# Why is the voltage drop across this 1 ohm resistor being neglected in this practice problem? (Thevenin's Theorem)

We are given the following circuit and told to calculate the equivalent voltage of everything to the left of the a and b terminals:

It omits the load resistor and lays the circuit out as such for mesh analysis:

It claims that voltage across terminals a and b, the voltage we are looking for, is the same as the potential in between the 4 ohm and 1 ohm resistor. Shouldn't that 1 ohm resistor cause the potential to change a little?

• what is 1 / (infinity)? – Brian Drummond Sep 29 '15 at 10:25

You're trying to find the Thevenin equivalent voltage of the network.

This is the voltage that the network produces when the output is open. When the output is open, no current flows out or in to the a terminal. Therefore no current flows through the 1 ohm resistor.

Since no current is flowing through the 1 ohm resistor, the voltage across it is 0, by Ohm's law.

To add a little bit to the accepted answer: in real life, the voltage after the 1 ohm resistor will not equal the voltage before it for only a split second during the charging of the circuit, when you first turn on the 32V source, or for the split second just after you ground the 32V source. The reason is because during these short transient times, "a" is charging up through the resistor (since it has a tiny bit of capacitance..as all wires and conductors do in real life), so a tiny current does exist. Once "a" is charged, and you reach the steady-state condition, the current ceases and the voltage on each side of the 1 ohm resistor is the same.

In theory, "a" has no capacitance, so you can ignore the above real-life situation. :)

The point, however, is that in this case, the real-life steady-state solution is the same as the theoretical one, though the real-life transient solution is momentarily different.

Now, the above is just some extra info. For the most direct and correct answer, see "The Photon"'s answer.

I think its an optical illusion and that's why it looks odd. the 4 ohm and 12 ohm resistors are whats important here. You connect the RL at that junction, so all that matters is that you get the 4 + 12 = 16, and 32/16 = 2A, confirmed by the 2A symbol there.

then its whatever voltage ratio you get. if E = IR then voltage is 2 amps x 12 ohms = 24v at your RL point there

the 1 ohm will cause the current to decrease but i think the voltage will be 24v. in real equipment they will only be able to supply so much current, so in real circuits you would need to watch the 1 ohm there, but in theory with no load defined, its assumed the other parts of this equation are perfect (right?)