Graph of V and I against time in AC

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The discussion centers on the graph of voltage (V) and current (I) plotted against time for a resistor in an AC circuit, noting that both curves are sinusoidal and in phase. A key point raised is the observation that the current curve has a higher amplitude than the voltage curve, which seems counterintuitive given Ohm's Law (V = IR). Participants clarify that the comparison of amplitudes is meaningless without considering the units on the scale of the graph. Additionally, it is highlighted that if the resistance is less than one ohm, the current can indeed appear greater than the voltage. Understanding the context of the graph and the units used is essential for accurate interpretation.
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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.
 
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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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