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Power and potental difference - battery

  1. Sep 20, 2011 #1
    I'm having trouble understanding how Ohm's Law applies to a battery that is being short-circuited. If the terminals are both connected by a low-resistance wire, then they should be at the same potential, since there's no resistor to cause a drop in potential. But the terminals of the battery are certainly not at the same potential, by definition. So what will happen to the voltage and current in this circuit? Will it simply be that the battery maintains its potential difference (say, 9V) and the current goes to infinity (with ~0 resistance)? Would the power then go to infinity also? Or is Ohm's law not valid in this case, as we're no longer operating in the intended regime for the battery? Does the internal resistance of the battery play a role here? Thanks.
     
  2. jcsd
  3. Sep 20, 2011 #2
    "connected by a low-resistance wire". That has a not null resistance, anyway. Ohm's law is valid always, just it may not tell you a lot if you are working with an ideal battery, with no resistance and which is connected by a superconductor. The power provided by the battery would be infinite, but of course this is just a mathematical curiosity.

    In reality, we can have wires and batteries with small (but not null) resistances. In this case, the current would be big (but finite), and thus the power provided P = i^2 R, which is some finite (probably big) number.
     
  4. Sep 20, 2011 #3

    xts

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    The battery has some internal resistance - not that small. So if you short-circuit it by low resistance wire, the voltage on terminals will be very small, and the current will be large, but limited by internal resistance rather than short-circuiting wire resistance. Most of the power will dissipate inside the battery rather than in your wire.
     
  5. Sep 20, 2011 #4

    Drakkith

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    Staff: Mentor

    The voltage (potential difference) between the terminals is whatever the battery is rated at. A very low resistance wire will cause a very large current until the battery is dead or damaged. Once the battery is dead the voltage between the terminals drops to 0, meaning that they are at the same potential now. Internal resistance of the battery will also come into play and will be the main limiting factor for current. Since power is just volts x amp = watts, peak power will be very high, but the duration will be very low, so your total power is about the same, disregarding losses and battery damage of course.
     
  6. Sep 23, 2011 #5
    OK thanks for the replies; this confirms some of my suspicions. I'm trying to examine the I-V curves for both a solar cell and a conventional battery and see how they differ. I have attached a picture of the solar cell I-V curve.

    images?q=tbn:ANd9GcQhH1CsK9fO6EEAF87n-CHLpOGzIKsBM9lCztV1Gaj0ekrIL_J5.jpg

    It has a y-intercept at the short-circuit current value and an x-intercept at the open circuit voltage. I'm having trouble picturing what the corresponding diagram would look like for a battery, however. It seems something similar should apply, with an open circuit voltage and zero current with a very large load, and a short circuit current and very low voltage with a low-resistance wire connecting the terminals. Is this the case? If so, does the I-V curve for a battery not differ much from that of a solar cell?
     
  7. Sep 23, 2011 #6
    Only if there is nothing connected to the battery. Otherwise the voltage is lower than the electromotive force of the battery (the rating of the battery). It decreases as the resistance of the load decreases (and the current increases).
    For short circuit the voltage between the terminals drops to (almost) zero.
     
  8. Sep 24, 2011 #7
    So then does the Voltage-vs-Current plot of the battery descend monotonically from the V_oc to 0 (or close to it) at I_sc?
     
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