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# Thevenin Equivalent - Urgent

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#### Micheal

##### New Member
Hi All,

Please refer the attach, if I need to use the Thevenin to solve this, what is the Vth and Rth?

Please advice how to start this as I see there is another current source there.

Thanks & Best Regards,
Micheal

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• untitled2.JPG
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The voltage supply and the two first resistors from left can be converted to a thevenin voltage at 2V and 1Ω.

Make note that those components I mentioned above, have no affect at all for the resulting thevenin equivalent.

It's a little tricky task, you see.

Last edited:
Hi,
The thevenin voltage is the voltage between A and B without any load .
Thus there is 2A flooding through the 2Ohm resistor on the right , then it's 4V .
The thevenin resitor is the equivalent resistor value between A and B after having replaced voltage source by wire and cut the wires linked to the current sources .
Thus for your scheme, after having cut the wires on both sides of the current source,
there is finally just one resistor between A and B, the right one .
The Rthevenin value is then 2 Ohm
Cédric

Last edited:
Well I am learning this stuff so I don't know the answer but so far we have two answers which are completely different! I wonder who is correct?

Maybe some third answer will come out ...

Personally, I think ced is correct.

Personally, I think ced is correct.
He is. I didn't completely answered the question

Hi All,

Thanks for help. That's mean the answer should be Vth = 4V, Rth = 2 Ohm, right?

I really confuse wiith the Thevenin theorem - when cut off the current source, left circuit need to be determined or right circuit need to be determined to get the Rth?

Seems like the right hand side circuit near the A-B need to be determined.

Any other comment?

Thanks & Best Regards,
Micheal

Hello,

Some things to remember are these:

1. A voltage source E in series with a resistance R is the same as a current source
in parallel to that same resistance R, with the current source equal to E/R.
2. A current source I in parallel to a resistance R is the same as a voltage source
in series with that same resistance R, where the voltage source is equal to I*R.

Using (1) with the voltage source in the circuit, we get a current source in
parallel to a resistance R, and when that R is in parallel to the center resistance
we get 1 ohm, but more importantly the new current source is equal to E/R
which equals exactly 2 amps, which happens to equal the original current source,
so the current at the top central node flows in and flows out through the
original current source leaving no current to flow through the center resistance
which means 0v at the central node. The result anyway is a 2 amp current
flowing through a 2 ohm resistor which means the voltage A to B is 4 volts,
and since a current source has infinite impedance the output resistance is
2 ohms. You can test this by theoretically injecting a small perturbation current
I into the output and noting how the output voltage changes. For example,
injecting 1ma forces the voltage to step up to 4.002v, where it originally was 4.000v,
and 0.002/0.001 is equal to 2 ohms.

Last edited:
Hi All,

Thanks for help. That's mean the answer should be Vth = 4V, Rth = 2 Ohm, right?

I really confuse wiith the Thevenin theorem - when cut off the current source, left circuit need to be determined or right circuit need to be determined to get the Rth?

Seems like the right hand side circuit near the A-B need to be determined.

Any other comment?

Thanks & Best Regards,
Micheal

It's a two-step process. First, you determine the output voltage at A-B. Since this port is fed with a currnet source ( the voltage is inconsequential in this particular curcuit ) then the voltage is calculated as the current times the resistance. The second step is to determine the ouput reisistance by setting sources in the circuit to zero. Then apply a source to the output port and measure the resistance. Since a zero current is an open circuit, everything to the left of the one in your curcuit does not contribute to the measured resistance. Had that been a voltage, then the result would be different.

Hi All,

Thanks for help. That's mean the answer should be Vth = 4V, Rth = 2 Ohm, right?

Right.

I really confuse wiith the Thevenin theorem - when cut off the current source, left circuit need to be determined or right circuit need to be determined to get the Rth?

Seems like the right hand side circuit near the A-B need to be determined.

Don't make it out to be harder than it really is. If you apply the KVL, the answer becomes pretty obvious. The only current through that 2R resistor across AB, is the two amps from that current source. That, right there, gives Vth. Given that the internal impedance of an ideal current source is infinite, it basically disconnects everything behind it from the circuit. That leaves just the one 2R resistor to determine Rth.

#### Attachments

• prob.jpg
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Thanks for the great explaination. I learn a lot from all of you.

Best Regards,
Micheal

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