In Fundamentals of Physics (HRW), an equation is derived for the current in a circuit in terms of its emf and resistance by the 'Energy method'; that is, deriving $\epsilon = i \cdot r$ and saying that the work done per unit charge ($\epsilon$) by the (ideal) battery is equal to the thermal energy dissipated per unit charge (i.r).

We know that the (ideal) battery will pump charges with the same emf and do the same work on the charges if or if not the resistor is in the circuit. So if the resistor is not in the circuit, where will the work done by the battery go if there is no resistive dissipation ? In other words, what will happen to the battery if its ends are connected across a super conducting wire ? Will it discharge like a capacitor ?If not, please tell me why the electrical circuit theory is meaningless without resistance.

Many teachers give an analogy between electric circuits and fluid flow in a circuit consisting of pipes and a motor as a battery. If possible, can anyone give me such an analogy with a circuit with zero resistance, internal and external ?

Please feel free to edit and add relevant tags to the question.

  • 1
    $\begingroup$ A basic tutorial for MathJaX can be found here. $\endgroup$ – ACuriousMind Oct 30 '14 at 15:57
  • $\begingroup$ Thanks! I was really looking for some site which gives step-by-step instructions. $\endgroup$ – Gaurav Oct 30 '14 at 16:08
  • $\begingroup$ The battery has an internal resistance which would discharge it even with a zero resistance wire between the terminals(from a quasi-real world standpoint). There is a big part of electrical circuit theory that works without resistors. Just think about a circuit with a sinus source and some inductors and capacitors. This is mainly used for processes where the resistance is much smaller than the reactance(search for it on the internet if you don't know what it is). $\endgroup$ – WalyKu Oct 31 '14 at 14:54

Your original text admitted three interpretations, and I'm leaving the answers here:

1: What happens with a toy model when there's a circuit with an ideal battery and no resistance?

All the charge moves around the circuit at one moment in time (infinite current). The energy must leave the system as Electromagnetic radiation - accelerating charges radiate, and while that radiation would happen at any bends in the wire, it would probably happen most at the terminals where the charge goes from stopped to moving (and visa versa). The effect of energy leaving a system via EMR is often ignored in circuits, but basically, we've made a single signal broadcast antenna.

A battery represents two separate reservoirs of charge, so if charge moved between them on the terminal side, the battery would be a rock. Therefore, the work done by the current must happen between the terminals on the wire side. Don't worry too much about this scenario - there's no such thing as ideal batteries. Their internal resistance is orders of magnitude smaller than a circuit's resistance, and orders of magnitude larger than that of the wire.

2: What happens when I connect the terminals of a real battery with an ideal (superconducting?) wire?

Real batteries have internal resistance, so the battery will heat up and quickly either run out of charge or burn/explode.

3: What happens when I connect the terminals of an ideal battery with a real wire?

We specify battery voltage in circuits, so the current running through the wire will be high enough that when you multiply it by the resistance, you get back the battery's voltage. Also, the wire will heat up quickly and radiate light and heat until the ideal battery runs out of charge.

Some of the other answers included other ideas.

First, any circuit is a loop, so it will have an inductance. Inductance slows down the current in a circuit, but does not effect the circuit in steady state (or provide a real (pun intended) voltage drop). In the ideal battery/wire case, the inductance would cause the current to grow over time - that's nonsense because infinite current can't grow (you also need non-zero resistance to find the time constant - I don't divide by zero).

Second, we sometimes think of batteries as chemical capacitors. An ideal battery is not a capacitor. But if it were and there were an inductance in the circuit, the charge would move from one side to the other with an angular frequency of $(LC)^{-1/2}$. In response to the edit question, 'Will it discharge like a capacitor?', the time constant for an RC circuit is $RC$, so zero in this case. The battery won't send charges back the other direction in the circuit though because ideal batteries are not capacitors.

Incidentally, a capacitor also slows down the movement of charge, but it reverses the polarity, so the phase moves in the opposite direction. Also of note, whereas an inductor slows down changes in current most when they change most, a capacitor allows the freest flow of current when it is uncharged.

Lastly, it seems pretty clear that you're talking about a closed circuit, but if you weren't, well, nothing happens on open ideal circuits (unless they were recently closed, or will be closed soon).

Responding to other edits:

"If not, please tell me why the electrical circuit theory is meaningless without resistance."

I'm not sure what you're asking, but can we build circuits with just transistors, inductors, capacitors and diodes? I guess, but it'd be a lot more difficult to keep the magic smoke in. Circuits would also be a lot more difficult because we often model speakers, lights, motors and almost every useful thing in a circuit as a resistance. LC circuits (which have no resistors, but non-zero resistance) have a few important applications, but even so, we often put resistors in to dampen non-frequency signals or manage the voltage (with a voltage divider for example).

"If possible, can anyone give me [a fluid flow] analogy with a circuit with zero resistance, internal and external?"

I refer you to the Waterfall, though I had hoped for an aquatic image shaped more like Ascending and Descending. Water does not have an easy analogy for electromagnetic radiation because they are different phenomena. In another direction, flow of fluids is tremendously resistive, so perhaps the analogy you're looking for is that as fluids (and circuits) get colder, resistance goes down. The behaviors of both of these systems are subject to laws that are very foreign to our understanding as warm intuitioned creatures, and they won't help you in your circuits class.

  • $\begingroup$ Thanks for the time and clearing my doubts question by question. I'm extremely sorry for the late reply. The answer explains the actuality pretty clearly and comprehensively. My original question was almost entirely based upon the first interpretation of your answer. You say that there is an 'infinite' current in the circuit. Can you define it in physical terms ? A battery is not a wire, so the current cannot pass across its terminals. "As such, the work done by the current must happen between the terminals on the wire side." What do you mean by that ? Can you elaborate ?.... $\endgroup$ – Gaurav Nov 3 '14 at 15:05
  • $\begingroup$ Also, I think current 'does' pass through the battery. The very reason it has to do work is because it has to make electrons pass from a lower potential to a higher potential as electrons go from - to + terminals within it. I think other than these, the answer does quite well to answer my question. Thanks. $\endgroup$ – Gaurav Nov 3 '14 at 15:10
  • $\begingroup$ When I ask: "Why is the electrical circuit theory meaningless without resistance ?", I mean why is not a current defined without a resistance ? People cite Ohm's Law, but it's just an experimental observation. If you think the same in the hydraulic analogy (roughly), the motor keeps pumping the fluid and once the flow reaches a particular speed, it stops doing work since it cannot accelerate it further. The fluid 'flow' doesn't become infinity. $\endgroup$ – Gaurav Nov 3 '14 at 15:33
  • $\begingroup$ Infinite current would be all of the charge on one side of the battery moving at one instant to the other side. Current is $dQ/dt$, so because $t$ approaches zero from the positive side, we call that infinite current. $\endgroup$ – user121330 Nov 3 '14 at 16:07
  • $\begingroup$ Current does not pass through a battery. A battery has two reservoirs of opposite signed charge. The charge cannot pass from one reservoir to the other without going through the terminals. If it could, the reservoirs would equalize and you'd have a rock, not a battery. $\endgroup$ – user121330 Nov 3 '14 at 16:11

The battery doesn't do any significant work if there's nothing connected to it.

You can think of a battery with nothing connected to it as being a battery with an extremely large resistor connected to it. That's even essentially physically accurate, instead of just being an idealization, because although the resistivity of air is extremely high, it's not infinite. The result of the extremely large resistance is that the current is extremely small, i.e., there are very few charged particles per second on which the EMF performs work. But since EMF is the work per unit charge, it's not inconsistent for the EMF to remain constant, even though the total work done by the EMF is extremely small.

  • $\begingroup$ I get your point that current will decrease with increase in resistance to keep the emf constant. But what does having zero resistance in the circuit mean ? And therefore, what does current becoming infinity mean ? $\endgroup$ – Gaurav Sep 20 '14 at 3:14
  • $\begingroup$ Although resistance can be "practically infinite", current can't. A battery, or any other source of EMF such as a generator, has a small, unavoidable internal resistance that can be treated as being approximately zero only as long as it's much smaller than the external resistance connected to the battery. If you short out the terminals of a battery with a wire (which has "practically zero" resistance), the current that flows will still be finite, due to being limited by the battery's internal resistance. $\endgroup$ – Red Act Sep 20 '14 at 18:51
  • $\begingroup$ One last question : Consider the hypothetical case where internal resistance is zero too. Using the loop rule, the emf in such a case will equal zero. But, the emf can't be zero because otherwise the battery will discharge like a capacitor, bringing both terminals to the same potential in no time. Or, will the magnitude of current increase progressively with time in such a case ? Thanks anyways ! $\endgroup$ – Gaurav Sep 21 '14 at 11:55

In a perfectly ideal circuit without any resistive losses (even internal to the battery), energy will flow to fill capacitance with charge, and be stored as an electrical field but the transient flow will also create magnetic fields in any inductance. Without loss, the energy will bounce back and forth between the inductance and capacitance as magnetic and electrical fields. That initial work is not lost, just bouncing back and forth between the fields.

  • $\begingroup$ Thanks for the answer! But I do not understand what you mean by "the energy will bounce back and forth between the inductance and capacitance as magnetic and electrical fields". Please can you elaborate ? What do you mean by "bounce back and forth" ? $\endgroup$ – Gaurav Sep 21 '14 at 12:02
  • $\begingroup$ Energy has an interesting behavior in that it can manifest itself in different 'forms'. Just as the energy in a pendulum flows periodically back and forth between kinetic and potential (due to gravity) so does the inductance-capacitor circuit, but in this case the energy is manifested as magnetic and electric fields. Without any means to 'lose' energy (friction & electrical resistance) the energy will perpetually flow back and forth. Your ideal circuit, without resistance traps an initial charge from the battery and the energy from this charge 'bounces' back and forth between the two fields. $\endgroup$ – docscience Sep 21 '14 at 15:47
  • $\begingroup$ Will inductance play any role in circuits with resistance ? $\endgroup$ – Gaurav Sep 21 '14 at 16:18
  • $\begingroup$ Also, what does the emf device do if it is still connected to the circuit ? I think the answer to these will be the answer to the main question. $\endgroup$ – Gaurav Sep 21 '14 at 16:34
  • $\begingroup$ Inductance will impede change in current that flows through the inductance. Capacitance will impede change in voltage applied across the capacitance. Real passive component circuits have to some degree capacitance, inductance and resistance, even if the only tangible components are resistors. An ideal voltage has zero output impedance so it does not load, nor dissipate the energy that is initially imparted - say to a series L-C circuit. So keeping it connected is required to keep the circuit closed and continue oscillations. For a parallel L-C circuit, the emf can be disconnected ... $\endgroup$ – docscience Sep 21 '14 at 22:03

In addition to the good answers above here is something to think about.

If the resistance is zero then for a current to flow there does not need to be a battery - the emf can be zero. No work is done as the current flows.

This may sound like a strange case, but it is how many strong magnets work in nmr (nuclear magnetic resonance) spectrometers. The have a 'loop' of superconducting 'wire' and no battery in the circuit. Magnetic inductance is used to start a current flowing in the loop. And the current keeps flowing in the loop. Forever. Provided that the superconductor remains a superconductor. In practice, this means keeping the 'wire' cold, frequently with liquid Helium. If the 'wire' stops superconducting then resistance goes up, energy is dissipated and it can be expensive.

If I remember correctly the problems with the large hadron collider and the hunt for The Higgs Boson was held up by similar damage to magnets / loss of liquid helium following an electrical fault where part of the problem was superconductors heating up and not being superconductors any more.

(Next part in response to comment below)

so what if a battery with zero internal resistance was connected...

... first I refer again to the other good answers to this question,...

... but I would like to add the point that when a superconducting ring of wire is 'charged up' with current by magnetic indcution it is a bit like a battery being connected to the circuit and the energy goes into the magnetic field generated. Normally, magnetic induction generates an emf or potential difference in a wire - with a superconductor the resistance of the wire is zero and thus a current starts to propagate immediately to counteract the magnetic induction so that in fact no emf or potential difference is induced.

Hope this is helpful to think about this question with respect to real physical superconducting magnets used today.

  • $\begingroup$ But what if you do connect an emf device with no internal resistance across the superconducting wire ? What does the battery do ? Since it is an emf device, it will not discharge like a capacitor. So will it progressively increase current ? $\endgroup$ – Gaurav Oct 31 '14 at 2:57
  • $\begingroup$ In response to this comment I have put an edit into the answer. NB some real batteries do discharge like capacitors, but they also have internal resistance. $\endgroup$ – tom Oct 31 '14 at 9:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.