Am I right that a wire (zero resistance) in parallel with a resistor (resistance R1) is equivalent to a wire (zero resistance)?

This follows trivially from the usual rule:
Req=(0*R1)/(0+R1)=0.

However, I have just run into a problem in which I have just such a case, and it seems that I am wrong! I have a circuit in which there is a voltage source in parallel with a resistor R1. In solving for the Thevenin resistance, however, the voltage source is supposed to be shorted, leaving a wire in parallel with R1. Hence, that resistor/wire combination should be made into a wire. However, I only get the right answer when I collapse this into a resistor with resistance R1. What is going on? (For the example in which I get the wrong answer, see page 118 of Hayt Kemmerly Durbin, practive problem 5.5: try doing a traditional Thevenin equivalent circuit for it!

This problem comes up with a very simple circuit: try to calculate the Thevinin equivalent of a battery V in parallel with a resistor R1 that both connect (in parallel) to some load resistance RL.

Following the usual Thevenin equivalent construction, we get:
1. Vth=V (trivially, as things are parallel).
2 Rth=0 (as when we remove V, we get R1 in parallel with V, which is just another piece of zero-resistance wire).

Or perhaps this is right. It is just saying that V is the voltage across Rload, and that we have an ideal voltage source because it will pass whatever current is needed to maintain V.

 

Originally posted by neuronet@Oct 18 2003, 09:32 PM

Or perhaps this is right. It is just saying that V is the voltage across Rload, and that we have an ideal voltage source because it will pass whatever current is needed to maintain V.

This is right. I was confused as I didn't realize how to "remove" current sources when finding the Norton/Thevenin equivalent circuit: to remove current source, you open the circuit there (i.e., "cut the wire") so no current can flow, right? This is exactly what you do when you apply superposition: replace V with a wire, I with an open circuit.


I am right, right???

 

Correct, for Norton/Thevenin equivalent circuits a voltage source is short circuited and for a current source an open circuit is made.

I am quite curious about this problem, have you tried simplifying the circuit resistances to try and remove the R1 resistor from being in parallel with the voltage source before replacing the voltage source with a short circuit. This may give a resistance in series with the voltage source which would work.

If you want maybe post a diagram so we can have a look.

 

A voltage source in parallel with a resistor is equivalent to just a voltage source. The resistor does absolutley nothing. Even if the resistor draws off some current, it will make no difference in calculating the other voltages/currents in the circuit. It does make a difference when calculating the power dissipated in the circuit.

Also a current source in series with a resistor is equivalent to just a current source, again the resistor does absolutley nothing.

 

Can we all say "Ohms Law" and leave Thevenin go hang ??

All Thevenins "theorems" do, i.m.h.o., is cloud the issue with unnecessary "equivalent" fol-de-rol.

It reminds me of the seemingly endless scrapping between Edison, Westinghouse and Tesla...............Who's da man ???

 

This post is 6 years old. Please don't respond to old posts!​

 

I suspect they deleted their post (you can do that here), leaving you standing all alone.