a potential source of confusion among participants. The discussion also touches on the concept of multiple batteries in parallel affecting current flow through a voltage source.

PainterGuy
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Hello, :smile:

Please have a look on the following linked diagram:
http://img32.imageshack.us/img32/1456/img0001vsi.jpg

While applying KVL it is not necessary to have the loop traversing direction and the current direction the same. e.g. the loop could be traversed CW and the current direction could be CCW. But in some situations having the directions for the both current and loop could make things easier such as in mesh analysis. Further, the direction of current defines the polarities of the resistors because the current flows from higher potential toward lower. If a resistor is traversed in the same direction as the current then the IR term would have -ve sign, otherwise it would be +ve.

In the linked diagram if we star traversing the loop from point "a", then the equation would be:

+E - IR1 -IR2 = 0

Question:-
In some cases I have seen when the current enters a voltage source from its positive terminal then the V is +ve otherwise it will be –ve. Why is so? Most often when the current enters the voltage source from its +ve terminal then the V is taken to be -ve. Please help.

Cheers
 
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Resistors don't have polarities, per say. The polarity that we ascribe to them is based upon the assumed direction of the current flowing through them. If our choice of current direction happens to be counter to the actual current flowing (which you might determine at the end of the analysis), then there's no problem -- the math takes care of everything.

Voltage sources are different. They most assuredly have a polarity. Whether you take the change in voltage as positive or negative depends solely on the direction in which you pass through the source while following your path around the loop. If you enter at the + terminal and exit at the - terminal, you have a voltage drop. If you enter at the - terminal and exit at the +, then you have a voltage rise.
 
Thanks, gneill.

I agree with what you say that if you enter at the + terminal and exit at the - terminal, you have a voltage drop. If you enter at the - terminal and exit at the +, then you have a voltage rise. But a book I have used do it differently.

In some cases I have seen when the current enters a voltage source from its positive terminal then the V is +ve otherwise it will be –ve. Why is so?

Cheers
 
As I conceptualize it, entering a voltage source from the positive terminal acts like a resistor. This can occur if you have several batteries wired in parallel, and there is enough EMF present to "force" current to go the other way through the voltage source. In any case, the voltage source will always impede current traveling through it in the reverse direction.

Does that answer your question?
 

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