Voltage change when resistance is zero

Biker · Jun 25, 2016

1. The problem statement, all variables and given/known data
Consider having a battery, its voltage lets say 20 volts. If you connect this battery to a wire. Its resistance is zero.

If you precisely measure the voltage at the end of the wire, Would it equal 20 volts or no?

2. Relevant equations
dV = I R

3. The attempt at a solution
Well using Ohm's law,
##\Delta V = I R ##
##\Delta V = I ~\text{x} ~ 0##
##\Delta V = 0##
Which means the voltage will be the same..

However, Shouldn't when a positive test charge move away from the positive terminal it lose potential energy? Why this result contradicts the previous one?
If it is the same, then you are exerting a force on an electron or ( a positive test charge) over a distance which means that you are creating energy from nothing

Or does the first result mean that you don't lose energy because of the wire? and Should I think it as a sea of electrons better to understand the perspective under this?

TSny · Jun 25, 2016

The assumption R = 0 is not realistic. Note ##I = \Delta V / R## would imply infinite current. Batteries always have some internal resistance ##r##. In some situations, ##r## can be neglected compared to the other resistances in the circuit. But if you are "shorting" the battery, then ##r## cannot be neglected.

See

Biker · Jun 25, 2016

TSny said: ↑

The assumption R = 0 is not realistic. Note ##I = \Delta V / R## would imply infinite current. Batteries always have some internal resistance ##r##. In some situations, ##r## can be neglected compared to the other resistances in the circuit. But if you are "shorting" the battery, then ##r## cannot be neglected.

See

Okay, Thanks for the important information.
There is a voltage drop if you connecting it to a loop, What if it is a straight line or an open circuit? Wouldn't the distance from the battery affect the voltage?

TSny · Jun 25, 2016

If you connect one end of a wire to one of the terminals of the battery but leave the other end of the wire disconnected (open circuit) then charge will quickly arrange itself along the wire to produce zero electric field inside the wire. There will then be zero current. The potential difference between any two points inside the wire will be zero. In particular, the potential difference between the two ends of the wire will be zero. However, the electric field outside the wire will not be zero and so different points outside the wire can be at different potential. I'm not sure if this helps with your question.

Biker · Jun 25, 2016

TSny said: ↑

If you connect one end of a wire to one of the terminals of the battery but leave the other end of the wire disconnected (open circuit) then charge will quickly arrange itself along the wire to produce zero electric field inside the wire. There will then be zero current. The potential difference between any two points inside the wire will be zero. In particular, the potential difference between the two ends of the wire will be zero. However, the electric field outside the wire will not be zero and so different points outside the wire can be at different potential. I'm not sure if this helps with your question.

Yes that is what I am asking for, So momentary there is a potential difference until the charges arranges them selves?

TSny · Jun 25, 2016

Biker said: ↑

Yes that is what I am asking for, So momentary there is a potential difference until the charges arranges them selves?

Yes. As you bring one end of the wire up to the terminal of the battery, charge is already rearranging in the wire to keep the electric field equal to zero inside the wire. So, at the instant you touch the end of the wire to the terminal, there probably won't need to be too much additional charge rearrangement except in the vicinity of the point of contact.

After one end of the wire is connected to a terminal of the battery, you could imagine changing the shape of the wire by bending it, etc. The charge on the wire will keep rearranging itself to maintain constant potential throughout the wire in the open circuit.

Biker · Jun 26, 2016

TSny said: ↑

Yes. As you bring one end of the wire up to the terminal of the battery, charge is already rearranging in the wire to keep the electric field equal to zero inside the wire. So, at the instant you touch the end of the wire to the terminal, there probably won't need to be too much additional charge rearrangement except in the vicinity of the point of contact.

After one end of the wire is connected to a terminal of the battery, you could imagine changing the shape of the wire by bending it, etc. The charge on the wire will keep rearranging itself to maintain constant potential throughout the wire in the open circuit.

Another question if you dont mind, If you have a really long parallel circuit ( basically a really long straight line that divides to branches at the end), Why is the voltage always the same? Why doesn't the distance from the battery matter as a matter of fact its potential energy comes from the battery ?

cnh1995 · Jun 26, 2016

Biker said: ↑

Why is the voltage always the same? Why doesn't the distance from the battery matter as a matter of fact its potential energy comes from the battery ?

Distance from the source does matter in very long circuits. Resistance of the wires is directly proprtional to their length. In small circuits like those in our houses or labs , this resistance is practically negligible and hence, we model the wires as ideal conductors. If the wires are long enough, their resistance is no longer negligible and there will be a voltage drop across the wires.

Biker · Jun 26, 2016

cnh1995 said: ↑

Distance from the source does matter in very long circuits. Resistance of the wires is directly proprtional to their length. In small circuits like those in our houses or labs , this resistance is practically negligible and hence, we model the wires as ideal conductors. If the wires are long enough, their resistance is no longer negligible and there will be a voltage drop across the wires.

Yea, of course.

But I didnt mean resistance from the wire. I mean as if the positive terminal of the battery is a point charge and the test charge as it moves away from the point of charge it loses it's potential energy because there is a force that the positive terminal exerts on the test charge or is this understanding of an electric circuit is wrong?

CWatters · Jun 26, 2016

It's not quite right.

The units of electric field are Volts per meter. Is there a voltage drop down your unconnected wire of zero resistance?

.

Biker · Jun 26, 2016

CWatters said: ↑

It's not quite right.

The units of electric field are Volts per meter. Is there a voltage drop down your unconnected wire of zero resistance?
.

About the unconnected wire case, Tsny pointed out that the electrons rearrange them selves to form a zero electric field so the change in voltage will be zero.

But the 2nd case that I am asking for is what I replied in my previous comment..

Edit: Is the change in voltage is negligible? because it shouldn't be..

CWatters · Jun 27, 2016

I'm not clear what you mean by the "2nd case" ?

A wire of zero resistance has zero voltage drop. It doesn't make a difference if the wire is open circuit/unconnected or if it's connected to a load.

CWatters · Jun 27, 2016

I suspect your question might be answered in your other thread..
https://www.physicsforums.com/threads/why-does-kirchhoffs-voltage-law-hold.876905/

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Voltage change when resistance is zero

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