Engineering Trying to find the short circuit current

[nnis13]

Homework Statement

i have a circuit like this

https://www.physicsforums.com/attachment.php?attachmentid=26017&stc=1&d=1274949363

and i want to reduce it to thevenin's equivalent as seen from the terminals of R5

Homework Equations

IF i short circuit the resistor R5 in order to find the short circuit current at its terminals,then
will this attempt also short circuit the 0.5A current source ?
And then can i remove it ?

The Attempt at a Solution

 

[vela]

It will short the current source, but you can't remove it.

 

[nnis13]

It will short the current source, but you can't remove it.

Can you please explain why i can't remove it?

I already know that if the voltage drop or rise across it terminals is 0V then there is no current flow.

Am i right?

So i will have to include the sourse in the equations?

 

[vela]

The current source will have a 0.5-A current flowing through it regardless of the voltage drop across its terminals. That's what a current source does by definition. If you want to find the short-circuit current, you'll have to include the current source in your calculations.

If you're trying to find the equivalent resistance of the circuit, you can set all the sources to zero -- that means removing all the current sources and shorting all the voltage sources -- and find the resistance of the remaining circuit.

 

[nnis13]

The current source will have a 0.5-A current flowing through it regardless of the voltage drop across its terminals. That's what a current source does by definition. If you want to find the short-circuit current, you'll have to include the current source in your calculations.

If you're trying to find the equivalent resistance of the circuit, you can set all the sources to zero -- that means removing all the current sources and shorting all the voltage sources -- and find the resistance of the remaining circuit.

OK thanks a lot for replying.
but If it is a resistor instead of a current source what should i do?

 

[vela]

What do you mean?

 

[nnis13]

What do you mean?

if i replace the current source with a resistor and then make a short circuit at it's terminals then should i remove the resistor?

 

[vela]

Why would you replace the current source with a resistor?

 

[Phrak]

Let me see. If I short a current source, then it inputs as much current into a node as it removes, and I can disregard it.

 

[vela]

Not if you're looking for the current flowing through the node.

 

[nnis13]

Why would you replace the current source with a resistor?

just an example because i am a little confused.

So if i short circuit a current source it continues to deliver current to the circuit ,but if i short circuit a resistor then no current flows throw it's terminals right?

 

[vela]

Right. If you short a resistor, the voltage across the resistor will be zero, so according to Ohm's law, the current through it will be zero.


Let's go back to your original question about shorting/removing a current source. I thought of a simple example of showing if you remove the current source, you'll get the wrong answer: Consider a current source I in parallel with a resistor R, and say you want to find its Thevenin equivalent (which, as you should know, is a voltage source V=IR in series with a resistor R). The Thevenin resistance is the ratio of the open-circuit voltage V_oc and short-circuit current I_sc. The open-circuit voltage will just be V_oc=IR. Now, if you say you can remove the current source because you're shorting it out, you'd then be left with just the resistor R connected to the short. The short-circuit current would therefore be I_sc=0, and the Thevenin resistance would be infinite. Clearly, that's not correct.

The correct analysis is: When you short the nodes, you can remove the resistor, and all the current from the source will flow through the short, i.e. I_sc=I. The Thevenin resistance will therefore be V_oc/I_sc=IR/I=R. The Thevenin equivalent will consist of a voltage source V_oc=IR in series with resistance R, as you'd expect.

 

[The Electrician]

One way to solve this is to remove the I2 source, solve the network for the short circuit output current, and then add 1/2 amp to that result.

 

[nnis13]

Right. If you short a resistor, the voltage across the resistor will be zero, so according to Ohm's law, the current through it will be zero.


Let's go back to your original question about shorting/removing a current source. I thought of a simple example of showing if you remove the current source, you'll get the wrong answer: Consider a current source I in parallel with a resistor R, and say you want to find its Thevenin equivalent (which, as you should know, is a voltage source V=IR in series with a resistor R). The Thevenin resistance is the ratio of the open-circuit voltage V_oc and short-circuit current I_sc. The open-circuit voltage will just be V_oc=IR. Now, if you say you can remove the current source because you're shorting it out, you'd then be left with just the resistor R connected to the short. The short-circuit current would therefore be I_sc=0, and the Thevenin resistance would be infinite. Clearly, that's not correct.

The correct analysis is: When you short the nodes, you can remove the resistor, and all the current from the source will flow through the short, i.e. I_sc=I. The Thevenin resistance will therefore be V_oc/I_sc=IR/I=R. The Thevenin equivalent will consist of a voltage source V_oc=IR in series with resistance R, as you'd expect.

Thanks a lot .I think i understood your point ,i will just make some more examples to make it clearer.

 

[Phrak]

Not if you're looking for the current flowing through the node.

OK, that makes sense, though current doesn't flow through a node, but in and out of a node to curcuit elements. And we are really taking the limit as R5--> its infinitesimal. We are looking for the current between two nodes as R5 becomes small.

 

[The Electrician]

I'm sure by now you have learned how to do a mesh (or loop) analysis. Look at your circuit--there are 4 obvious meshes; one of them includes R5. If you were to perform a loop analysis, you could calculate a current in a loop that includes R5. You could then replace R5 in the schematic with a straight line, representing a wire. That same loop would still be there, and the current in the loop would be the current in the wire (short) that replaced R5.

 

[The Electrician]

You don't even need to short the R5 terminals to get the Thevenin eqivalent.

Remove R5 and caculate the open circuit voltage at those terminals.

Short the voltage source, open the current sources and calculate the equivalent resistance seen from the R5 terminals, and you now have the Thevenin equivalent.