# A couple of basic questions about resistance

#### xox

Joined Sep 8, 2017
597
Suppose you were to connect together two 1K resistors in series and attach one free end to a 5V source and the other to ground. Between those two resistors you get a voltage divider of 2.5V. Now obviously, running a wire from that point to ground does not make a short because the current is clearly passing through a 1K resistor. But had you connected a 2.5V source directly to ground you would have a short.
1. Does that mean that if you used a 2.5V battery with an internal resistance of 1K then attaching both ends to a loop of wire would not result in a short?
2. I notice that when I connect the voltage divider output to ground in the simulator, the current in the second resistor drops to zero. Does that mean that in actuality absolutely no current flows through it, or just a very small current?

#### DickCappels

Joined Aug 21, 2008
7,829
1. Your definition of a short appears to be a case in which there is zero ohms across a voltage source, so if that is your definition and you separate the voltage source of the battery from its internal resistance then there is no short. My definition of a short is a case in which a current path with some impedance is bridged by a zero ohm path (or a much lower impedance than the unshorted path), in other words, a "shorter" path for the current, thus bypassing the load. I would call your case #1 a short across the load.

2. The simulator assumes that the "wires" have no resistance, so in the simulator no current flows through the shorted out components. In real live wires have some resistance so you have two current paths that share the current as a function of their resistances.

• tranzz4md and xox

#### xox

Joined Sep 8, 2017
597
Okay, so technically a short, but nonetheless in case #1 only 5 milliamps passes through the wire, correct?

#### DickCappels

Joined Aug 21, 2008
7,829
(some text removed for clarity)
1. Does that mean that if you used a 2.5V battery with an internal resistance of 1K then attaching both ends to a loop of wire would not result in a short?
Not 5 ma. 2.5V/1k = 2.5 ma.

• xox

#### tranzz4md

Joined Apr 10, 2015
310
People use terms like "short", "ground," "power" so often and effectively that some meaning, I should say precision of meaning, gets lost.

Dick went straight to it.

Never forget or doubt: e=IR and a circuit is always present.

#### xox

Joined Sep 8, 2017
597
Not 5 ma. 2.5V/1k = 2.5 ma.
Oh yeah! Well, the simulator shows 5ma flowing through the wire. Maybe it's because there isn't a load being connected to the output of the divider, the result being that the voltage of the wire drops to 0V (ground) and so the second resistor essentially "disappears" (due to the "short out" effect you were describing earlier) effectively leaving just a 5V source connecting a 1K resistor to ground (and thus 5 ma of current)?

EDIT: It looks like a 500 ohm resistor connected to the voltage divider is required to get the current down to 2.5 ma. Perhaps that's just a peculiarity of voltage dividers in general; they don't "provide" a voltage so much as "present" one, that is, the actual voltage changes based on what type of load is connected to it? I actually noticed this effect when I was trying to split 12V into smaller voltage branches using dividers. Connecting a load to one branch threw off all the other branches. I ended up connecting each branch to an op-amp voltage buffer and that seemed to solve the problem (owing to the fact that the op-amp is ideally a zero-current pulling device).

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#### WBahn

Joined Mar 31, 2012
26,398
To back up with DickCappels is saying, the term "short" is ambiguous and means different things in different contexts. In MOST situations we can model a "short" as a zero impedance path between the points being "shorted". In your first example, what was being shorted was not the supply, but rather just one of the 1 kohm resistors.

As Dick pointed out, a true 0 ohm short is very rare (but IS possible with something like a superconductor). But let's say that the shorting wire had a resistance of 1 ohm. In parallel with 1 kohm that would mean that about 99.9% of the current would flow through the "short" and only about 0.1% through the shorted resistor.

To put that into further perspective, a #24 solid hookup wire (typical of solderless breadboarding wire) has a resistance of about 25 mohms per foot. So the wire needed to short across a typical resistor would be about 2.5 mohms. Hence something like 99.9998% of the current will flow through the shorting wire and the amount flowing through the resistor (something on the order of about 10 nA) would require significant efforts to even detect reliably.

• xox

#### WBahn

Joined Mar 31, 2012
26,398
Oh yeah! Well, the simulator shows 5ma flowing through the wire. Maybe it's because there isn't a load being connected to the output of the divider, the result being that the voltage of the wire drops to 0V (ground) and so the second resistor essentially "disappears" (owing to the "short out" effect you were describing earlier) effectively leaving just a 5V source connecting a 1K resistor to ground (and thus 5 ma of current)?
It sounds like you are using the original 5 V source in your simulation and not the 2.5 V source in series with 1 kohm like you asked about.

The Thevenin equivalent of your voltage divider is a 2.5 V source in series with 500 ohms (the two 1 kohm resistors are seen as being in parallel from the perspective of an external circuit), so that would give you 5 mA if you shorted that source.

• xox