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.