Transfer function and magintude known, unsure how to appoarch

1. Jul 7, 2012

cybhunter

1. The problem statement, all variables and given/known data

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

(omega not given)

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

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

3. 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?

2. Jul 8, 2012

Staff: Mentor

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.

3. Jul 8, 2012

rude man

Is Zline 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"!

4. Jul 8, 2012

Staff: Mentor

I believe OP is saying ω is not specified.

5. Jul 8, 2012

rude man

I believe you're right!

6. Jul 9, 2012

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:

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

7. Jul 9, 2012

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.

8. Jul 9, 2012

Staff: Mentor

That's handy.
I'm with you there.
That's where you are going wrong. You have determined what XC must be connected in parallel with the Thevenin impedance, but now you have decided to connect that XC 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.