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I just started taking classes on circuit analysis. have a question regarding this Thevenin circuit. My end goal is to find the maximum power that can be obtained at A and B. I managed to find the potential difference across A and B, to be 10V, and current I, to be 2A, in the circuit using Kirchhoff’s voltage law. However I’m unable to resolve the resistors in a way that gives me the effective resistance across A and B and I’m not too sure how to move on from there to calculate the maximum power. Any advice would be greatly appreciated.

Edit: additional question, I was wondering in which direction would current flow when it passes the 6v cell. Based on my calculation, the resulting current flowing through the 4 ohm resistor is 1.5A

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For thevenin resistance-

In 2nd fig. ,

4ohm resistor is shorted so remove it and you'll get 1ohm parllel to 2ohm so equivalent resistance would be 2/3ohm and that is thevenin resistance

For maximum power -

there are 2-3different methods but You can connect a resistor of value equal to thevenin resistance (2/3) between point a and b (in 1st fig.) And calculate current through that resistor and then you can calculate power by formula P = I² × R.

  1. direction and value of current can be found using mesh,nodal , superposition theorem whatever you like you can use to calculate it
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  • \$\begingroup\$ to find maximum power, what’s the point of connecting a resistor of equal value to Thevenin resistance? Why can I just connect them straight? \$\endgroup\$ – Ffriends Aug 30 '20 at 16:36
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    \$\begingroup\$ You have one independent variable (resistor between ab ) and dependent variable (current) which changes as you change the value of resistor between ab ,and then you have power which is function of both current and resistor ,since you can manipulate value of resistor directly Because it is independent variable ,so at some point there must be a value of resistor at which power becomes maximum and that value is equal to thevenin resistance and that theorem is called maximum power transfer theorem , you'll see that theorem in you circuit course \$\endgroup\$ – user215805 Aug 30 '20 at 16:49
  • \$\begingroup\$ Ohh I see. Okay thanks a lot for you clarification! \$\endgroup\$ – Ffriends Aug 30 '20 at 17:05
  • \$\begingroup\$ It seems a bit counter-intuitive at first that you get maximum power transfer by loading the source to half the open circuit voltage, In real world applications you would never load a supply down that much as it is wastes energy, typically your maximum load would drop your supply only 10% , so 10% wasted in the source and 90% used. , but for solar and wind power and radio antenna's you do want to operate at maximum power transfer point. (solar and wind are non-resistive so MPP is not at half Voc.) \$\endgroup\$ – BobT Aug 31 '20 at 21:39
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To find the Thevenin resistance you have to find the equivalent resistance seen through the terminals, when all independent voltage sources are shorted, and all independent current sources are open circuits. Your resulting circuit will then look like this: -

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Now, we look through the terminals. Looking into a, we see that current can flow two ways - either through the 1ohm or the 2ohm resistor. Both currents will end up at b regardless. When looking into the terminals, try redrawing the circuit you progressively, like this: -

enter image description here

Now it's very easy to find the equivalent resistance, and thereby the Thevenin resistance.

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For more complicated circuits , you can also calculate the open circuit voltage e.g 10v , and then calculate the short circuit current e.g. 12A+3A = 15A . The Thevenin resistance is simply Rt = Voc/Isc or 10/15 or 0.6666ohm.

Max power transfer is Rload = Rsource = 0.6666ohm.

The 4ohm resistor , and the 2A internal current flow are just red herrings.

Oh, and the additional question , 2A flows into the 6v battery, and 1.5A comes out of the battery into the 4 ohm resistor, so the difference of 0.5A is flowing into the battery, i.e. it is charging the battery. If the resistor was changed to 3 ohm then there would be zero amps flowing into the battery, so the same voltage as if the battery was not there, however if you were to remove the 6v battery at this time, all the voltages would stay the same, but the thevenin equivalent would be different. Voc is the same , but Isc will be less.

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