Thevenin's theorem dillemma

[Bangalibabu]

While finding thevenin's voltatge, I opened the load resistance in a circuit ,but it was seen that two current source of differnt rating were in series. How to solve this kind of problem?

 

[A.Anand Srinivasan]

it is just that we are subtituting a infinite impedance at that point when we are removing the load resistance... so the difference of the two currents pass through the resistance,.... we can find the thevenin resistance by shorting the voltage sources and open circuiting the current sources....
we know Iab=Vab/Rth....

try this....

 

[barath_87]

can you please post the circuit diagram.....

 

[the_edge]

Why do you use Thevenin in this case? From your problem it is obvious that you can find voltage on that resistance "V=R*(I1+I2)"... But it would be good to upload that circuit, so we could help you...

 

[marquitos]

I think that that depends on what points you are using to do the measuring and what kind of meter because if the impedance of your meter is bigger on one point and lower on the other then you are going to see something different but would be great if you can send the schematic for a better analysis.

 

[TwinT]

If you post the schemma I try to answer

 

[barry14]

Two current sources in series with different values of current violate the definition of a constant current source. This could only be remedied if one or both of the sources have an associated equivalent parallel resistance (which would make then non-ideal current sources).

 

[princeiut]

Circuit diagram plz.

 

[IBO]

kcl current law is not satisfied by the two series current sources.
a circuit diagram would be appreciated