Impedance of different devices (derivations)

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Impedance defines the ratio of AC voltage amplitude to AC current amplitude, as well as the phase by which the voltage leads the current (even though this may sometimes have a negative value). Although the idea can be extended to define the relationship between the voltage and current of any arbitrary signal, this article deals with only sinusoidal signals, since any arbitrary signal can be approximated as a sum of sinusoids through Fourier Analysis. What follows below is a derivation of impedance for each of the three basic circuit elements, the resistor, the capacitor, and the inductor.

1 Resistor
2 Capacitor
3 Inductor
4 See also

[edit] Resistor

For a resistor, we have the relation:

$v_\mathrm{R} \left( t \right) = {i_\mathrm{R} \left( t \right)}R$

Which is simply another way of stating Ohm's Law.

Considering the voltage signal to be

$v_\mathrm{R}(t) = V_p \sin(\omega t) \,$

It follows that

$\frac{v_\mathrm{R} \left( t \right)}{i_\mathrm{R} \left( t \right)} = \frac{V_p \sin(\omega t)}{I_p \sin \left( \omega t \right)} = R$

Which tells us that the ratio of AC voltage amplitude to AC current amplitude across a resistor is $R$ , and that the AC voltage leads the AC current across a resistor by 0 degrees.

This result is commonly expressed as

$Z_\mathrm{resistor} = R \,$

[edit] Capacitor

For a capacitor, we have the relation:

$i_\mathrm{C}(t) = C \frac{dv_\mathrm{C}(t)}{dt}$

Considering the voltage signal to be

$v_\mathrm{C}(t) = V_p \sin(\omega t) \,$

It follows that

$\frac{dv_\mathrm{C}(t)}{dt} = \omega V_p \cos \left( \omega t \right)$

And thus

$\frac{v_\mathrm{C} \left( t \right)}{i_\mathrm{C} \left( t \right)} = \frac{V_p \sin(\omega t)}{\omega V_p C \cos \left( \omega t \right)}= \frac{\sin(\omega t)}{\omega C \sin \left( \omega t + \frac{\pi}{2}\right)}$

Which tells us that the ratio of AC voltage amplitude to AC current amplitude across a capacitor is $\frac{1}{\omega C}$ , and that the AC voltage leads the AC current across a capacitor by -90 degrees (or the AC current leads the AC voltage across a capacitor by 90 degrees).

This result is commonly expressed in polar form, as

$Z_\mathrm{capacitor} = \frac{1}{\omega C} e^{-j \frac{\pi}{2}}$

or, more simply, through the help of Euler's formula, as

$Z_\mathrm{capacitor} = \frac{1}{j \omega C}$

[edit] Inductor

For the inductor, we have:

$v_\mathrm{L}(t) = L \frac{di_\mathrm{L}(t)}{dt}$

This time, considering the current signal to be

$i_\mathrm{L}(t) = V_p \sin(\omega t) \,$

It follows that

$\frac{di_\mathrm{L}(t)}{dt} = \omega V_p \cos \left( \omega t \right)$

And thus

$\frac{v_\mathrm{L} \left( t \right)}{i_\mathrm{L} \left( t \right)} = \frac{\omega V_p L \cos(\omega t)}{V_p \sin \left( \omega t \right)}= \frac{\omega L \sin \left( \omega t + \frac{\pi}{2}\right)}{\sin(\omega t)}$

Which tells us that the ratio of AC voltage amplitude to AC current amplitude across an inductor is $ω L$ , and that the AC voltage leads the AC current across an inductor by 90 degrees.

This result is commonly expressed in polar form, as

$Z_\mathrm{inductor} = \omega L e^{j \frac{\pi}{2}}$