Graph of V and I against time in AC

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SUMMARY

The discussion centers on the graphical representation of voltage (V) and current (I) over time in an alternating current (AC) circuit, specifically using Serway's College Physics 7th edition as a reference. Participants clarify that the amplitude of the current curve can exceed that of the voltage curve due to the units being plotted on the same axis, which can lead to misleading interpretations. The relationship defined by Ohm's Law (V = IR) does not directly dictate the amplitude comparison when different scales are used, especially with resistances below one ohm.

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While revising on A.C, I came across the graph of V and I across a resistor plotted against time on the same axis. The book is Serway's College Physics 7th edition. Both curves are sinosoidal and are in phase which I understand, but why does the curve of I is always at a higher amplitude than V? As I know, V = IR, wouldn't the voltage curve have a higher amplitude, that is the peak value of voltage would correspond to a lower peak current for a given resistor? Why in this case is it the other way round?

Thanks for any inputs.
 
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look at the units on that scale. They are not the same. Your question is meaningless. You will see what I mean once you figure out the solution.
 
Last edited:
As Cyrus said, when different units are plotted on the same axis, comparison is meaningless without a scale.

But even if the axis is specifically drawn such that 1A (one ampere) and 1V (one volt) have the same segment length on the vertical axis, the I curve can be higher than the V curve. Think about what happens when you have a resistance that is less than one ohm.
 

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