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My book defines impedance as the ratio between the amplitude of voltage to that of the current (Vmax to Imax), I just wanted to make sure whether its also true for the instantaneous values of voltage and current.

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  • $\begingroup$ It is not even true for dc voltages and currents. $Z=dV/dI$ not $Z=V/I$. $\endgroup$
    – CuriousOne
    Apr 8, 2016 at 8:27
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    $\begingroup$ I thought that too at one stage, but thats not actually correct. It is indeed V/I. dV/dI is not consistent with devices such as capacitors, and only works for resistors (because they are linear with respect to V and I). This is by convention. $\endgroup$ Apr 8, 2016 at 10:48

4 Answers 4

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Wikipedia states

In quantitative terms, it is the complex ratio of the voltage to the current in an alternating current (AC) circuit.

https://en.m.wikipedia.org/wiki/Electrical_impedance

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Impedance expresses a linear response relation between current and voltage, but is usually considered only for fourier-transformed voltages and currents. I.e., it is a complex function $Z(\omega)$ so that: $$V(\omega) = Z(\omega) I(\omega).$$

The proper real-time analogue would not be the instantaneous ratio of voltage to current, but rather a relation that is nonlocal in time: $$ V(t) = \int_{-\infty}^{+\infty} dt'\, Z(t-t') I(t'). $$ This $Z(t-t')$ function would be the fourier transform of the $Z(\omega)$ function. And so you can say "If a current pulse of duration $dt'$ is applied at time $t'$, what is the voltage at another time $t$." The ratio of the voltage to the current is this nonlocal impedance $Z(t-t') dt'$.

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It is indeed the ratio of $V_{max}$ to $I_{max}$ but only when you are talking about a sinusoidal voltage source and a "linear" component. For example, consider the charging curve of a capacitor:

Capacitor Charge

It would be incorrect to assume the impedance of this circuit is just $V_{max}\over I_{max}$ as this circuit behaves in a little bit of a more complex fashion.

How to think about this

It is probably easiest to understand this with a rotating phasor, for devices such as capacitors, the voltage and time at any instant have to be related by:

$ I = C {dV\over dt}$

You can get the impedance easily by imagining your circuit with a sinusoidal voltage input. If you imagine voltage as a rotating vector in the complex plane, then the current will also be another rotating vector in the complex plane. Here, impedence is the ratio of the lengths of these vectors.

Now since any fourier transform can be thought of as a sum of complex phasors rotating at different frequencies, then you're fine to just consider the case of a single rotating phasor.

Summary

This is therefore not true in general, the instantaneous value of the voltage over the current does not always give you impedence. For example, the charging capacitor above certainly shouldn't have zero impedance as time approaches infinity :)

Hope that helps.

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In electrical circuits impedance $Z = \dfrac {V_{\text{peak}}}{I_{\text{peak}}}$ or $\dfrac {V_{\text{rms}}}{I_{\text{rms}}}$

If $I = I_{\text{peak}} \sin \omega t$ and $V_L = L \dfrac {dI}{dt}$ then $V_L = \omega L I_{\text{peak}}\cos \omega t = V_{\text{peak}} \cos \omega t$

This gives $Z = \omega L$

However look at the ratio of the instantaneous values one gets

$\dfrac {V_L}{I} = \omega L \cot \omega t$

which is certainly not constant.

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