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How would one calculate the Thévenin voltage and Thévenin resistance for the circuit above?

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  • \$\begingroup\$ You're much more likely to receive better help and answers if you show your attempts and specific problems you face during the process. \$\endgroup\$
    – Wesley Lee
    Oct 22, 2016 at 20:12
  • \$\begingroup\$ Ive tried so many times, i just dont know how to do it \$\endgroup\$
    – A.Glen
    Oct 22, 2016 at 20:15

2 Answers 2

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Thevenin Voltage

The Thevenin voltage e used in Thevenin's Theorem is an ideal voltage source equal to the open circuit voltage at the terminals.

The 2 Ohm resistor is "open" and the 4 and 6 Ohm resistors form a voltage divider, therefore:

Vt = (50V * 6 Ohm) / ( 4 Ohm + 6 Ohm) =

300 / 10 = 30V

Explanation: The voltage at the node between the 3 resistors is the same proportion between the 6 Ohm resistor (which is the one attached to 0V) and the total resistance presented by the voltage divider (6 + 4 Ohm), so 6/10, 60%. 60% of 50V = 30V.

This can be better visualized this way:

schematic

simulate this circuit – Schematic created using CircuitLab

Thevenin Resistance

The Thevenin resistance r used in Thevenin's Theorem is the resistance measured at terminals AB with all voltage sources replaced by short circuits and all current sources replaced by open circuits.

So:

Rt =

(6 Ohm in parallel with 4 Ohm) + 2 Ohm =

(6 * 4)/(6 + 4) + 2 = 24/10 + 2 = 4.4 Ohm

This can be better visualized this way:

schematic

simulate this circuit

Therefore:

schematic

simulate this circuit

All the info I used can be found here. The calculator on the website gives the same results.

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I think they wanted you to first convert your voltage source and series resistance (Thevenin voltage source) into a Norton current source equivalent. Then perform the merge of the two resulting parallel resistors. Then convert that back to a Thevenin voltage source. Then combine the remaining resistors ... and you are done. So:

schematic

simulate this circuit – Schematic created using CircuitLab

Which then becomes:

schematic

simulate this circuit

Which then becomes:

schematic

simulate this circuit

And then you get back your Thevenin as:

schematic

simulate this circuit

And now for the final result:

schematic

simulate this circuit

I believe those are the steps you are supposed to have performed, and in that order. Note that I'm assuming you know how to convert between Thevenin and Norton. If you don't, you need to first acquire that. But I think it is reasonable you know these things already, given the type of problem you were given and the goal you were asked to achieve. Usually, you are exposed to both Thevenin and Norton equivalents first.

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  • \$\begingroup\$ Why not applying Thevenin directly? I don't see any point of performing the transformation. \$\endgroup\$
    – CroCo
    Oct 25, 2016 at 1:58

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