Does my laptop/cell phone battery charge up like a capacitor?

[LogicX]

This would be fun to know. So, if I need to leave my house quickly, is it worth it to throw my dead phone on the charger for 10 minutes because that initial charge time is the most efficient, or is charge a linear relation with time?

 

[phinds]

No, it's a battery. A battery is not a capacitor. I don't know the charge curve for a battery but I assure you it is NOT the exponential curve of a capacitor.

 

[Drakkith]

I'm 90% sure that it is worth it to put your phone on the charger for a few minutes. I KNOW my phone doesn't take anywhere close to the same amount of time to charge as it does for the battery to be used. At WORST you would be barely getting those 10 minutes if the phone didn't charge at all, as you would be powering it from the wall for that amount of time.

 

[cjameshuff]

This would be fun to know. So, if I need to leave my house quickly, is it worth it to throw my dead phone on the charger for 10 minutes because that initial charge time is the most efficient, or is charge a linear relation with time?

The charge curve's not exponential, though it may resemble an exponential curve for some types. An ideal battery would have a flat curve, voltage being determined entirely by the chemistry, power going into causing a reversible set of chemical reactions at a constant rate. In the real world, it's something specific to the battery type that depends on how the resistance of the plates and electrolyte, concentrations of compounds, surface area, etc change during the charging process. Likely fairly constant for most batteries that aren't almost totally dead or mostly charged, however.

As for your dead phone, some devices charge faster than they use their battery. 10 minutes on the charger could mean a couple extra (short) calls.

 

[Borek]

Nernst equation is what describes the voltage when you refer to the chemistry only.

For the reaction

Red^m <-> Ox^(m+n) + ne

cell potential is

$$E = E_0 + \frac {RT}{nF} \ln \frac{a_{Ox}}{a_{Red}}$$

where F is a Faraday constant, a_Ox and a_Red are activities of the reduced and oxidized form. Expression under the logarithm may take more complicated form for more complicated reactions. Besides, there are always two reactions taking place, and the observed potential is a difference between potentials of both half cells, but the total voltage will be still described by the Nernst equation, just the ln part will change. Activities change almost linearly with the charge level.

 

[pa5tabear]

Nernst equation is what describes the voltage when you refer to the chemistry only.

For the reaction

Red^m <-> Ox^(m+n) + ne

cell potential is

$$E = E_0 + \frac {RT}{nF} \ln \frac{a_{Ox}}{a_{Red}}$$

where F is a Faraday constant, a_Ox and a_Red are activities of the reduced and oxidized form. Expression under the logarithm may take more complicated form for more complicated reactions. Besides, there are always two reactions taking place, and the observed potential is a difference between potentials of both half cells, but the total voltage will be still described by the Nernst equation, just the ln part will change. Activities change almost linearly with the charge level.

In a real world battery, which factors do we need to think about in addition to the chemistry?

 

[Borek]

Mostly internal resistance. At least that's what I know, could be I am missing something.