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This could easily be merged with the article on Duality_(electrical_circuits). The current article is more detailed but less comprehensive than that one. 84.227.254.143 ( talk) 06:28, 30 March 2014 (UTC)
Jordsan, what you have written, "the dual of Z is the admittance Z'=Y", is not correct, or at least adds no meaning. Z' is not an admittance, it is an impedance. The admittance that is numerically equal to Z' has an impedance of Z. Thus, it is equivalent to Z, not the dual of it. Take the case of an inductance, L. The impedance of this one-port is Z=jωL. The dual impedance of this is Z'=1/(jωL), which is in the form of the impedance of a capacitor. The capacitance of this capacitor is numerically equal to L. The admittance of this dual circuit is Y'=1/Z'=jωL. Y' is still representing a capacitor and is still the dual of Z. Thus, the admittance Y' (the dual of Z) is numerically equal to the impedance Z. Spinning Spark 14:37, 29 September 2015 (UTC) @ Jordsan: Spinning Spark 14:38, 29 September 2015 (UTC)
@Spinningspark edits are incorrect and confusing. They are also not consistent.
In the beginning of the page, taking all equations together, it clearly states Z'=1/Z=Y=I/V and Y'=1/Y=Z=V/I.
In general, introducing a "reverse impedance and admittance" is silly when we already have the relation Z=1/Y. The whole page should just use impedance and admittance as is standard in electrical engineering. That would also fit with /info/en/?search=Duality_(electrical_circuits)
In dual circuits:
V <-> I
Z <-> Y
etc.
Now, aside from this unnecessary algebraic inverse, there are clear physical errors in the table, specifically the figure captions. One cannot simply equate R=G and Z=Y, no matter how you arrange it. By the pages own definition at the top, Z=V/I and Y=I/V, impedance and admittance cannot be the same, even if you normalize and scale it. While the (inverse) impedances are correct, the figure captions for the dual passive elements are imprecise/misleading and should simply read:
Resistor R <-> Conductor G ... G!=R but is actually the inverse. A 5 Ohm resistor must be depicted by a 0.2 S conductor as its dual, not 5 S as the current caption suggests. Both value and unit of the element change! Conductor G <-> Resistor R ... same as above, 1 mS conductor is dual of 100 Ohm resistor
Inductance L <-> Capacitance C ... here the numerical value stays the same, but the unit still changes. Capacitance C <-> Inductance L ... a 10 mF capacitor becomes a 10 mH inductor, so C!=L since a physical variable is value + unit!!
And now come the serious errors with the series and parallel connections:
1st caption says: Y=Z1+Z2 (wrong!) while Z'=1/(Z1+Z2) (=Y)
Both caption and figure are actually wrong, since a series connection of impedances becomes are parallel connection of admittances! But Y=Y1||Y2=Y1Y2/(Y1+Y2) = 1/(Y1/Y1Y2 + Y2/Y1Y2) = 1/(1/Y2 + 1/Y1) = 1/(Z1+Z2), so the dual impedance is not wrong.
2nd figure and caption says: 1/Y=1/Z1+1/Z2 (wrong!)
It's again parallel imp. become series admittances with NOT the same values (normalized/scaled or not). Y=Z'=Y1+Y2=1/Z1+1/Z2
Simple proof with DC and 2 series resistors only:
R1=Z1=400 Ohm
R2=Z2=100 Ohm
Z1+Z2 (series impedance)
input voltage/drop source 10 V DC
Clearly the voltages split into 8 V over R1 and 2 V over R2 with a current of 20 mA flowing.
In the dual circuit, current and voltage shift role (voltage "flows" and current "drops").
Taking the (incorrect) approach of G=R and going to a parallel of the impedances as the pictures and captions suggest:
G1=Z1=400 S
G2=Z2=100 S
Z1||Z2
input "flow" source 10 V DC
Voltage now "flowing" 2 V into G1 and 8 V into G2, with a parallel current "drop" of 800 A. Hmm, that doesn't seem match up, no matter how you put or scale it ...
However, if we do it correctly with G=1/R and parallel admittances:
G1=Y1=2.5 mS
G2=Y2=10 mS
Y1||Y2 (parallel admittance)
input "flow" source 10 V DC
Voltage "flow" divided in 8 V into G1 and 2 V into G2, with parallel current "drop" of 20 mA. Oh look at that, a duality!
Similarly, the active elements in the end of the table are imprecise. One cannot state V=I, which is only numerically correct, not in the units.
And before the normalization/scaling is pointed at, it doesn't make sense either, or rather it is physically and mathematically incorrect. Beginning of page says Z'=1/Z. That is a pretty clear cut formula. Unit of impedance is Ohms, [Z] = 1 Ohm. So [Z']= 1/Ohm.
Then the scaling formula says Z'/R0 = R0/Z. So in terms of units we have 1/(Ohm*Ohm) = 1 Ohm/Ohm ... which makes the equation physically invalid. Scaling in ANY equation means multiplying BOTH sides with a scaling value. One can't just multiply one side and divide the other ... that's elementary algebra. — Preceding unsigned comment added by 192.38.90.64 ( talk) 12:00, 24 August 2020 (UTC)
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This could easily be merged with the article on Duality_(electrical_circuits). The current article is more detailed but less comprehensive than that one. 84.227.254.143 ( talk) 06:28, 30 March 2014 (UTC)
Jordsan, what you have written, "the dual of Z is the admittance Z'=Y", is not correct, or at least adds no meaning. Z' is not an admittance, it is an impedance. The admittance that is numerically equal to Z' has an impedance of Z. Thus, it is equivalent to Z, not the dual of it. Take the case of an inductance, L. The impedance of this one-port is Z=jωL. The dual impedance of this is Z'=1/(jωL), which is in the form of the impedance of a capacitor. The capacitance of this capacitor is numerically equal to L. The admittance of this dual circuit is Y'=1/Z'=jωL. Y' is still representing a capacitor and is still the dual of Z. Thus, the admittance Y' (the dual of Z) is numerically equal to the impedance Z. Spinning Spark 14:37, 29 September 2015 (UTC) @ Jordsan: Spinning Spark 14:38, 29 September 2015 (UTC)
@Spinningspark edits are incorrect and confusing. They are also not consistent.
In the beginning of the page, taking all equations together, it clearly states Z'=1/Z=Y=I/V and Y'=1/Y=Z=V/I.
In general, introducing a "reverse impedance and admittance" is silly when we already have the relation Z=1/Y. The whole page should just use impedance and admittance as is standard in electrical engineering. That would also fit with /info/en/?search=Duality_(electrical_circuits)
In dual circuits:
V <-> I
Z <-> Y
etc.
Now, aside from this unnecessary algebraic inverse, there are clear physical errors in the table, specifically the figure captions. One cannot simply equate R=G and Z=Y, no matter how you arrange it. By the pages own definition at the top, Z=V/I and Y=I/V, impedance and admittance cannot be the same, even if you normalize and scale it. While the (inverse) impedances are correct, the figure captions for the dual passive elements are imprecise/misleading and should simply read:
Resistor R <-> Conductor G ... G!=R but is actually the inverse. A 5 Ohm resistor must be depicted by a 0.2 S conductor as its dual, not 5 S as the current caption suggests. Both value and unit of the element change! Conductor G <-> Resistor R ... same as above, 1 mS conductor is dual of 100 Ohm resistor
Inductance L <-> Capacitance C ... here the numerical value stays the same, but the unit still changes. Capacitance C <-> Inductance L ... a 10 mF capacitor becomes a 10 mH inductor, so C!=L since a physical variable is value + unit!!
And now come the serious errors with the series and parallel connections:
1st caption says: Y=Z1+Z2 (wrong!) while Z'=1/(Z1+Z2) (=Y)
Both caption and figure are actually wrong, since a series connection of impedances becomes are parallel connection of admittances! But Y=Y1||Y2=Y1Y2/(Y1+Y2) = 1/(Y1/Y1Y2 + Y2/Y1Y2) = 1/(1/Y2 + 1/Y1) = 1/(Z1+Z2), so the dual impedance is not wrong.
2nd figure and caption says: 1/Y=1/Z1+1/Z2 (wrong!)
It's again parallel imp. become series admittances with NOT the same values (normalized/scaled or not). Y=Z'=Y1+Y2=1/Z1+1/Z2
Simple proof with DC and 2 series resistors only:
R1=Z1=400 Ohm
R2=Z2=100 Ohm
Z1+Z2 (series impedance)
input voltage/drop source 10 V DC
Clearly the voltages split into 8 V over R1 and 2 V over R2 with a current of 20 mA flowing.
In the dual circuit, current and voltage shift role (voltage "flows" and current "drops").
Taking the (incorrect) approach of G=R and going to a parallel of the impedances as the pictures and captions suggest:
G1=Z1=400 S
G2=Z2=100 S
Z1||Z2
input "flow" source 10 V DC
Voltage now "flowing" 2 V into G1 and 8 V into G2, with a parallel current "drop" of 800 A. Hmm, that doesn't seem match up, no matter how you put or scale it ...
However, if we do it correctly with G=1/R and parallel admittances:
G1=Y1=2.5 mS
G2=Y2=10 mS
Y1||Y2 (parallel admittance)
input "flow" source 10 V DC
Voltage "flow" divided in 8 V into G1 and 2 V into G2, with parallel current "drop" of 20 mA. Oh look at that, a duality!
Similarly, the active elements in the end of the table are imprecise. One cannot state V=I, which is only numerically correct, not in the units.
And before the normalization/scaling is pointed at, it doesn't make sense either, or rather it is physically and mathematically incorrect. Beginning of page says Z'=1/Z. That is a pretty clear cut formula. Unit of impedance is Ohms, [Z] = 1 Ohm. So [Z']= 1/Ohm.
Then the scaling formula says Z'/R0 = R0/Z. So in terms of units we have 1/(Ohm*Ohm) = 1 Ohm/Ohm ... which makes the equation physically invalid. Scaling in ANY equation means multiplying BOTH sides with a scaling value. One can't just multiply one side and divide the other ... that's elementary algebra. — Preceding unsigned comment added by 192.38.90.64 ( talk) 12:00, 24 August 2020 (UTC)