What Happens to the I-V Graph When Voltage Polarity Is Reversed?

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SUMMARY

The discussion centers on the behavior of the I-V graph when voltage polarity is reversed, specifically in the context of Ohm's Law. Participants clarify that while reversing voltage results in negative values for both current (I) and voltage (V), the slope of the graph, representing resistance (R), remains positive. The standard practice is to plot current as a function of voltage (I(V)), which is more practical for controlling voltage in electrical components. Additionally, the linearity of the I-V relationship is contingent upon constant resistance, as seen in components like light bulbs where resistance varies with temperature.

PREREQUISITES
  • Understanding of Ohm's Law
  • Familiarity with I-V characteristics of electrical components
  • Knowledge of linear vs. nonlinear graphing
  • Basic concepts of resistance and conductivity
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  • Study the implications of reversing voltage polarity in circuit analysis
  • Learn about the I-V characteristics of nonlinear components
  • Explore the relationship between resistance and temperature in materials
  • Investigate practical applications of I(V) graphs in electronic design
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Electrical engineers, physics students, and anyone interested in understanding the principles of circuit behavior and graphing electrical relationships.

AngelofMusic
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A very quick question about Ohm's Law:

When I graph I (x-axis) vs. V (y-axis), I would get the standard linear relationship specified by Ohm's Law.

One question asked: What would happen if the voltage were reversed? (I think it means what would happen if the polarity was reversed.)

I think that the graph should remain the same, because then V would become negative and so would the current. But some others are saying that this would result in a negative slope for the graph.

Can anyone help?
 
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The slope of the graph is R, the resistance. Therefore, in a simple case such as this, it cannot be negative. You are correct then, both V and I would be "reversed".
 
I should also mention, just for general knowledge, that it is more common to draw a graph of I as a function of V, i.e I(v) rather than V(I) that you drew. The graph I(v) of an electric component describes the current through the component as a function of the potential on it, and its slope is the conductivity of the component (which is 1/R). It is a more useful graph because usually you control the V on the component, and not the current I through it. :smile: Also, I(v) (or V(I)) are not always linear, they are only linear when the resistance of the component is constant. In light bulbs, for example, the resistance changes as the heat grows, so the graphs of it will not be linear at all.
 
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Thanks a lot for your clarification! And upon more careful reading, I was asked to plot the I-V curve, which indeed has I as a function of V. So thanks for your reminder as well! Caught my mistake! :-)
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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