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Curry

  • one year ago

Easy question about circuits.

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  1. Curry
    • one year ago
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  2. Curry
    • one year ago
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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.

  3. radar
    • one year ago
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    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.

  4. radar
    • one year ago
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    What are you using for frequency?

  5. Curry
    • one year ago
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    20

  6. Curry
    • one year ago
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    sorry i mean 50.

  7. Curry
    • one year ago
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    Well I thought (10 + 20j ) || -j/50C = 30.

  8. Curry
    • one year ago
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    Cause i'd use the R thevenin.

  9. Curry
    • one year ago
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    the answer the gave was 0.8mF. and when i plug that into the equation I typed above, i got 50. :/

  10. anonymous
    • one year ago
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    hey @radar can you tell me what is Z here i know it is used for power factor but what is this 10+j20

  11. radar
    • one year ago
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    It represents a real (resistor) value of 10 and a series inductive reactance of 20 Ohms.

  12. anonymous
    • one year ago
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    ok if it would have been a capacitor then Z=10-20j right?

  13. Curry
    • one year ago
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    yes

  14. Curry
    • one year ago
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    because the equation is -J/wc

  15. anonymous
    • one year ago
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    what is -J/wc and inductor lags behind the current wrt resistor so shouldn't it be -(negative) and opposite for capacitor

  16. radar
    • one year ago
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    The value of the capacitor will cancel out the inductive reactance allow matching for maximum transfer. Yes Z=10 -j20 would represent capacitance.

  17. anonymous
    • one year ago
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    yeah i get that for the max power but um kinda confused with the signs

  18. Curry
    • one year ago
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    So wait, sorry, back to my question, where am I going wrong?

  19. radar
    • one year ago
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    The conjugate surd for a int impedance of 10 + j20 would be 10 - j20 just like when you rationalizing a denominator.

  20. radar
    • one year ago
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    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

  21. Curry
    • one year ago
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    hmm, i think there's a small confusion. So, we're not multiplying by the frequency. But rather 2pi(frequency) = 50.

  22. Curry
    • one year ago
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    In other words, the inductor's value would have been 2/5.

  23. Curry
    • one year ago
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    when we have the equation Acos(wt+b), the value for an inductor would be jwL and for capacitor it'd be -j/wC

  24. Curry
    • one year ago
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    Hope that helps.

  25. radar
    • one year ago
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    I still come up with a unreasonable omega and then frequency. Need to do some research. Presently I am using 50 as omega.

  26. Curry
    • one year ago
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    yes, that is correct.

  27. Curry
    • one year ago
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    so in other words your value for the capicatance impedance is -J/50C. We just have to solve for C.

  28. radar
    • one year ago
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    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.

  29. radar
    • one year ago
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    That is much larger 1,000 micro farad vs .8 microfarad.

  30. radar
    • one year ago
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    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.

  31. Curry
    • one year ago
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    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.

  32. ybarrap
    • one year ago
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    |dw:1433647505784:dw|

  33. Curry
    • one year ago
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    @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?

  34. ybarrap
    • one year ago
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    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.

  35. ybarrap
    • one year ago
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    Here's a better picture -

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  36. Curry
    • one year ago
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    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

  37. ybarrap
    • one year ago
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    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.

  38. ybarrap
    • one year ago
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    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.

  39. Curry
    • one year ago
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    well the answer he posted was 0.8mF. so, i'm guesing he derived an acutal value.

  40. Curry
    • one year ago
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    @sidsiddhartha

  41. radar
    • one year ago
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    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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