Transfer function and magintude known, unsure how to appoarch

[cybhunter]

Homework Statement

Vs=7200V 0 degrees
ZLine=2Ω+20jΩ
ZLoad=138Ω+460jΩ
XC=unknown capacitance

(omega not given)

Find the capacitive reactance when, place in parallel with ZLoad will make the load appear purely resistive

Homework Equations

conj(Zline) =2Ω-20jΩ (for max power transfer to the load)

Zparallel=[(Zload)(XC)]/[Zload+XC] =Zline

The Attempt at a Solution

I multiply the numberator and denomator by the conjugate of (138+j460-jXC) to get

[[(460XC-j138XC)*(138-i(460-XC)]/(138^2+(460-XC)^2)]=2-20j

when I rearrange, I end up with:
68XC^2+240XC^2-9800XCj+980XC+210447i=230644
as you can see, I end up with a complex number as a solution, which does not make sense

How should I approach these kinds of problems where the transfer function (sans a component value) and the resulting vector is known?

 

[NascentOxygen]

ZLoad=138Ω+460jΩ
XC=unknown capacitance

(omega not given)

Find the capacitive reactance when, place in parallel with ZLoad will make the load appear purely resistive

I don't think maximum power transfer need be brought into this. You are just asked to change 138Ω+460jΩ into 138+j0

If your examiner intended you do more than this, then the question fails to show it, as far as I can see.

 

[rude man]

Is Z_line the characteristic impedance Z₀ of the line, or is it a series impedance at the source? It's a very strange value to be Z₀ of the line. Typical Z₀ of a line is 50 + j0 ohms all the way to 300 + j0 ohms.

BTW Ω is not a parameter, it stands for "Ohms"!

 

[NascentOxygen]

BTW Ω is not a parameter, it stands for "Ohms"!

I believe OP is saying ω is not specified.

 

[rude man]

I believe OP is saying ω is not specified.

I believe you're right!

 

[cybhunter]

after racking my brain, going through practically a ream of paper and almost pulling my hair out for a couple days, apparently someone placed the problem on yahoo answers over a year ago:

http://answers.yahoo.com/question/index?qid=20110523124504AA7phCg

When I product over sum the impedance values (by theoretically shorting the independent voltage source), I end with an interesting conundrum: the Zth value is
2.06816Ω+19.1948jΩ (from (2+20j)*(138+460j)/(140+480j)). By plugging in XC= -19.1948j into the load resistance I end up with a j value that equals:
(-19.1948j)(138+460j)/(138+440.8052j) equals 0.2383-19.95j. When added to the the 2+20j (Line impedance), the resulting value is 2.2383+0.044j (2.239∠1.13) . I may be splitting hairs here but that 1.13 is driving me nuts

 

[rude man]

Something strange here. You need to know the electrical length θ = βx where β = phase constant and x = actual line length.

β = ω/v = ωT = 2π/λ

so you need to know ω and also delay time/unit length T, or just wavelength λ. Or L and C, inductance & capacitance per unit length.

These parameters cannot be deduced from Z0 alone.

 

[NascentOxygen]

after racking my brain, going through practically a ream of paper and almost pulling my hair out for a couple days, apparently someone placed the problem on yahoo answers over a year ago:

http://answers.yahoo.com/question/index?qid=20110523124504AA7phCg

That's handy.

When I product over sum the impedance values (by theoretically shorting the independent voltage source), I end with an interesting conundrum: the Zth value is
2.06816Ω+19.1948jΩ (from (2+20j)*(138+460j)/(140+480j)).

I'm with you there.

By plugging in XC= -19.1948j into the load resistance I end up with a j value that equals:
(-19.1948j)(138+460j)/(138+440.8052j)

That's where you are going wrong. You have determined what X_C must be connected in parallel with the Thevenin impedance, but now you have decided to connect that X_C in parallel with only part of what comprises the Thevenin impedance. Connecting a capacitor in parallel with the load (viz., 138+j460) is not equivalent to connecting that same capacitor in parallel with the Thevenin impedance.