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I've learned that a resistor converts some electrical energy into heat energy while the current flows through it and thus causes a power loss, but what if there's not any resistor in a circuit. Will current still flow?

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    $\begingroup$ The wires in a circuit have resistance, as does the battery supplying the potential difference (this is known as internal resistance). Usually in calculations which involve resistors, we neglect the resistance of the wire because the resistor's resistance is so much larger. Thus, the current will flow and heat will be dissipated from the wire into the surroundings. What do you think would happen in the theoretical case of no wire resistance and also no internal resistance? $\endgroup$ – Will Jul 3 '13 at 18:07
  • $\begingroup$ That's the point. The concept of a resistor is not getting clear in my mind. Can you explain the use of a resistor? $\endgroup$ – Syed Sahl Jul 3 '13 at 18:12
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    $\begingroup$ Have you tried reading the wikipedia article? en.wikipedia.org/wiki/Electrical_resistance $\endgroup$ – Will Jul 3 '13 at 18:24
  • $\begingroup$ Physics cannot provide an answer to a non-physical situation. Zero resistance does not exist. Your question has no answer. I'd vote to close if there were no bounty. $\endgroup$ – garyp Aug 1 '16 at 16:05
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    $\begingroup$ @garyp zero resistance doesn't exist now? Someone tell Kamerlingh Onnes... $\endgroup$ – CR Drost Aug 2 '16 at 13:01
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Short version: Yep. That's what a short circuit is.

Typically, you will get current to flow as long as it has a path with finite resistance (even zero), a voltage difference, and a supply of charge carriers (e.g., electrons). If there really were no resistance in the circuit, the electrons would go around the circuit, and arrive back at the beginning of the circuit with as much energy as the potential difference (the voltage). That final energy is usually what is dissipated as heat or other types of energy by the circuit. But without resistance (or inductance) it won't have a chance to lose the energy, and will return to the voltage source with lots of energy, which will typically screw up the voltage source. This is essentially what a short circuit is.

However, in any realistic circuit (including a short circuit) and with any realistic voltage source, you will always have some resistance, even if you do not have something specifically designed to be a "resistor". For example, even a normal wire has some resistance. That resistance is so low that we usually ignore it, because other things in the circuit usually have far larger resistances, so it's usually a good approximation to ignore the wire. But when it's just the wire, you can't ignore its resistance. Current will flow, and since the resistance is low -- though not zero -- you will just get a really large current. This will heat up the wire just like any resistor (as you know). And typically that will cause problems like melting the wire or its insulation, or just starting a fire.

But suppose you used a perfect superconducting wire. Well even then, any realistic voltage source still has what's called "internal resistance". You need to add the voltage source's internal resistance to the resistance of everything else in the circuit to get the total. Again, this is frequently so low that we just ignore it -- but we can't when the circuit is just a superconducting wire and a voltage source. Of course, any realistic voltage source will also have a limit to the amount of current that it can supply, as Wikipedia will tell you. But if you really minimize the resistance of the total circuit, then you will typically maximize that current.

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  • $\begingroup$ You are saying achieving 0 resistance is not practical. What if i theoretically take a 0 resistance wire and have a potential difference across it ? $\endgroup$ – Shubham Sep 15 '15 at 15:54
  • $\begingroup$ Reread the last paragraph of my answer. Even with a "0 resistance wire", you still have internal resistance in your voltage source, which means that a finite amount of current flows. The electrons convert electrical potential energy at one end of the wire into kinetic energy at the other end. $\endgroup$ – Mike Sep 15 '15 at 23:00
  • $\begingroup$ @Shubham Let's consider a theoretical Ohmic circuit with zero resistance in the wire and zero internal resistance in the voltage source. Then Ohm's law requires that for a finite voltage, the current goes to infinity. Clearly such an example is infeasible in practice since we can't have infinite current, but luckily we don't need to worry about such a case since circuits with exactly zero resistance don't exist! However, analyzing this limiting case gives us insight that circuits with small resistances have large currents. This makes sense. With less resistance, the electrons can flow faster. $\endgroup$ – Ian Jul 31 '16 at 18:23
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Okay, there are a lot of interesting questions packed in here; I can definitely see why @rpfphysics put a bounty on this one!

Are resistors necessary in a circuit?

This answer is an emphatic no. Resistors are one component in our "ideal theoretical objects which help us model circuits," and they are a component which happens to be very important and necessary for modeling real-world objects, as each real-wire can usually be modeled (if there's not so much current that it e.g. melts) as an ideal-wire in sequence with an ideal-resistor with some small resistance. But we can certainly use models where there are no resistors.

Here's a simple example:

a voltage source at top left, +5V, feeds into one of two selection terminals of a 3-pole selector switch S. The constant terminal of S connects to a capacitor C which connects to an inductor L, both at right. The inductor connects to a wire which is held at ground by a ground terminal; this wire then connects back around to the other selection terminal of S: the switch can either break the loop and connect +5V to ground through L and C, or else the switch can close the loop and remove the +5V source from the system altogether.

Here you see a capacitor $C$ in line with an inductor $L$ , held with a constant voltage that can be connected and disconnected with a switch $S$. Suppose that before the voltage is connected there is no charge or current in the system; everything will therefore be at the ground voltage. Now if you connect the voltage source, current will want to flow in, opposed initially only by the inductor, which resists changes in current but not the currents themselves. The relevant differential equations are that $I = C \dot V$ for the voltage across the capacitor while $V = L \dot I$ for the voltage across the inductor, where dots are time derivatives; these can be solved to find out that $I(t) = I_0 \sin(t/\tau)$ where this time constant for the oscillation is $\tau = \sqrt{LC}$ while the current constant is $I_0 = +5\text{V} \cdot \tau / L$. The basic physics here is that when the current across the capacitor has built up 5V of potential across it, so that the capacitor doesn't "want" to charge anymore, the inductor sees a current across it. Not liking changes in current, the inductor keeps the current constant and this pulls even more current out from the voltage source, overcharging the capacitor while the inductor slows down. The capacitor then discharges back into the voltage source, running the inductor in reverse.

In the following image I have animated the above analysis. Arrows for current point in the direction of the current; arrows for voltage point in the direction of increasing voltage.

Animated version of the prior image where we see a current going like the sine of t, and both the inductor and the capacitor have voltages going like the cosine of t, specifically 5 (1 - cos t) for the capacitor and 5 cos t for the inductor, so that they always sum to +5V.

So you see, we can analyze these sorts of things.

Now you can imagine what happens when we flip the switch back! If there is any voltage left on the capacitor, this thing will just cycle that charge back and forth, back and forth, around the loop forever, with the capacitor now picking up negative voltage, then positive voltage, and so on. In practice if you build this with real components those little resistances will eventually kill this perfect sinusoidal signal, of course, but we can easily imagine the resistances going to 0 and the resonator being this perfect eternal sine wave. No contradictions here.

So why were there contradictions when we replace $L$ and $C$ with a straight wire?

Resistance stops paradoxes

Stop me if your high-school classmates asked you this one before: "What happens when an unstoppable force meets an immovable object?" Usually it's not clear what "unstoppable force" means, but it could be defined as "an object travelling with finite velocity and infinite mass, so that it cannot be accelerated." What happens when this meets an object which cannot move?

The answer of course is that these are physical idealizations, and that they are by definition incompatible. If you are modeling unstoppable forces, then by the definition of "unstoppable force", none of the objects in your model can be truly immovable. If you are modeling immovable objects, then by the definition of "immovable object", none of the forces in your model can be unstoppable. You have to pick one or the other for the model to make any sense at all, or else you need to make absolutely sure that they never come in contact without something else sitting between them which can absorb the paradox!

Well that seems pretty obvious when we're talking kinematics, but you've hit on the exact same thing in terms of circuitry! It turns out the definition of "ideal voltage source" is "unstoppable force": no matter what, I am going to transfer electrons from this point over here marked "voltage source" over to this point over here marked "ground" with an unstoppable force that will raise their potential energy by 5 electron-volts. And it turns out the definition of "ideal wire" is "immovable object": "this ideal wire will make sure that whatever voltage is on its one side is also mirrored exactly on its other side." So you can't combine them without something -- be it a resistor or an $LC$-pair -- to absorb that paradox. Resistors are just one way of putting something in between these so that the paradoxes don't break the equations.

If you don't do this, then the answer is in principle very simple, "an infinite current flows over the wire." That's simply what the equations demand, because $V = I R,$ in order to model $R=0$ with $V\ne 0,$ needs $I = \infty.$

In practice what happens? Well, these machines that we call "voltage sources" are not perfect; they weaken their voltage the moment you start pulling current off of them. This is called the "impedance" of the source line and it will limit the amount of current flowing across the wire. Furthermore the real wire will offer a little resistance. Finally, the real wire may well heat up to the point where it melts and the resistance thereby could go to infinity. The world fails to be ideal. But in the ideal model, where neither thing breaks, the current probably would just go to infinity.

And this sort of stuff becomes important for superconductors!

Superconductivity, by definition, is a current flowing without resistance. This is only possible if the voltage across the junction is 0, otherwise there would necessarily be both current and voltage and therefore an effective resistance $R = V / I.$ Probably the biggest stumbling block for undergraduates is the tendency to think of a voltage which drives the current; in this case the cause of the current is some "superconducting phase" which has nothing to do with the voltage! So when you put a voltage across the junction, it must necessarily drive some number of electrons+Cooper-pairs that it no longer acts superconducting and instead has a resistance. Superconductivity is not some magical $R=0$ cure to all our ills; rather it is more like an $R(I)$ function which for certain small values of the current $I$ happens to be 0. These things will cease to be ideal really fast when you start putting significant voltages on them, because that is the definition of 'resistance'.

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    $\begingroup$ The "unstoppable force" is perhaps better described as an "irresistible force". $\endgroup$ – sammy gerbil Nov 23 '18 at 12:15
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but what if there's not any resistor in a circuit. Will current still flow?

For simplicity, let's work in the context of ideal circuit theory where ideal wires have precisely zero resistance.

Here's a simple circuit with only a current source and a wire.

enter image description here

In this circuit, there is a 1A current circulating clockwise around the circuit.

So, the answer is yes, there can be a current in a circuit without a resistor.

The concept of a resistor is not getting clear in my mind. Can you explain the use of a resistor?

The concept of a resistor is very simple: it is a circuit element where the voltage across is proportional to the current through.

Resistors are used in a variety of ways including:

(1) developing a voltage proportional to current

(2) limiting current through a circuit

(3) dropping voltage from a higher level to a lower level (see "Voltage divider")

(4) dropping current from a higher level to a lower level (see "Current divider")

There are many others but these examples should get you started.

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  • $\begingroup$ If suppose I have a 5V battery and a wire (like the circuit above).What will be the maximum current in the circuit ?(Greater than 5A?).Is there any theoretical limit?I am neglecting inductance,resistance of the wires and the battery. $\endgroup$ – Karan Singh Dec 9 '15 at 16:34
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    $\begingroup$ @KaranSingh, for ideal circuit elements (ideal 5V voltage source, ideal wire), the KVL equation for the circuit is inconsistent, i.e., $5V = 0V$ for finite current through. Put another way, the current is 'infinite' (arbitrarily large). However, physical batteries have finite short circuit current and, excepting superconductors, physical conductors have non-vanishing resistivity. We model this by adding an 'internal' resistance $r_i$ in series with the ideal battery and another resistance $r_s$ in series with the ideal conductor to model the non-zero resistance. $\endgroup$ – Alfred Centauri Dec 10 '15 at 0:20
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Short answer yes. The current will flow through the wire. We ignore this when there is a resistor because the resistance in the resistor is much higher than through the wire.

Long answer.

There are a couple things at work here.

In reality it is not clear what you mean by "let's say there isn't any resistor" because then it is unclear what the battery is connected to. Thus I am going to go on a tangent and explain these things so you understand better.

Firstly When we say current flows through a wire and into a resistor we abstract away all the unnecessary information and focus just on what helps us solve the problem. The problem being how does a resistor affect the voltage, current and power in a wire. Since under normal operating conditions this is pretty simple we use ohms law (and KCL,KCI) to solve for this system.

Secondly All things tend to prevent the flow of electrons from a sink to a source. For instance let's say you have a battery. There is no current flowing from its positive to its negative end because both the air and the internal insulation of the battery are preventing current flow.

Back to your example. Let's say you have an wire connected to a positive and negative end of a battery. The current will flow through the wire to the negative end of the battery. Since the resistance of the resistor is much higher than the circuit with just the wire and since resistance in series add, most of the voltage drop will be across the resistor so we do not have to worry about the voltage drop across the wire. Thus we never talk about current without a resistor even though it would be there and in fact the wire would act like a resistor.

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Current flows from high electric potential to low electric potential which lies at two ends (terminals) of the cell because of difference in no. of electrons accumulated at two ends. If they (ends) are connected by an ideal conductor wire (zero resistance theoretically), it forms a close loop. It's quite certain that current flows through a closed loop as current simply is the flow of electrons and electrons possess the tendency to flow from high to low electric potential. As there is no resistance to oppose the flow, the flow of electrons will be vigorous, what we call short circuit.

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  • $\begingroup$ Suppose I apply 10 V. What will the current be? $\endgroup$ – garyp Aug 2 '16 at 2:18
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yes Current will flow. to be very simple ...

V=IR when where R is rsistance, I is current , V is potential

I=V/R if V is finite and R is 0 , current will be infinite or not defined

thus the situation itself is not defined or could not be attained or tested but as theoretically current will be infinity

even heat released out H=(R)(I^2) cannot be defined as (I^2) is infinity and (R) is 0 so the product of infinity and 0 is not defined

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protected by Qmechanic Oct 5 '14 at 8:31

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