Admittance, Conductance and Susceptance

[k2keylargo]

I'm confused by Camara's EERM. I've always blown off admittance because I can always take the recripocal of R and X and get them if a problem requires. However, I'm looking at the formulas in chapter 27, AC circuit fundamentals and it gives the formulas:

G=1/R

B=1/X These I use, as I said above.

Then it goes on to say

Y=G+jB = R/(R^2 + X^2) -j [X/(R^2 +X^2)]

Z= R + jX = G/(G^2 + B^2) -j [b/(G^2 + B^2)]

I'm confused. These two sets of equations don't jive. Can anyone explain? :blink:

 

[mudpuppy]

I agree that something doesn't look right. Plugging in numbers for R and X, I do not get the same result from the equations. Have you checked "the other board"'s errata? I will try to remember to check my EERM at work tomorrow.

Personally, I don't worry too much about it; I take Y= 1/Z and let the calculator deal with the complex arithmetic.

 

[k2keylargo]

I did check the errata - nothing there. Wikipedia gives the same equations for Y in terms of R and X, but Wiki's explanation doesn't start with G = 1/R and B = 1/X.

Please help.... I tried a simple problem to check myself - trying to calculate:

Admittance of a 30 ohm resistor in parallel with a 4 H inductor, with freq.= 10 radians/sec, so X = 40.

using R = 30 and X = 4 x 10 = 40,

calculating Y = 1/Z = 1/R -j(1/X)

adding 1/30 - j (1/40) by converting to polar,

I get that Y = 1/Z = 41.7 x 10^-3 angle is -36.9 degrees

If I use the formula Y = R/(R^2 + X^2) -j[X/(R^2 = X^2)] I get that Y = 20 x 10^-3 angle is -53.1 degrees.

Can anyone tell me what I'm doing wrong here? I'm really pulling my hair out staring at this.... :smileyballs:

 

[mudpuppy]

Ok I am starting to see the problem. The equations G = 1/R and B = 1/X are not always true. The correct equations are given on the wikipedia page for admittance, specifically, G = Real(Y) = R/(R2+X2)). If X is zero, then this simplifies to G = 1/R. The equation for B is similar. 1/R and 1/X work fine if you are working with pure resistances or pure reactances, but they don't work when you try to combine them into an impedance/admittance.

This is a subtle point, but an impedance, Z is a complex number in and of itself so you can't substitute in 1/R and 1/X for G and B without taking the complex arithmetic into account (and I'm too rusty in the fundamentals of complex numbers to explain it better).

I did check the errata - nothing there. Wikipedia gives the same equations for Y in terms of R and X, but Wiki's explanation doesn't start with G = 1/R and B = 1/X.
Please help.... I tried a simple problem to check myself - trying to calculate:

Admittance of a 30 ohm resistor in parallel with a 4 H inductor, with freq.= 10 radians/sec, so X = 40.

using R = 30 and X = 4 x 10 = 40,

calculating Y = 1/Z = 1/R -j(1/X)

adding 1/30 - j (1/40) by converting to polar,

I get that Y = 1/Z = 41.7 x 10^-3 angle is -36.9 degrees

If I use the formula Y = R/(R^2 + X^2) -j[X/(R^2 = X^2)] I get that Y = 20 x 10^-3 angle is -53.1 degrees.

Can anyone tell me what I'm doing wrong here? I'm really pulling my hair out staring at this.... :smileyballs:

The second one is correct. The problem in your first method is in this equation: Y = 1/Z = 1/R -j(1/X). This should be Y = 1/Z = 1/(R+jX).

In other words, 1/(R+jX) is not the same as 1/R - j/X.

I hope this helps; I'm having a hard time putting the point into words.

Also, I'm not sure if you misspoke, but if the problem is an inductor in parallel with a resistor you can't just directly add the resistance and reactance (that would be true for a resistor and inductor in series). It should be Z1*Z2/(Z1+Z2).

 

[k2keylargo]

mudpuppy,

thanks for the explanation, but I'm still puzzled :reading:

Is it true that we can put everything in terms of impedances, that is

resistors = R

inductors = jwL

capacitors = -j/wC = 1/jwC

then we can add any impedance in series by just Ztotal = Z1 + Z2 + Z3, and we can add in parallel by using Ztotal = 1/[1/Z1 + 1/Z2 + 1/Z3],

regardless if Z1,Z2 and Z3 are resistors, inductors or capacitors?

If this is true,

let Z(total) = Zt,

Z(resistor) = Zr

Z(inductor) = Zi

then Zt = ZrZi/(Zr + Zi)

this should be the same as

Zt = 1/[1/Zr + 1/Zi] shouldn't it? To get from one to the other, this is my logic:

if you take the denominator in the above formula DEN= [1/Zr + 1/Zi]

simplify it by multiplying the first term by Zi/Zi, and multiplying the second term by Zr/Zr,

then the denominator becomes:

DEN = [(Zi/ZiZr) + (Zr/ZiZr)]

so now you can combine these two terms and the denominator becomes:

DEN = (Zi + Zr)/ZiZr

and going back to the original equation = Zt, (the numerator was just 1)

so Zt = the inverse of the denominator = ZiZr/(Zi + Zr) which is the other equation!

This all doesn't fit with the formulas for Y that I originally posted ... and I still don't understand why.... I'm thinking that the formula for Y came about because a complex conjugate was used to multiply a denominator.... I'm trying to think about this as I write, but I'm starting to burn out... :hung-037:

 

[mudpuppy]

mudpuppy,
thanks for the explanation, but I'm still puzzled :reading:

Is it true that we can put everything in terms of impedances, that is

resistors = R

inductors = jwL

capacitors = -j/wC = 1/jwC

then we can add any impedance in series by just Ztotal = Z1 + Z2 + Z3, and we can add in parallel by using Ztotal = 1/[1/Z1 + 1/Z2 + 1/Z3],

regardless if Z1,Z2 and Z3 are resistors, inductors or capacitors?

Yes

If this is true,
let Z(total) = Zt,

Z(resistor) = Zr

Z(inductor) = Zi

then Zt = ZrZi/(Zr + Zi)

this should be the same as

Zt = 1/[1/Zr + 1/Zi] shouldn't it?

Yes

To get from one to the other, this is my logic:
if you take the denominator in the above formula DEN= [1/Zr + 1/Zi]

simplify it by multiplying the first term by Zi/Zi, and multiplying the second term by Zr/Zr,

then the denominator becomes:

DEN = [(Zi/ZiZr) + (Zr/ZiZr)]

so now you can combine these two terms and the denominator becomes:

DEN = (Zi + Zr)/ZiZr

and going back to the original equation = Zt, (the numerator was just 1)

so Zt = the inverse of the denominator = ZiZr/(Zi + Zr) which is the other equation!

This all doesn't fit with the formulas for Y that I originally posted

This is where I disagree. . . I think it does fit with the equations for Y.

In your first example with the 30 ohm resistor and 40 ohm inductor in paralell:

Zt = Zr*Zi/(Zr+Zi)

Zt = 30*j40/(30+j40) = 19.2 + j14.4

Yt = 1/Zt = 1/(19.2 + j14.4) = 0.0333 - j0.025

Using the other method:

Zt = 1/(1/Zr + 1/Zi)

Zt = 1/(1/30 + 1/j40)

Yt = 1/Zt = 1/1/(1/30 + 1/j40) = 1/30 + 1/j40 = 1/30 - j1/40 = 0.0333 - j0.025

It looks like where the problem lies is the equation from the book assumes the R and X are in series, so let's try this again:

Y = R/(R2+X2) - jX/(R2+X2)

So for the total admittance:

Yt = Rt/(Rt2+Xt2) - jXt/(Rt2+Xt2)

So first we have to calculate Zt to get Rt and Xt:

Zt = Rt + jXt = Zr*Zi/(Zr + Zi) = (as calculated before) 19.2 + j14.4

So Rt = 19.2, Xt = 14.4

Yt = 19.2/(19.22+14.42) - j14.4/(19.22+14.42)

Yt = .0333 - j0.025

Not to be confusing, but yet another method: with admittances in parallel, you can add the admittances directly:

Yt = Y1 + Y2

Yt = 1/Zr + 1/Zi

Yt = 1/30 + 1/j40

Yt = .0333 - j0.025

Let me know if any of this makes sense.

 

[k2keylargo]

YOU'RE RIGHT!!!!! arty-smiley-048: The equations given in the book are for SERIES R AND X!!!! hOLY S--T! I didn't recognize that. I can't believe I kept looking at the formula and didn't see that.

Y = R/(R^2 + x^2) -jX/(R^2 + x^2) FOR SERIES R AND X.

The book starts the paragraph off with "Admittance is useful in analyzing parallel circuits..." I kept assuming the formula for Y was then for a parallel R and X.

Thanks for all your help here, mudpuppy! :thankyou:

 

[benbo]

OK. I think that's more than enough algebra for one day.

 

[Guest]

^^^ What's wrong with algebra??!?!

JR

 

[mudpuppy]

Thanks for all your help here, mudpuppy!

Glad I could help!

OK. I think that's more than enough algebra for one day.

 

[3dB down]

This is the kind of support that makes this site such a valuable asset. Mentoring cannot be under estimated especially when an engineer is out there pretty much on his or her own. Thanks.

 

[Guest]

^^^ How did you end up in Nebraska when you hail from California?

JR

 

[Wolverine]

I'm guessing that, like many people I know, he sold his 700 sqf bungalow in Yorba Linda, moved east, and bought a county in Nebraska. Now he decides his own tax rate. Am I close?