I need to do this circuit but I can't do it properly, I don't know if I done well the nodes analysis, the answers are Rth= 79.66 ohm and Vth= 5.29 V, thanks for your help.
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\$\begingroup\$ A very warm welcome to the site. Have corrected your question title. 'Thevenin' comes from Mr Thevenin. It's not 'Thevening', as if its been Thevened :-) \$\endgroup\$– TonyMCommented Aug 30 at 10:25
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\$\begingroup\$ Notice that V1 will short-out R5 and R1 resitors as well. \$\endgroup\$– G36Commented Aug 30 at 11:05
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\$\begingroup\$ Elisa, Are you required to solve this in some particular way? Or are you allowed to use any approach that gets you the right answer? You mention nodal. Which is why I ask. \$\endgroup\$– periblepsisCommented Aug 30 at 12:30
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\$\begingroup\$ @periblesis No it is not necessary do it with nodal. \$\endgroup\$– ElisaCommented Aug 30 at 12:44
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1\$\begingroup\$ @Elisa Then I added a small note to help out. Otherwise, I think Andy has also done a lot to help you. \$\endgroup\$– periblepsisCommented Aug 30 at 12:50
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3 Answers
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To find the Thevenin voltage at A and B you should try and simplify things.
The first thing you should note is this: -
This means that the circuit reduces to: -
And, the 9 volt source and the 5 volt source combine to be a single 14 volt source in series with R4.
You can also short out R7 because it is in series with a current source and plays no role: -
Can you take it from here?
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\$\begingroup\$ I will try do it by that way, thank for your answer Andy aka \$\endgroup\$– ElisaCommented Aug 30 at 12:50
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You missed just one detail in getting the right Thevenin resistance:
\$R_1\$, \$R_2\$, \$I_1\$, \$V_2\$, and \$R_5\$ can't alter anything as seen from A and B. What this subnet may alter is the current in \$V_1\$. But otherwise \$V_1\$ spans across that subnet and therefore bypasses it from the global perspective of terminals A and B.
Since you can use any method to get the Thevenin voltage, then I think Andy has set you up for that. You can either Nortonize the \$14\:\text{V}\$ source and the \$100\:\Omega\$ resistor, add that current source to \$I_2\$ and then Thevenize the resulting Norton. Or use some other approach.
The voltage at B will be more positive than the voltage at A, though. So be careful about specifying the resulting Thevenin voltage.
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\$\begingroup\$ Thank you for your answer periblepsis \$\endgroup\$– ElisaCommented Aug 30 at 12:54
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The Thevenin voltage \$V_{th}\$ would be negative as the Norton current source generated with the 14V voltage source when added to 2A current source becomes a 0.6A current source towards ground.After this, convert the Norton current source back to a Thevenin voltage source.The Thevenin resistance is (82 || (2200+100+500)) which is 79.66Ω. Be slightly careful in your circuit analysis.
