I have had a problem with Thevenin's theorem

Thread Starter

asdf arfw

Joined Sep 9, 2017
43
http://tinyurl.com/y868jetr <-- this is the circuit
I have had a problem with the bottom circuit
I tried to use Thevenin's theorem to find out, what is the value of current flows past through the 3ohm resistor
I used it with the top circuit and it worked , but when i tried to use it with the bottom circuit , it doesn't work at all.
I dont know what I've made a mistake.
I got Vth = 4.8 V from KVL ---> 8 - 6I - 4I = 0 ( the 3ohm resistor was shorted ) and Rth = 3
So 4.8 - 3I - 3I = 0 so the current should be 0.8A
but when I wrote the circuit into the program, the current that was calculated from the program is 0.888..A
and then I used mesh method to solve it and I got the same answer as the program.
I cant find out what I've made a mistake, Could you tell me pls.

p.s. sorry for my bad english, It's my 2nd language.
 

Jony130

Joined Feb 17, 2009
4,974
But Rth is 2.4Ω and I = 4.8V/(2.4Ω + 3Ω) = 0.8888A

If you want to find Rth resistance remove 3 Ohm's resistor from the circuit (open circuit) and short the voltage source. And finnaly find the resistance seen from 3 ohm's resistor terminals.
 

Thread Starter

asdf arfw

Joined Sep 9, 2017
43
thank you a lot , but can I ask you one more thing
I'm wondering , how did you get the value of Rth = 2.4 Ohm.
I removed 3 Ohm's resistor from the circuit already and then I shorted the voltage source.
so the circuit should have 4,2 and 6 Ohm's resistor.
but 4 Ohm's resistor and 2 Ohm's resistor are in series , so the solution should be 6 Ohm.
and then 6 Ohm's resistor is parallel to another 6 Ohm's resistor , so the solution of Rth should be 2 Ohm, Isn't it?
 

Thread Starter

asdf arfw

Joined Sep 9, 2017
43
I don't know how to edit message, but I mean the solution should be 3 Ohm not 2 Ohm. Sorry for that.
 

Thread Starter

asdf arfw

Joined Sep 9, 2017
43
wait, so I only have to calculate R2 parallel to R4 right?
why we don't need to think about the another R1 resistor, Is it because it has been shorted?
I never knew that before. Thanks a lot.
 

Jony130

Joined Feb 17, 2009
4,974
why we don't need to think about the another R1 resistor, Is it because it has been shorted?
Yes, because R1 is shorted via very low resistance wire (0Ω). This is why all R4 current will flow through the short circuit and R1 current is 0A.

RT = 1/(1/R1 + 1/Rs) = 1/(1/2Ω + 1/0.0001Ω) = 0.000099995Ω
 

WBahn

Joined Mar 31, 2012
24,692
wait, so I only have to calculate R2 parallel to R4 right?
why we don't need to think about the another R1 resistor, Is it because it has been shorted?
I never knew that before. Thanks a lot.
The ideal voltage source isolates the two parts of the circuit from each other and prevents them from interacting. Because the voltage is fixed at that point, changing the circuit on either side of it can only affect the current draw from the source, but since the voltage remains fixed the other side of the circuit can't see any effect.
 
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