Find load impedance / (Thevenin / Norton too) with RLC & AC circuit?

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  • #1

Homework Statement




[Broken]


For the circuit shown, what value of ZL results in maximum average power transfer to ZL? What is the maximum power in milliwatts?


Homework Equations



P max = (VTh)2/4RL

Thevenin, Norton procedures, voltage division, current division, etc etc

The Attempt at a Solution



This is the exact question; kind of confusing if it means find ZL through the inductor or ZL through the load. But I assume it means find the total impedance through the load which includes RL and the capacitor.


Really not sure how I would start for this though, so assuming it's the formula above I need to use, then I guess find the Thevenin Resistance here, and even more so for VTh.

Assuming finding RTh is the same way too (ie, if I'm supposed to find the resistance from A to B with the V source shorted and the load cut out),

then would RTh be (3000 + 4000jΩ)? (Or ZTh because it's impedance?). Not sure about what VTh might be either though.


Any tips on where to start here would be helpful, thanks.
 
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Answers and Replies

  • #2
gneill
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Looks like a situation where you want to apply the Maximum Power Transfer Theorem. Look it up :smile:
 
  • #3
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This circuit at nodes A and B is already in Thevenin form: a voltage source of 10 V + source impedance of 3 + j4 kOhm...

How would you choose the capacitance to make max power to be delivered to the load ZL = R + jXc? Hint: you want the max current to flow in this circuit, and use the fact that Xc is always negative.
 
  • #4
So is the formula for it supposed to be

PL = (1/2)|Vs|2(RL)/(Rs + RL)2

XL = -Xs

RL = Rs

?


So

Vs = 10V + 0j

Rs = 3000

I take it XL = -j4000Ω


So for the load is it

3000 - j4000Ω ?
 
  • #5
gneill
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Do you want us to confirm a guess?
 
  • #6
My apologies, but I'm trying to find out how this works.


It seems way easier than it looks if that's the way then.
 
  • #7
gneill
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My apologies, but I'm trying to find out how this works.


It seems way easier than it looks if that's the way then.
It's a matter of understanding the Maximum Power Transfer theorem (MPTT), and in particular, how it applies to impedances (as opposed to simple resistances for the source and load). Once that's done, your "guess" should become a confident statement :smile:

Yes, you want the load reactance to be the negative of the source reactance, because then the current and voltage will be in phase, so I*V will be maximized in the load. And the usual proofs associated with the MPTT tell you what the load resistance should be (that's where real power is delivered: into resistances where power is dissipated rather than stored and returned).
 
  • #8
Plugging the first formula in to find maximum power I got:

PL = (1/2)102*(3000/60002)

PL = 0.00416 W

= 4.16 mW
 
  • #9
gneill
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Plugging the first formula in to find maximum power I got:

PL = (1/2)102*(3000/60002)

PL = 0.00416 W

= 4.16 mW
The result is not correct. Can you explain the "1/2" in that formula? Perhaps you should try to derive it symbolically.
 
  • #10
I would need help with that. Not sure what you mean, unless it's really the wrong formula.
 
  • #11
gneill
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Suppose that the reactances of the source and load cancel each other. They effectively disappear from the circuit. You're left with the source and load resistances and the source voltage. Can you write an expression for the power dissipated by the load resistor?
 
  • #12
I'm thinking it's

[V/(RS + RL)]2*RL then?

So

PL =- (10/6000)2*3000

PL = 0.00833W

= 8.33mW
 
  • #13
gneill
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I'm thinking it's

[V/(RS + RL)]2*RL then?
I get the impression that you're not confident about that expression. Did you find it somewhere or did you derive it for the circuit? Being able to derive required expressions from basic principles is an essential skill for circuit analysis; the circuits are not always so straightforward as the one here, and "canned" formulas just won't be available to memorize.

Anyways, it so happens that your formula above is correct for the present situation :smile:

So

PL =- (10/6000)2*3000

PL = 0.00833W

= 8.33mW
Yup. That's fine.
 
  • #14
Honestly, I got it from somewhere. The first one formula I posted is for average maximum power, I think, and kept to a formula like that because you still said to disregard the reactive impedance.


I am looking at it again though and it seems to make sense since P = I2R and I = V/R and the R which is actually the real impedance only is Rs + RL

so P = (V/Rsource+Rload)2*RL.


Thank you again anyway though, you have helped a lot.
 

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