In an ideal circuit, How can there be a current b/w points a & b, when there is no potential difference and thus no electric field between a & b? If there is no current, then where does current across the resistor come from, because that means no charge is coming from the battery (?).
Electricity – How Can There Be a Current and an Electric Field in an Idealized Wire with No Voltage Drop?
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- Q1 If there is a potential difference between the leads of the battery, why there has to be a voltage drop somewhere in the circuit (wire, resistor, load etc.) in order for the current to be flowing? In a perfect conductor, shouldn't the electrons be constantly accelerating therefore infinte current? or at least a current proportional to the total number of electrons available in the wire?
- A1: As you know, we got Ohm's law, $U = R \cdot I$. Now, we take the limit $R \to 0\Omega$, while $U=const$. You see, that the current $I$ must diverge, $I\to \infty$. So, if we assume that the battery is a perfect voltage supply, the current must diverge. Alternatively, we can ask the question which voltage the battery supplies, if the current is limited by some value $I_{max}$. Now we get $U\to 0 V$ if $R\to 0\Omega$. So somehow we have to decide, which picture we like to use. For me, I prefer the first one, because I like to think of a battery as a perfect voltage supply.
- Q2: ... I don't know.
- Q3: I take this as your question: What are we really measuring when we hook up a voltmeter to measure voltage across the resistor?
- A3: There is not only one working principle but many working principles exists. Have a look here. I think, at least some of their names are self explaining. So I just copy them here: Parmanent Magnet Moving coil Voltmeter, Moving Iron Voltmeter, Electro Dynamometer Type Voltmeter, Rectifier Type Voltmeter, Induction Type Voltmeter, Electrostatic Type Voltmeter, Digital Voltmeter.
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.
Best Answer
An electric field isn't necessarily required to sustain a current. Remember electric charge is accelerated by an electric field.
In the case of an ideal conductor, which is assumed to connect the source to the resistor, the current can be any value and the voltage across the conductor is identically zero.
This isn't a contradiction. Consider the motion of an object in the absence of friction. No force is required to sustain that motion (only to change it).
Analogously, in the absence of resistance in the ideal conductor, no electric field is required to sustain a current through.
If it helps, consider a non-ideal conductor with some total resistance R. The voltage across to sustain a current $I$ through is:
$$V = I\cdot R$$
Now, let $R$ go to zero and see that, for any value of $I$, the voltage across is zero.
I reluctantly add this because, after some discussion in the comments, I think there is some confusion over the meaning and purpose of ideal circuit theory.
When the OP opens the question with "In an ideal circuit", he sets the context as ideal circuit theory which is a well known, well understood, widely used branch of electrical engineering. Perhaps the OP isn't aware of this context. Perhaps some of those that answered and/or commented aren't aware. Thus, this addendum.
What needs to be made clear is that ideal circuits and circuit elements are used to model physical circuits and physical circuit elements. The ideal circuit elements are meant to correspond to mathematical terms in the equations for the solution of the circuit. They do not represent physically realizable electric circuit components.
Thus, any answer along the lines of "there are no ideal circuits" entirely misses the point.
And, any complaint along the lines of "there must a voltage across because of Ohm's Law" entirely misses the point.
The confusion lies, I think, with the distinction between a physical schematic or, if you will, a "wiring diagram", and an ideal circuit schematic.
What's the difference?
The first represents the physical components and their connections. Useful for technicians, test engineers, etc. etc. but not for calculations and/or simulations.
For that, an ideal circuit schematic is used either explicitly or implicitly to translate the physical circuit into a mathematical model that can be used to calculate and simulate.
For example, here's the schematic symbol for an ideal transformer with the secondary connected to a load:
Unlike a real, i.e., physical transformer, the ideal transformer is lossless and has infinite bandwidth. How would one calculate or simulate a real transformer? By augmenting the ideal circuit schematic with additional ideal circuit elements that model the non-ideal characteristics.
For example, an ideal circuit model of a real transformer looks like this:
Note that every circuit element in that diagram is ideal and thus, isn't physically realizable but the entire ideal circuit corresponds to a good mathematical model of a real transformer that can used for calculations and simulations.
To further drive this point home, let's consider the OPs schematic as a "wiring" diagram for a physical battery connected with wires to a physical resistor.
Since this a DC circuit, a simple model of a battery is an ideal voltage source in series with some small value ideal resistor. A simple model of a physical wire is small value ideal resistor. Thus:
But, and again, each circuit element above is ideal including the wires that connect the ideal circuit elements.
And, again, for the ideal wire, there is no voltage across for any value of current through. This defines the ideal wire and that's really all that needs to be said about this.