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Does the voltage division rule work for reactive loads?

https://en.wikipedia.org/wiki/Voltage_divider

It says the voltage ration H of the circuit is Z2/Z1+Z2 which looks the same as replacing the simple resistive voltage division rule R2/R1+R2 with Z's.  But is impedance at a given frequency a simple number?

Given a device with constant output impedance of 10 ohms, can its effect on frequency response of attached headphones be then simplified as follows?

Let Z1 be the impedance of the earphones at 1kHz and Zf be the impedance at a given other frequency in question.
V1 = Z1/Z1+10
Vf = Zf/Zf+10
Vf/V1 = Zf(Z1+10)/Z1(Zf+10)
Frequency response at f/ Frequency response at 1kHz difference in dB= 10 * log {[Zf(Z1+10)/Z1(Zf+10)]^2} dB

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Re: Does the voltage division rule work for reactive loads?

Reply #1
https://en.wikipedia.org/wiki/Voltage_divider

It says the voltage ration H of the circuit is Z2/Z1+Z2 which looks the same as replacing the simple resistive voltage division rule R2/R1+R2 with Z's.  But is impedance at a given frequency a simple number?

Impedance of a reactive load is a complex number, but for voltage transfer you only care about the magnitude.  The phase will introduce problems with power transfer (since the reactive component of a load cannot absorb power), but for a voltage divider that doesn't matter.  You can simply take the magnitude of the impedance in ohms for a given frequency and use that. 

Let Z1 be the impedance of the earphones at 1kHz and Zf be the impedance at a given other frequency in question.
V1 = Z1/Z1+10
Vf = Zf/Zf+10
Vf/V1 = Zf(Z1+10)/Z1(Zf+10)
Frequency response at f/ Frequency response at 1kHz difference in dB= 10 * log {[Zf(Z1+10)/Z1(Zf+10)]^2} dB

You can calculate the difference in voltage transfer that way.  That won't exactly be the frequency response (since the sensitivity of the headphones may also vary with frequency), but you have the right idea in terms of thinking about impedance.

 
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