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Superconductors and Energy Conservation

  1. Aug 21, 2009 #1
    A classic problem in freshman electrodynamics is as follows: We have a capacitor (capacitance C1) charged up w/ charge Q, which means the energy in the system is:

    [tex]U_1 = \frac{Q^2}{2C_1}[/tex].

    We then disconnect the battery, and connect a second capacitor (capacitance C2) in parallel w/ the first. Since the capacitance of capacitors in parallel adds arithmetically, the new system has a total energy of:

    [tex]U_2 = \frac{Q^2}{2(C_1+C_2)}[/tex]

    Or in other words, [tex]U_2 < U_1[/tex].

    The question is, where did the "lost" energy go?

    The answer is that if we model the circuit as having a non-zero resistance, and we calculate the power dissipated in the resistor as charge moves from one capacitor to the other, we derive an expression that exactly accounts for the "lost" energy (I won't go through the math here). So there's no real big mystery here.

    My question is: what happens if this circuit were "super-conducting"?

    That is, what if the wire connecting the capacitors were truly resistance-less? Then do we genuinely have a paradox, in the sense that conservation of energy is violated? Or do even super-conductors have some small, tiny non-trivial resistance, through which the charge discharges, and which accounts for the "lost" energy?
     
  2. jcsd
  3. Aug 21, 2009 #2
    There are actually two wires connecting the two capacitors. They will still have mutual inductance, even if the resistance is zero. So you will have an LC (resonant) lossless circuit. As soon as you start exchanging stored energy between capacitors and inductors, the energy will continue oscillating until the energy is radiated away.
     
  4. Aug 21, 2009 #3

    f95toli

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    Also, superconductors are only lossless at DC. "Suddenly" disconnecting something means that you have a transient current, i.e. a current with high frequency components.
    So, a real circuit might oscillate for a while but these oscillations will eventually die out, mostly due to resistive losses in the circuit (radiation losses are usually negligible even for superconductors).
     
  5. Aug 21, 2009 #4

    Vanadium 50

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    And superconductors have a maximum current they can carry. Having a voltage across one when the resistance drops means the current shoots up, and if does too much of this the superconductor will go normal.
     
  6. Aug 21, 2009 #5
    I made a LTSpice model (see thumbnail) of the capacitor C1 (2uF, 10 volts), discharging into capacitor C2 (1 uF, 0 volts), through a 10 nsec long, 300 ohm transmission line, representing a superconducting cable connecting C1 and C2. The 300 ohms represents the two superconducting cables laid out in close proximity on a table. The delay length is long, just to observe the cable reflections. The green trace is the current leaving C1 (use right scale). The red trace is the current entering C2 (right scale), and the black trace is the voltage on C2 (left scale).
    at t=20 nsec, C1 is discharged into the transmission line. Because the voltage on C1 is 10 volts, and the transmission line impedance is 300 ohms, the initial current pulse (green) is about 33 milliamps. At 30 nsec, the pulse (red) arrives at C2. At 40 ns, the reflection arrives back at C1 (green), etc., etc.,. The black curve is the voltage buildup at C2. At 200 nsec, it is 55 mV, and appears to be rising quadratically. After many microseconds, the voltage pulse begins to plateau at C2, and eventualy the energy returns to C1.
    There is no current surge, primarily because the transmission line impedance is 300 ohms, which limits the current from a 10 volt source. No conductor carrying any current was disconnected (from the battery), so there is no voltage spike. At these currents, there is no flux penetration into the superconductor. At hundreds of amps, the magnetic field would begin penetrating into the cable (Meissner effect in type 2 superconductor), which would probably be asociated with resistive losses.
     

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    Last edited: Aug 21, 2009
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