Easy question about circuits.

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Easy question about circuits.

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just for reference, R load = Rt, for max power transfer. I calculated it, but not getting the right answer. it should be 0.8mF.
the conjugate surd would allow maximum, so you need 10-j20 for a load. Find a value of C that has a reactance of 20 Ohms at the operating frequency.

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What are you using for frequency?
20
sorry i mean 50.
Well I thought (10 + 20j ) || -j/50C = 30.
Cause i'd use the R thevenin.
the answer the gave was 0.8mF. and when i plug that into the equation I typed above, i got 50. :/
hey @radar can you tell me what is Z here i know it is used for power factor but what is this 10+j20
It represents a real (resistor) value of 10 and a series inductive reactance of 20 Ohms.
ok if it would have been a capacitor then Z=10-20j right?
yes
because the equation is -J/wc
what is -J/wc and inductor lags behind the current wrt resistor so shouldn't it be -(negative) and opposite for capacitor
The value of the capacitor will cancel out the inductive reactance allow matching for maximum transfer. Yes Z=10 -j20 would represent capacitance.
yeah i get that for the max power but um kinda confused with the signs
So wait, sorry, back to my question, where am I going wrong?
The conjugate surd for a int impedance of 10 + j20 would be 10 - j20 just like when you rationalizing a denominator.
What I am having a problem converting the (50t) to a frequency, I don't see pi there so for a frequency I am coming up with 8 Hz which is way to low, I suspect that frequency is suppose to be 50 Hz
hmm, i think there's a small confusion. So, we're not multiplying by the frequency. But rather 2pi(frequency) = 50.
In other words, the inductor's value would have been 2/5.
when we have the equation Acos(wt+b), the value for an inductor would be jwL and for capacitor it'd be -j/wC
Hope that helps.
I still come up with a unreasonable omega and then frequency. Need to do some research. Presently I am using 50 as omega.
yes, that is correct.
so in other words your value for the capicatance impedance is -J/50C. We just have to solve for C.
That was my plan, -j/(50C) = 20 Giving me C= .001 Farad Much larger than the .8 Micro farad that you say was the approved solution.
That is much larger 1,000 micro farad vs .8 microfarad.
The internal impedance provided by the 10 + j20 is 22.36 at 63 degrees ohms. Since the load is 30 Ohms resistor a capacitor of a value will be needed when connected in parallel that will reduce the impedance to 22.36 with a -63 degrees. I am going to have to put on a bigger thinking cap to solve that. I will try, but right now I have other things to do. Good luck with it. I don't think it is so simple.
yes, i thought it was more complicated than the professor made it to be. He said, this should be a minute problem... Because this is the first question on the practice final... :/ Being a Comp Eng major is too hard.
|dw:1433647505784:dw|
@ybarrap Wait, i'm looking at that diagram, but i'm not able to derive the solid answer, what should i be looking at in there?
Here's a simulation of the circuit. I did a sweep of various C to show that the lower the value of C, the capacitance, the more power is seen across \(R_L\). I used a phase shifter to simulate the impedance \(Z=10+20j\). Since this phase shifter is not a function of frequency, there is no impedance matching that can be done using it. So, we make the Capacitance as small as possible, to make the magnitude of the capacitor's reactance as large as possible to minimize the current through it. I think that the impedance Z was actually meant to be frequency-dependent.
Here's a better picture -
1 Attachment
hmm, max power transfer requires taht R load = R total. So, Z ccan actually come to a one solid answer, and would not be frequency dependent. Well it would be frequency dependant. But we're assuming just for that frequency
Yes, rload = rsource typically, but here the rsource has no dependence on frequency, it's just z=10+20j, a phase-shift with some attenuation. I think the problem is that your prof meant for z to be a function of frequency -- not sure.
The problem would be more straightforward if \(z=10+jwL\) or something like that, with L=20. Then we would set \(X_C||R_L=z^*\) for maximum power transfer.
well the answer he posted was 0.8mF. so, i'm guesing he derived an acutal value.
Here is a link that discusses Z matching. I noticed the first example using the L network to obtain maximum power transfer. It does not show how the equations were derived, but shows a simple method: http://electronicdesign.com/communications/back-basics-impedance-matching-part-2 Here is the circuit:|dw:1433682448727:dw| With that value of Q and using the equation \[X _{c}=R _{L}/Q=15\sqrt{2}=21.2\]That is in Ohms. Now if that is correct using the value of omega you can now calculate the value of C.

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