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# Norton & Thevenin eq. practice

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#### Karkas

##### Member
Hello, excuse me if this is out of forum but i wasn't allowed to enter to the electronic theory forum, maybe because i'm still too new.

I'm working on a lab report of this equivalent circuits, i'm almost done but there's a quiestion about the consumed power for different values of load.
The practice is this, yo have a circuit and you determine the Thevenin equivalent & the Norton equivalent theorically and then you put three different values of load one by one, they are 0.5 kΩ, 1k and 1,5kΩ, to calculate the power consumed by them. after that, you do the same but in the project board, practically.
This is what happens:
When I change the load from 0.5k to 1kΩ the power consumed increases, but when I change it again to 1,5kΩ from 1kΩ it decreases, not so much but it decreases, and the question that says "What can you say about the power consumed for different values of load?" is getting a little tough for me to answer, because I don't know why the power behaves like that.

I hope you can help me out with this.

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#### kchriste

##### New Member
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Show us a schematic of the circuits you are having problems with.

#### Karkas

##### Member
This is the cricuit, the Vth is 6.6 V and the In is 5,97 mA, Rth=Rn= 1,1 kΩ
i don't understand why that happens with the power in the load.

The power consumed by the load using the thevenin equivalent (this happens theorically and practically with any of the equivalents) is:

0.5 kΩ 8.75mW
1 kΩ 10.11 mW
1.5 kΩ 9.81 mW.

There you can see what i said.

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#### kchriste

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By "carge" I assume you mean the load at A & B? If the load is equal to Rth of the circuit then the max power transfer occurs. If the load is higher or lower than Rth then less power is transferred to the load. Impedance matching.
If you draw the Thevenin equivalent of the circuit above you'll see why that is so. I think your mW calculations are a bit off. How did you arrive at them?

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#### Karkas

##### Member
Well, first i have to ask you to accept my apologies, for using a very wrong word. Yes that's the load.

I got to that power by this P=V²/R and P=I²*R, and P=V*I,
maybe I'll have to make some other tests with another values of load to check that. Rth is almost 1.1kΩ, and when i take a load of 1kΩ, it can aproach to the max power transfer right?

Thanks for the answer.

#### kchriste

##### New Member
Forum Supporter
Rth is almost 1.1kΩ, and when i take a load of 1kΩ, it can aproach to the max power transfer right?
Yes, that is correct. 1K is the best out of the three values. (0.5 kΩ, 1 kΩ, or 1.5 kΩ)

#### Karkas

##### Member
Well thanks, the lab report is done, next time i will post it in the homework sub-forum.

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