# How can current pass through (resistanceless) connecting wires?

This is not a duplicate of: Will current pass without any resistance?. I read it but my question isn't answered there.

I'm a physics tutor for high school students and this is my understanding of how current flows:

Across any resistor if there is a potential difference, there will be electric field across that element from the point of higher potential to the point of lower potential. Now since resistor (conductor) contains free electrons they flow (drift) in the direction opposite to electric field and thus we have current.

It implies that, if there is no potential difference between two points, there can't be electric field between them, so no drifting of electrons, hence no current. I have been using this logic to explain why we remove certain resistors in circuits (eg. in a balanced Wheatstone bridge).

Question: Consider the following circuit: if I apply Ohm's law between point a and point b then

$$V_a - V_b = \Delta V = IR_{ab} = 0$$ since $$R_{ab} = 0$$

which implies $$V_a = V_b$$. So from above mentioned logic, there shouldn't be any current flow between them. Then how is the current flowing?

What exactly is wrong with my thinking? How can current pass through connecting (resistance less) wires?

• Currents flow through superconductors just fine. Your logic error is in assuming that because there is no voltage drop across some small section of 'ideal' wire (or superconductor) that there can't be a voltage drop across a circuit containing that small section. Sep 7 '17 at 19:51
• Possible duplicates: physics.stackexchange.com/q/45040/2451 , physics.stackexchange.com/q/69919/2451 and links therein. Sep 7 '17 at 20:10
• Sep 8 '17 at 3:13

You're right. I'll work my way up to an answer. First, for current to flow, you need a force to push charges around the loop. The force that does this is the electric field. There's an electric field within the wire and it follows the shape of the wire around the circuit. This electric field is responsible for giving the charges of the wire a net drift direction ("Friction" is what keeps the pushing force from accelerating the charges to infinity - so we obtain a nice average drift speed of charges).

## $E$ Field in Conductors

There's nothing wrong with an electric field in a conductor. We usually think that electric fields have to be 0 within a conductor, but this is only in the static case. Imagine applying an electric field to a conductor. You probably already know this. But the electric field is definitely not zero in the conductor. It's only zero after everything has settled down and we are in the regime of electrostatics. However, before we got to electrostatics, there was definitely an $E$ field in the conductor. Likewise for circuits, the wire is a conductor but it is never able to reach electrostatics (the details of which are a little nuanced - the battery essentially prevents the wire from reaching a static situation). Therefore having an $E$ field within the wire is completely fine. Where does this $E$ field come from? This is getting a little bit off topic but the battery has an electric field. It's this electric field plus the electric field of induced/piled up charges along the surface boundary of the conducting wire that shapes the field within the wire. Anyways, the point is that an $E$ field exists and does the pushing. There's one other thing you have to know: for many substances $J = \sigma E$ where $J$ is current density and $\sigma$ is conductivity. This is Ohm's law from which $V = IR$ can be derived. For copper, $\sigma$ has an order of magnitude of $10^7$. The point is that $E$ is very small in a conductor.

## Potential $V$

Wires do have a potential drop $V = - \int \vec{E} \cdot d\vec{l} < 0$ as $\vec{E}$ isn't zero and points in the same direction as $d\vec{l}$ (assuming we are integrating in that direction). So start at terminal $a$ of the battery and move to the start of the resistor, point $b$. The voltage drop is $V(b) - V(a) = - \int_a^{b} \vec{E} \cdot d\vec{l}$, which is miniscule because $E$ is so small. In the resistor, the electric field $E$ becomes really large (low conductance $\sigma$). Therefore the voltage drop $V(c) - V(b) = - \int_b^{c} \vec{E} \cdot d\vec{l}$ is large because $E$ is large. Then on the other side of the wire, because you still have your $E$ field pushing, you'll have a tiny voltage drop again. These voltage drops in the wire can be neglected.

• I just love the way you've explained. Especially till the point leading upto the answer. This is getting a little bit off topic but the battery has an electric field. Please go bit off the topic. Sometimes its fine to go bit off the topic :). Sep 7 '17 at 20:43
• $J = \sigma E$ you said. So you are using current density (which is assuming equivalent current in the wire) $\sigma$ to explain the low value of Electric field. Isn't there another way to prove that E is very small? Sep 7 '17 at 20:45
• I can't really say much more on how an $E$ field gets into/is set up in the wire. All I can say is that the battery has something to do with it along with surface charge in the wire which 'shapes the direction' of the $E$ field to match the direction of the wire. Surface charges pile up in bends of the wire (just as charges pile up at sharp points in a conductor generally speaking). It's these concentrations that help shape the field. However, I can say that there does have to be an $E$ field in the wire, else charges would be zigzagging with no net direction Sep 8 '17 at 4:06
• (In my opinion, saying that the wires are superconducting makes things confusing in a certain sense, it's also less satisfying and doesn't match reality. Wires burning up and energy loss in wires are real and valid concerns). I believe Ohm's law $J = \sigma E$ is the best way to see that an electric field, for a reasonable current density $J$, is small in a conductor. Yeah current $I$ through a surface is $I = \int\int\vec{J}\cdot d\vec{a}$. For uniform $\vec{J}$, $I = JA$. So $I/A = \sigma E$ Sep 8 '17 at 4:09

Your quoted statement (while practically true) is hiding a few things. You seem to be reading it as "potential difference causes current".

Instead, potential difference (and an electric field) causes charge acceleration. It's just that in the steady-state limit, this charge acceleration is exactly balanced by the process within the resistor so that current is constant.

If you think of your car driving down the road, you need constant engine power to keep the car moving at a particular speed. If you don't, friction and air resistance bring it to a stop. In the wire, you need to have a potential difference to "push" the charge through the resistor at a particular rate.

But if we take the car into a frictionless vacuum, no engine is necessary for it to roll down the road. While it does need an initial push, we can turn off the engine and let it drift indefinitely. In your circuit, if A to B is an ideal conductor (superconductor, $R=0$)we can get the charge to begin moving, no electric field or potential difference is necessary for it to continue flowing in the zero-resistance portion. So in the steady-state, you're correct that there will be no potential difference.

If instead A to B is a normal wire, then $R$ is not zero, but just smaller than we need to worry about normally. That means it has a finite resistance, and there will be a (small) electric field within.

• It's just that in the steady-state limit, this charge acceleration is exactly balanced by the process within the resistor so that current is constant. There is a "steady-state limit" for a current through a resistor ? I'm only aware of steady state in a RC circuit. Can you kindly give link to read more on this? Sep 7 '17 at 20:54
• Yes. The current through a resistor is zero when the circuit is open, and changes over time when the circuit is closed. In the limit as $t \rightarrow \infty$, the resistance and voltage difference are equal, so the current remains constant. It's identical to RC (or RLC), but because there is no explicit capacitor, the time constant is much shorter. Sep 7 '17 at 21:01
• #1. So the case is like in newton's laws? A body doesn't need force to keep moving if it was already moving at a constant velocity? But if I accept this logic, then it would mean that #2. all the points in the zero-resistance wire are at same potential hence Electric field is zero in it but not zero where there is resistance. Implying, Electric field in a conducting loop is discontinuous. Is that the case? Also #3. Why is your answer contradicting DWade64's answer: who is saying that there is indeed an Electric field but very less for a resistance less wire? Sep 7 '17 at 21:02
• The electric field goes to zero when the circuit is at steady state (which is where you are using Ohm's law). It is not zero in transition because the inductance of the wire is non-zero. Resistance gets you voltage difference in constant (non-zero) current. Inductance gets you voltage difference in non-constant current. Sep 7 '17 at 21:06
• What you are saying is mindblowing! I don't know what to google to read more on this. Can you kindly give some links or references for detailed reading? Sep 7 '17 at 21:09

$I=\Delta V/R$. If $\Delta V$ and $R$ are both zero, then $I=0/0$, or in other words, this equation doesn't tell you anything about the current anymore and it can be anything. You have to find it from the rest of the system.

Current, $$I$$ depends on the resistance of the overall circuit, that dictates the amount of electrons flowing in the circuit. This will be true even when the connecting wires have zero resistance. $$V_a = V_b$$ implies $$R_{ab} = 0$$, but doesn't necessarily mean $$I = 0$$; current is still calculated by circuit R divide V. I believe this is the confusion.

The electrons drift in the conducting wires in the direction opposite to the direction of electric field, in closed loop. The mistake lies in considering two points a and b in the wire. I suggest you to apply Kirchoff's voltage law. This eliminates the misconception.