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

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    Law Ohm's law
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Reversing voltage polarity in an I-V graph affects both current and voltage values, leading to negative values for both. The slope of the graph, representing resistance, remains positive, as resistance cannot be negative. It is more common to graph current as a function of voltage (I(V)), which provides a clearer understanding of how current responds to voltage changes. In cases where resistance varies, such as in light bulbs, the I-V relationship may not be linear. Understanding these concepts is crucial for accurately interpreting electrical behavior in components.
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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! :-)
 
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