# Series-Parallel Resonance

Discussion in 'Homework Help' started by shub, May 1, 2009.

1. ### shubNew Member

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Hello, I am trying to complete an exercise in my Linear Circuits class. I figured out how to do Series Resonance and I've mostly grasped Parallel resonance (just inverting some numbers), but I am thrown for a loop on this Series-Parallel combination. My textbook is making my head spin!

I have an image of the circuit and an accompaying word doc for the table I am supposed to fill in. If anybody would be willing to complete a sample row using one of frequencies with a description of the forumulas, so I can finish the rest of the table on my own, that would be wonderful.

Basically, we change the frequency of Vs in accordance with the table values. The resistor above the coil is supposed to represent the winding resistance. I don't know, but I am assuming the resistor that is in series with the parallel combination at the start of the circuit is a Thevenin resistance?

I tried making an equivalent parallel circuit of the coil with its resistance as an Leq and Req after I found the Q of the coil. But I didn't know what to do from there.

The word doc:http://www.geocities.com/wrmiro/Presonant.doc

Thanks in advance for any voluteers that come forward.
Bill

Last edited: May 1, 2009
2. ### Miles ProwerMember

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Whenever dealing with circuits that combine series and parallel branches, always remember:

If f= 500Hz,

Xl= jwL= j(2pi X 500)10E-3= j31.52 (Bl= 1 / j31.52= -j0.0317)

Xc= 1 / jwC= 1 / j(2pi X 500 X 100E-9)= -j3.18E3 (Bc= 1 / -j3.18E3= j314.47E-6)

Btot= Bc + Bl= j314.47E-6 - j0.0317= -j0.0314

Xc || Xl= 1 / Btot= 1 / -j0.0314= j31.847

For the total impedance, find the admittance of that series branch:

Z= 4.7 + j31.52

Y= 1 / Z= 1 / (4.7 + j31.52)= 4.628E-3 - j31.036E-3

Bc= jwC= j(2pi X 500 X 100E-9)= j314.159E-6

Ytot= Y + Bc= 4.628E-3 - j31.036E-3 + j314.159E-6= 4.628E-3 - j30.722E-3

Zt= 1 / Ytot= 1 / (4.268E-3 - j30.722E-3)= 4.44 + j31.934

Output voltage is simply a voltage divider problem here:

Vo= ViZ / (Rs + Z)=

80(4.44 + j31.934) / (1000 + 4.44 + j31.934)=

(355.2 + j2554.72) / (1004.44 + j31.934)= 1.34 + j2.53 Vrms

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I think my brain just exploded.

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5. ### Hero999Banned

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It looks harder than it really is.

The above calculations are fairly simple, expecially if your calculator can do complex numbers.

6. ### Miles ProwerMember

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Any decent scientific calculator these days should be able to do complex arithmetic, and it takes a helluvalot longer to type up the post than it does to bang out the answers. Still helps to actually know how complex arithmetic works for a better understanding of what it is that your calculator is telling you.

7. ### MikebitsWell-Known Member

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But I can't find j on my calculator

8. ### shubNew Member

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Thank you Miles. This Assignment is due Monday, I am at work right now, but I will be following your calculations closely this evening. I highly appreciate the breakdown. I wasn't familiar with the w variable (but have realized it is 2pif) and I also wasn't sure what the B letter was for, does this mean for "Branch B"? I'll update when I have gotten a few frequency rows done on the chart.

Did you notice on that word document that there is a place for Vr and Vtank? Would the Vout you calculated be the Vr? Would that mean I need a separate calculation for the Vtank? I don't understand how to measure a tank- is it the voltage sum of the cap and the coil in that parallel branch?

Last edited: May 3, 2009
9. ### Miles ProwerMember

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No, 'B' is the symbol for susceptance, which is the reciprocal of reactance, whose symbol is 'X'.

B= 1 / X or X= 1 / B

It's the imaginary part of the admittance: Y= G + jB as X is the imaginary part of impedance: Z= R + jX, and, of course:

Y= 1 / Z

Z= 1 / Y

That was simply sloppy. Vo is Vtank. They want the voltage appearing across the parallel RLC part of the circuit. Figuring that is simply a matter of applying the usual voltage divider rule. As for "Vr", I don't know what "R" they're referring to. However, you can easily find it once the voltages are known.

10. ### Miles ProwerMember

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All depends on what brand of calculator you have. Mine doesn't have a 'j' either. It uses the ordered pair format (Re,Im) (i.e. like FORTRAN). Other calcs might use the symbol 'i' instead of 'j'. In electronics, j= sqrt(-1) since 'i' was already taken for current.

11. ### shubNew Member

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I've got half the figures into the spreadsheet and it's going great. Is the resonant frequency for this circuit, the resonant frequency of the tank circuit? I am not sure how to do it because of that 1k series resistor... I think my brain is gettin tired...

12. ### Miles ProwerMember

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^
|
|

Yes, the resonant frequency is that of the "tank".

w= (LC)^-.5

f= 5.032KHz

13. ### shubNew Member

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How did you inverse this Complex number. I can't get my calculator to do this. Nor can I get excel to do it, it is very finicky with it's COMPLEX number function.

14. ### Miles ProwerMember

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First off, for any complex number C, there is what's called the "conjugate", C*. This is done by reversing the sign of the imaginary part.

C= a + jb D= c - jd

C*= a - jb D*= c + jd

To do the division with the complex numbers in Cartesian form, simply multiply the numerator and denominator by the conjugate of the denominator. (Don't forget that 1= 1 + j0)

A= a + jb
B= c + jd

A/B= [(a + jb)(c - jd)] / [(c + jd)(c - jd)]

You can also convert to the Steinmetz form:

R= sqrt(a^2 + b^2)

Theta= arctan(b/a)

C= R /_ Theta

A= R1 /_ Theta1
B= R2/_ Theta2

A/B= R1/R2 /_ Theta1 - Theta2

Leave it that way, or convert back to the Cartesian:

a= Rcos(Theta)
b= jRsin(Theta)

To get up to speed with complex number arithmetic, go Here, scroll about half way down the page, and start with "Complex Addition". Be sure to follow the links at the end of the articles for the next article.

Last edited: May 5, 2009