Question

Easy question about circuits.

Answer 1

just for reference, R load = Rt, for max power transfer. I calculated it, but not getting the right answer. it should be 0.8mF.

Answer 2

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.

Answer 3

What are you using for frequency?

Answer 4

20

Answer 5

sorry i mean 50.

Answer 6

Well I thought (10 + 20j ) || -j/50C = 30.

Answer 7

Cause i'd use the R thevenin.

Answer 8

the answer the gave was 0.8mF. and when i plug that into the equation I typed above, i got 50. :/

Answer 9

hey @radar can you tell me what is Z here
i know it is used for power factor but what is this 10+j20

Answer 10

It represents a real (resistor) value of 10 and a series inductive reactance of 20 Ohms.

Answer 11

ok if it would have been a capacitor then Z=10-20j right?

Answer 12

yes

Answer 13

because the equation is -J/wc

Answer 14

what is -J/wc
and inductor lags behind the current wrt resistor so shouldn't it be -(negative) and opposite for capacitor

Answer 15

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

Answer 16

yeah i get that for the max power but um kinda confused with the signs

Answer 17

So wait, sorry, back to my question, where am I going wrong?

Answer 18

The conjugate surd for a int impedance of 10 + j20 would be 10 - j20 just like when you rationalizing a denominator.

Answer 19

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

Answer 20

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

Answer 21

In other words, the inductor's value would have been 2/5.

Answer 22

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

Answer 23

Hope that helps.

Answer 24

I still come up with a unreasonable omega and then frequency. Need to do some research. Presently I am using 50 as omega.

Answer 25

yes, that is correct.

Answer 26

so in other words your value for the capicatance impedance is -J/50C. We just have to solve for C.

Answer 27

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.

Answer 28

That is much larger 1,000 micro farad vs .8 microfarad.

Answer 29

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.

Answer 30

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.

Answer 31

@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?

Answer 32

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.

Answer 33

Here's a better picture -

Answer 34

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

Answer 35

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.

Answer 36

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.

Answer 37

well the answer he posted was 0.8mF. so, i'm guesing he derived an acutal value.

Answer 38

@sidsiddhartha

Answer 39

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:
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.