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I had a quick question on the electric potential between two points on a short circuit. I apologize if this question does not make sense or if I am using terms like, "electric potential" incorrectly. Let me know where I am incorrect in my thinking. I am trying to learn this material and it is becoming a challenge.

From what I understand, there exists a potential of sorts between the anode and the cathode terminal of a battery and that I am able to measure this potential with a Voltmeter. I have not been able to test this scenario in real life, since it involves a short circuit, but from what I understand, when the leads of the voltmeter are placed close to the battery terminals, there should be a reading indicating a presence of voltage. This example is shown in the first picture, and I am assuming that there would be a potential of 12 V present.

schematic

simulate this circuit – Schematic created using CircuitLab

My question is, as we move the voltmeter test leads closer to one another, would there be an observable change in the potential?

schematic

simulate this circuit

I am making the assumption that as the test leads get closer together, the potential between these two points in space is getting lower and I should be able to measure this change in my voltmeter. Perhaps I do not fully understand what exactly a Voltmeter is measuring and how it relates to potential. Any help in understanding this scenario would be appreciated. Cheers.

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    \$\begingroup\$ This is a little odd, Adam. Your diagrams show a battery with a short-circuit across it. I can't fathom what learning value that circuit has for you but your voltmeter will read near-enough 0 V and the battery would get hot, along with the wire shorting it out, then something nasty could happen to one or both. Please can you explain (a) your background and (b) what it is you're trying to learn here. Thanks. \$\endgroup\$ – TonyM Jan 17 '18 at 17:44
  • \$\begingroup\$ Hello, no worries, I was thinking that when I connect two ends of a battery together I get a current flow, and that the highest potential would exist between these two points, and that a lower potential would exist as I move the voltmeter leads closer together. I was just wondering if this was correct. \$\endgroup\$ – Adam Bak Jan 17 '18 at 17:53
  • \$\begingroup\$ You can happily short-circuit alkaline batteries. They are consumer-proof, and will just get hot. No fire. \$\endgroup\$ – user2497 Jan 17 '18 at 18:02
  • \$\begingroup\$ Your assumption is correct. Unfortunately using a weak alkaline battery shows only millivolts if you use short (say max. 1 meter) and reasonably thin (say 0,5...1 mm) copper wire. The battery voltage is pulled onto its knees due the low resistance of your wire and the much higher internal resistance of the battery .If you use a normal 12V car battery you can see probably several volts, but your experiment will be disastrous. In a couple of seconds your wire is red hot and burns its way into your flesh. If you use a lithium battery, it probably explodes. DO NOT try them. Learn Ohm's Law! \$\endgroup\$ – user287001 Jan 17 '18 at 18:23
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Your schematics have omitted the battery's internal resistance.

schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. The real battery can be modeled as an ideal voltage source with internal series resistance.

From what I understand, there exists a potential of sorts between the anode and the cathode terminal of a battery and that I am able to measure this potential with a voltmeter. The potential is real.

I have not been able to test this scenario in real life, since it involves a short circuit, ...

A wise choice.

... but from what I understand, when the leads of the voltmeter are placed close to the battery terminals, there should be a reading indicating a presence of voltage. This example is shown in the first picture, and I am assuming that there would be a potential of 12 V present.

Due to the internal resistance of the battery the terminal voltage will collapse to almost zero when a short circuit is applied. I say 'almost' because your best short circuit will have some resistance.

My question is, as we move the voltmeter test leads closer to one another, would there be an observable change in the potential?

It's simply a matter of the ratio of the internal resistance to that of your short circuit.

I am making the assumption that as the test leads get closer together, the potential between these two points in space is getting lower and I should be able to measure this change in my voltmeter.

You are correct. Let's do a quick example.

  • 12 V battery.
  • Battery short-circuit current is 20 A.
  • \$ R_{internal} = \frac {V_{OC}}{I_{SC}} = \frac {12}{20} = 0.6 \ \Omega \$ where OC is open-circuit and SC is short-circuit.
  • Let's put a 0.001 Ω "short" across the terminals.
  • Terminal voltage will drop to \$ V_T = V_{OC} \frac { R_{SC} }{R_{INT}+R{SC}} = 12 \frac { 0.001} {0.001+0.6} = 0.0017 =\mathrm V = 1.7 \ \mathrm {mV} \$.

How good is your meter?

Meanwhile, what's happening in your battery?

\$ P = I^2R = 20^2 \times 0.6 = 240 \ \mathrm W \$. This may be dangerous.

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  • \$\begingroup\$ Thank you for the explanation. Would this mean that, when you connect a piece of wire between two points of differing potential, like in a short circuit scenario, that this potential equalizes, and your voltmeter would read 0 V across the terminals of a battery when short circuited together? \$\endgroup\$ – Adam Bak Jan 17 '18 at 18:02
  • \$\begingroup\$ I've shown how to do the calculations. What part do you not understand? \$\endgroup\$ – Transistor Jan 17 '18 at 18:09
  • \$\begingroup\$ Not sure, I'm new to electrical engineering, and probably just a lack of knowledge, I guess I'm having a hard time understanding why I am not able to measure the voltage of a battery when the battery is short circuited. Several answers mention that the terminal voltage would collapse to zero when short circuited, but yet the battery is still producing an electric potential between two points in space causing current flow. \$\endgroup\$ – Adam Bak Jan 17 '18 at 18:17
  • \$\begingroup\$ Yes, but most of the potential is lost on the internal resistance. There is very little left for the external "resistance" (your short-circuit). In my example there is only 1.7 mV available. \$\endgroup\$ – Transistor Jan 17 '18 at 18:20
  • \$\begingroup\$ So if I were able to put a voltmeter inside of the battery, I would be able to measure around 12 Volts, but on the metal wire connecting the terminals, there would be little voltage measured. Understood, I think I am going to have to hit these physics books. Thanks for the help. (: \$\endgroup\$ – Adam Bak Jan 17 '18 at 18:23
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First of all, as @TonyM said, do not experiment with this circuit in real life, as it involves short circuit of the battery, and depending on battery compound and its charge you get fire, explosion, burn or whatever bad can happen to chemistry and physical components and you due to overcurrent.

The learning point here for you must be an understanding that every conductor is having its resistance, and total current flowing through it is equal to I=U/R, where U is voltage of the battery (ideal voltage source), and R is total resistance of your circuit, in your diagram the resistance of the wire.

When you place two probes of voltmeter as on your pics, you measure voltage of the wire between the probes, given that we assume voltage meter resistance to be infinity (if not, its resistance comes to play).

Thus in general voltage should read higher if you have larger distance between probes. But again, on your picture resistance of the wire should be almost zero, and current flowing through it may be huge, thus you will read V=I*R, and it is up to the scale of these I and R if your voltmeter will sense resulting voltage.

If you take "special" wire (not just copper one) having several kOhms, then you should easily notice the difference. Finally, you can make a chain of resistors, and measure on different points, you will learn this simple Ohm law.

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On the schematics you drew, no. Because schematic wires are ideal all points on that wire or net are the same potential and you have a direct short.

However, in the real world if you were to wire up this configuration with a real non-ideal wire, you would be able to measure different potentials on the wire. Why? because wires have reistance, usually measured in Ω/(unit length).

For example: 24 gauge wire is 25.67Ω/1000ft so you would get around 0.025Ω with one foot of wire and around 0.002Ω for 1 inch of wire. And you would be able to measure a potential across this, but it would be really small and you might have a difficulty doing this with a regualar meter

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schematic

simulate this circuit – Schematic created using CircuitLab

In the simple case an ideal short circuit reduces the voltage source to 0V.

Consider:

  • All voltage sources have some effective series resistance, ESR (inverse to its power capability.)
  • All short circuits have some resistance ( inverse with its wire diameter.)
  • Thus with any measured current source, one can trace a short circuit by locating where the voltage drop does not change from source to short circuit.
  • This is also a convenient way of using a length of wire to calibrate it as a current sense R in milli-ohms using a load in series with the short wire.
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I have not been able to test this scenario in real life, since it involves a short circuit

In a theoretical circuit batteries and wires have no resistance, so your scenario is even less testable because it is impossible. Why? Ohm's Law says that I = V / R. If the resistance is 0 then the current is infinite, which is an invalid solution. There can be no voltage across the wire, and yet there is!

In practice both the battery and wiring do have resistance, so the current will be large but not infinite. You can simulate this scenario by adding the resistances into the circuit:-

schematic

simulate this circuit – Schematic created using CircuitLab

In this circuit R_batt is the internal resistance of the battery, and POT1 and POT2 are potentiometers formed by the resistance of the wires. The total series resistance in the circuit is 3Ω so the 'short circuit' current will be 12V / 3Ω = 4A. The voltage read by the voltmeter will vary from 0 to 8V depending on the positions of the wipers on the 'potentiometers'.

If you use a powerful battery and wire with sufficient resistance you can test this scenario in real life. A practical test might use a car battery and thin nichrome wire, which has much higher resistance than copper.

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