Finding the maximum power delivered using Thevenin's theorem

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  • Thread starter jisbon
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  • #1
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Homework Statement:

As shown below

Relevant Equations:

-
1592017853060.png

Using mesh analysis, my simultaneous equations seem to be wrong and I can't figure out why. Any stuff that I should take note of?

Attempt:

1592017835584.png
 

Answers and Replies

  • #2
cnh1995
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I prefer the node voltage method. I find branch-currents much less confusing than the mesh-currents.:smile:

Assume the negative terminal of the battery to be the ground i.e. Vb=0.
Call the leftmost node c and assume it to be at potential Vc w.r.t. ground. You know Va=1V.
Apply KCL at node c to obtain Vc and so on..

You are free to use any method of your choice, but it looks much easier to me this way.
 
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  • #3
gneill
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As @cnh1995 mentions, the node voltage method looks promising here. Note that the 5i controlled voltage source forms a supernode, effectively reducing the number of nodes that you have to consider. In fact, this circuit can be analyzed with a single node equation! (although you will need to provide a single auxiliary equation to account for the supernode behavior).
1593130941064.png

From the above figure, clearly

##V_c = V_o - 5 i## where ##i = \frac{V_c}{5}##

That's enough information to find ##V_c## in terms of ##V_o##, so you have adequate supernode auxiliary equation information. In particular you find that ##V_c = \frac{V_o}{2}##.

Usually when there are controlled sources present in a circuit where you want to determine the Thevenin equivalent you are instructed to replace the load with a fixed source (current or voltage) and proceed to determine the impedance that the source "sees" looking into the circuit. That's a tried-and-true method.

I've found that, if your algebra skills are reasonable, you can solve for both the Thevenin resistance and Thevenin voltage at the same time for the vast majority of these sort of circuits by simply leaving the load resistance "R" as an unknown variable and solving for ##V_o##.

The idea is that a Thevenin equivalent with a load ##R## is nothing but a simple voltage divider with a source voltage and two resistances in series: ##R_{th}## and ##R##.

1593132333019.png


This has a simple solution for ##V_o## namely the standard voltage divider equation:

##V_o = V_{th} \frac{R}{R + R_{th}}##

So, if you can solve your circuit for ##V_o## keeping ##R## as a variable, and manipulate the result into the form of a voltage divider as shown above, then you can pick out both the Thevenin voltage and resistance from that result by inspection.
 
  • #4
I would use node analysis instead of mesh, as you got only 2 nodes, but 5 meshes.
There are two ways to use the node analysis. 1. Consider Vb as ground and write KCL for both A and C nodes. Also write the equation for Va and Vc (which is Va - Vc = 5i where i=Vc/5). 2. Or you can easily consider Vc as ground, then we have: Va = 5i and Vb, which gives you one less equation than the first way.

In order to find the thevenin circuit, all you need to do is to replace the the R resistor with an independent Voltage/Current source, with Vtest/Itest volatge/current (Independent current source would be easier as you are writing KCL).
Then you have to solve the equations and find something like: Vtest = Rth*Itest + Vth.
Now you got the thevenin circuit with Vth, Rth and R.

Now you have to calculate the power of R resistor, P=IV. You want to calculate the maximum power, so you have to differentiate the Power function (P=VI) and find where it can be zero (P' = 0).
If you do that, you will see if you pick a resistor which is equal to Rth, you will deliver the maximum power to the R resistance.
 

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