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Curry
 one year ago
Easy question about circuits.
Curry
 one year ago
Easy question about circuits.

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Curry
 one year ago
Best ResponseYou've already chosen the best response.0just for reference, R load = Rt, for max power transfer. I calculated it, but not getting the right answer. it should be 0.8mF.

radar
 one year ago
Best ResponseYou've already chosen the best response.7the conjugate surd would allow maximum, so you need 10j20 for a load. Find a value of C that has a reactance of 20 Ohms at the operating frequency.

radar
 one year ago
Best ResponseYou've already chosen the best response.7What are you using for frequency?

Curry
 one year ago
Best ResponseYou've already chosen the best response.0Well I thought (10 + 20j )  j/50C = 30.

Curry
 one year ago
Best ResponseYou've already chosen the best response.0Cause i'd use the R thevenin.

Curry
 one year ago
Best ResponseYou've already chosen the best response.0the answer the gave was 0.8mF. and when i plug that into the equation I typed above, i got 50. :/

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0hey @radar can you tell me what is Z here i know it is used for power factor but what is this 10+j20

radar
 one year ago
Best ResponseYou've already chosen the best response.7It represents a real (resistor) value of 10 and a series inductive reactance of 20 Ohms.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok if it would have been a capacitor then Z=1020j right?

Curry
 one year ago
Best ResponseYou've already chosen the best response.0because the equation is J/wc

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0what is J/wc and inductor lags behind the current wrt resistor so shouldn't it be (negative) and opposite for capacitor

radar
 one year ago
Best ResponseYou've already chosen the best response.7The value of the capacitor will cancel out the inductive reactance allow matching for maximum transfer. Yes Z=10 j20 would represent capacitance.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeah i get that for the max power but um kinda confused with the signs

Curry
 one year ago
Best ResponseYou've already chosen the best response.0So wait, sorry, back to my question, where am I going wrong?

radar
 one year ago
Best ResponseYou've already chosen the best response.7The conjugate surd for a int impedance of 10 + j20 would be 10  j20 just like when you rationalizing a denominator.

radar
 one year ago
Best ResponseYou've already chosen the best response.7What 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

Curry
 one year ago
Best ResponseYou've already chosen the best response.0hmm, i think there's a small confusion. So, we're not multiplying by the frequency. But rather 2pi(frequency) = 50.

Curry
 one year ago
Best ResponseYou've already chosen the best response.0In other words, the inductor's value would have been 2/5.

Curry
 one year ago
Best ResponseYou've already chosen the best response.0when we have the equation Acos(wt+b), the value for an inductor would be jwL and for capacitor it'd be j/wC

radar
 one year ago
Best ResponseYou've already chosen the best response.7I still come up with a unreasonable omega and then frequency. Need to do some research. Presently I am using 50 as omega.

Curry
 one year ago
Best ResponseYou've already chosen the best response.0so in other words your value for the capicatance impedance is J/50C. We just have to solve for C.

radar
 one year ago
Best ResponseYou've already chosen the best response.7That 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.

radar
 one year ago
Best ResponseYou've already chosen the best response.7That is much larger 1,000 micro farad vs .8 microfarad.

radar
 one year ago
Best ResponseYou've already chosen the best response.7The 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.

Curry
 one year ago
Best ResponseYou've already chosen the best response.0yes, 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.

Curry
 one year ago
Best ResponseYou've already chosen the best response.0@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?

ybarrap
 one year ago
Best ResponseYou've already chosen the best response.0Here'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 frequencydependent.

ybarrap
 one year ago
Best ResponseYou've already chosen the best response.0Here's a better picture 

Curry
 one year ago
Best ResponseYou've already chosen the best response.0hmm, 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

ybarrap
 one year ago
Best ResponseYou've already chosen the best response.0Yes, rload = rsource typically, but here the rsource has no dependence on frequency, it's just z=10+20j, a phaseshift with some attenuation. I think the problem is that your prof meant for z to be a function of frequency  not sure.

ybarrap
 one year ago
Best ResponseYou've already chosen the best response.0The problem would be more straightforward if \(z=10+jwL\) or something like that, with L=20. Then we would set \(X_CR_L=z^*\) for maximum power transfer.

Curry
 one year ago
Best ResponseYou've already chosen the best response.0well the answer he posted was 0.8mF. so, i'm guesing he derived an acutal value.

radar
 one year ago
Best ResponseYou've already chosen the best response.7Here 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/backbasicsimpedancematchingpart2 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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