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

In summary, solving for the capacitance required to provide maximum power to the load results in a value of j4000Ω.
  • #1
Color_of_Cyan
386
0

Homework Statement

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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  • #2
Looks like a situation where you want to apply the Maximum Power Transfer Theorem. Look it up :smile:
 
  • #3
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 = -j4000ΩSo for the load is it

3000 - j4000Ω ?
 
  • #5
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
Color_of_Cyan said:
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 into find maximum power I got:

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

PL = 0.00416 W

= 4.16 mW
 
  • #9
Color_of_Cyan said:
Plugging the first formula into 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
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
Color_of_Cyan said:
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.
 

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

How do I find the load impedance for an RLC circuit?

The load impedance for an RLC circuit can be found by using the following formula: Zload = Rload + j(XL - XC), where Rload is the resistance of the load, XL is the inductive reactance, and XC is the capacitive reactance. You can also calculate the load impedance by finding the total impedance of the circuit and subtracting the source impedance (Thevenin or Norton equivalent).

What is the Thevenin equivalent of an AC circuit?

The Thevenin equivalent of an AC circuit is a simplified circuit that represents the AC circuit as a voltage source in series with a single impedance (ZThevenin). The voltage source is equal to the open-circuit voltage of the original circuit, and the impedance is equal to the total impedance of the original circuit seen from the load terminals.

How do I calculate the Thevenin equivalent of an RLC circuit?

To calculate the Thevenin equivalent of an RLC circuit, follow these steps:

  1. Remove the load from the circuit.
  2. Find the open-circuit voltage at the load terminals.
  3. Find the total impedance of the circuit seen from the load terminals.
  4. The Thevenin voltage is equal to the open-circuit voltage, and the Thevenin impedance is equal to the total impedance.

How is the Norton equivalent different from the Thevenin equivalent?

The Norton equivalent is similar to the Thevenin equivalent, but instead of representing the AC circuit as a voltage source in series with an impedance, it is represented as a current source in parallel with an impedance (ZNorton). The current source is equal to the short-circuit current of the original circuit, and the impedance is equal to the total impedance of the original circuit seen from the load terminals.

Can I use the Thevenin and Norton equivalent circuits for any type of AC circuit?

Yes, the Thevenin and Norton equivalent circuits can be used for any type of AC circuit, as long as the circuit is linear and contains resistors, inductors, and capacitors. However, the calculations for the Thevenin and Norton equivalent may be more complex for circuits with multiple sources or non-linear elements.

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