Power and potental difference - battery

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Discussion Overview

The discussion revolves around the application of Ohm's Law to batteries, particularly in scenarios involving short-circuiting. Participants explore the implications of internal resistance, potential difference, current behavior, and the relationship between voltage and current in both batteries and solar cells. The scope includes theoretical considerations and comparisons of I-V curves.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant questions how Ohm's Law applies when a battery is short-circuited, noting that the terminals should theoretically be at the same potential with low resistance but are not.
  • Another participant asserts that while Ohm's Law is always valid, the scenario with an ideal battery and superconductor leads to infinite power, which is a mathematical curiosity rather than a practical reality.
  • It is mentioned that the internal resistance of the battery limits the current in a short-circuit scenario, leading to a small terminal voltage and large current.
  • One participant states that the voltage across the terminals is the battery's rated voltage until it is short-circuited, at which point the voltage drops significantly, and the internal resistance becomes the primary limiting factor for current.
  • A participant expresses interest in comparing I-V curves for solar cells and batteries, noting the characteristics of short-circuit current and open-circuit voltage for both.
  • Another participant clarifies that the voltage across the battery terminals decreases as the load resistance decreases, approaching zero in a short-circuit condition.
  • A follow-up question is posed regarding whether the Voltage-vs-Current plot for a battery descends monotonically from open-circuit voltage to near zero at short-circuit current.

Areas of Agreement / Disagreement

Participants express varying views on the behavior of voltage and current in short-circuit conditions, particularly regarding the role of internal resistance and the implications for power. There is no consensus on the exact nature of the I-V curve for batteries compared to solar cells, indicating ongoing exploration and debate.

Contextual Notes

Participants highlight the importance of internal resistance and the conditions under which Ohm's Law applies, but there are unresolved aspects regarding the exact behavior of voltage and current in extreme scenarios.

ajace
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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.
 
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"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.
 
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.
 
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.
 
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?
 
Drakkith said:
The voltage (potential difference) between the terminals is whatever the battery is rated at.
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.
 
nasu said:
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.

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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