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Batteries usually use electro-chemical reactions to store energy. These reactions have a limit to how fast they can transfer that energy. For example, a typical lead acid car battery can only draw so much energy; after a certain point it begins to break down, producing hydrogen gas which then can ignite with free oxygen in the air. An analogy would be a ...


41

The defibrillator requires a high voltage to do its job. ordinarily this would require a very large battery stack (hundreds of individual cells) to achieve the voltage requirement. Instead, defibrillators use a smaller battery pack to drive a chopper circuit that steps the voltage up through a transformer, after which the result is rectified, filtered, and ...


39

Yes, the total mass of a battery increases when the battery is charged and decreases when it is discharged. The difference boils to Einstein's $E=mc^2$ that follows from his special theory of relativity. Energy is equivalent to mass and $c^2$, the squared speed of light, is the conversion factor. I would omit the scenario I. If the lithium is leaking from ...


30

There will be an increase of the mass of the battery when you charge it, though that increase is going to be undetectably small. I would do the calculation in reverse i.e. start with a fully charged battery and calculate how much it decreases in mass when you run it down. The mass decreases because the battery does work $E$ on the electrons that flow ...


29

Sometimes it is easier to understand circuitry in the context of water. What you're imagining is two tanks of water of equal size linked together by a pipe that has been sealed off. If one tank holds 5% water and the other holds 35% water, when you remove the seal, the tanks equalize and you end up with 20% in both tanks. What you're forgetting is that ...


25

It is energetically unfavourable to split a water molecule into the two ions $\text{H}^+$ and $\text{OH}^-$ i.e. you need to put in energy to do it. However at room temperature water molecules have a range of energies and there are always a few molecules with enough energy to ionise. So any sample of pure water at everyday temperatures always contains a few $...


24

Yes Sam, there definitely is electric field reshaping in the wire. Strangely, it is not talked about in hardly any physics texts, but there are surface charge accumulations along the wire which maintain the electric field in the direction of the wire. (Note: it is a surface charge distribution since any extra charge on a conductor will reside on the surface.)...


24

The key here is the voltage of both the batteries. The battery in the phone is generally at a voltage of 3.7V. The battery pack has a higher voltage or a circuit which gives a voltage of 5V to your phone. So, as long as the voltage with which you charge the phone is higher than that of the battery, the percentage of power in it doesn't matter and the phone ...


24

Connecting your phone to the battery pack doesn't directly connect the cells in parallel. I assume this is where your guess of an equilibrium with equal voltage -> equal charge percentage comes from. Shorting lithium-ion / lithium-polymer (LiPo) cells together like that would likely cause one or both to literally catch fire from the high currents, or from ...


23

The short answer is that although capacitors do not hold as much total energy as a battery the same size, they can release energy faster than batteries can. In a portable defibrillator (or a taser!) a battery charges a capacitor, then the capacitor releases the the charge into the subject much, much faster than it could have been supplied directly from the ...


23

The ability to deliver energy relatively quickly is basically the distinction between a "capacitor" and a "rechargeable battery". This isn't a physics factoid so much as just what the words mean. For example, in the below plot:$ {\require{color}} {\definecolor{capacitor}{RGB}{255,10,10}} {\definecolor{lightCapacitor}{RGB}{255,131,131}} {\definecolor{...


21

There are many reasons for this situation. Power produced is non-adjustable. The battery produces power at nearly constant rate (slowly decaying with time). It cannot be increased and if not consumed (or stored) the power is lost. (Mentioned by DumpsterDoofus) low power density. ${}^{63}\text{Ni}$ for instance produces ~5 W/kg (and kg here is just mass of ...


19

Batteries do not behave in such an ideal way across all conditions. The simplest model of a battery as a circuit element is the one you describe - a pure voltage source. A slightly-more sophisticated model is as a voltage source connected to a fixed resistor, called the battery's internal resistance. A typical battery has an internal resistance of between 1 ...


17

In ideal circuit theory, the parallel connection of two voltage sources results in an inconsistent equation, e.g., a 3V and 2V source connected in parallel, by KVL, gives the equation: 3 = 2. In the real world, batteries are not ideal voltage sources; batteries can supply a limited current and the voltage across the battery does, in fact, depend on the ...


16

A battery generates a voltage by a chemical reaction. There is a class of chemical reactions called redox reactions that involve the transport of electrons, and you can use the reaction to drive electrons through an external circuit. This is the basis of a battery. The battery will continue to provide power until all the reagents have been used up and the ...


13

First, your camera is not designed to work with batteries below a certain voltage. When it detects an excessively low battery voltage it turns itself off. That circuit stays in the "off" state until voltage is completely removed from the circuit. When you operate your camera, the current required by your camera varies according to what you do with it. So ...


13

You should not connect different batteries in parallel. If you do, the battery with the highest voltage will discharge into the other one, until they end up with equal voltages. If the second battery (the lower voltage one) is a rechargeable, then it will be charged by the first one, again until the two have the same voltage. In this case the end voltage ...


13

Mostly the problem is that in batteries, current flow is not by electrons as in something like a copper wire, but by physical movement of ions. Only so many ions can migrate to the right place and perform the right chemical reaction over some fixed time.


12

The resistance of water, even with ions and minerals and such, is still fairly high. So, a tiny current flowed through the water, but not very much. Additionally, the heating effect that often destroys them when short circuited would also be nullified by the cooling water.


11

Electrons that reach the positive terminal indeed remain there. The potential difference between the two terminals pushes electrons from the negative anode toward the positive cathode. When an electron reaches the cathode, it stays there to equalize the original charge imbalance between the two nodes. When electrochemical redox reaction sustaining the ...


11

Heat. Batteries have internal resistance and so produced heat when current flows through them (Joule heating). Also, the heat generated increases by the square of that current. E.g, doubling the charging current causes the heat produced to be increased 4 times. Ultracapacitors are a different technology that can be used like batteries--they have very very ...


11

One aspect that hasn't been covered in the other answer is what is really needed to make a defibrillator work reliably and safely. Defibrillating a heart isn't simply "OK, let's electrocute the patient"! In order not to damage the heart, a very careful application of energy is needed. That means that the defibrillator must create a "well behaved" electrical ...


11

Short answer: Potential is defined depending on the choice of a origin (e.g. ground). The positive plate of a capacitor has potential $Q/C$ greater than negative plate of the same capacitor. We do not know its potential compared to anything else, unless we know how they are connected in a circuit. Here the points a and b are connected by an ideal conducting ...


10

Once the battery is fully charged it will not accept any more energy (current) from the charger, since all the energy levels that were depleted when empty are now at their highest level. For example in a Lithium ion battery when all the ions have arrived at the proper electrode the resistance to more current becomes very large, but not infinite since there ...


10

The key thing is that there is NO electric field within the perfect wire. So, there is no force acting on the electron, and thus no work done on it (while it's in the perfect wire). This goes back to the definition of a perfect conductor (which the perfect wire is). Within a perfect conductor, there is no electric field. Instead, the charges (which have ...


10

I understand also that there would be a tiny minuscule resistive loss through the wire, but really it's not enough to say anything about. On the contrary, it's crucial. Assuming an ideal voltage source (can supply unlimited current) of voltage $V_S$, an ideal resistor of resistance $R$, and an ideal uncharged capacitor of capacitance $C$, are suddenly ...


10

The ions are converted into gases $H_2$ and $O_2$ at the electrodes, so water is gradually being removed from the container - but it requires a very very large amount of charge to flow in order to convert all of the water into gases. The $H^+$ ions exist as hydronium ions $H_3 O^+$. There is a reversible equilibrium in the water : $2H_2O <=> H_3O^+ + ...


9

Real batteries have a finite energy storage capacity. Adding additional cells adds additional capacity (this is why I would add them, you haven't really specified any context so it's hard to say what you are looking for). Also, it's worth noting: I'm assuming that all the cells are identical, and internal resistance is negligible. This is an important ...


9

I've just sacrificed an AA manganese alkaline battery to the cause of physics. When I first shorted the battery it produced a current of about 9.5 amps, which I thought was actually pretty impressive. However over the course of 30 seconds the current dropped to around 5 amps. The battery got pretty warm, though I don't think it would have set fire to ...


9

Consider for a moment, a cell that is not connected to a circuit, i.e., there is no path for current external to the cell. The chemical reactions inside the cell remove electrons from the cathode and add electrons to the anode. Thus, as the chemical reactions proceed, an electric field builds between the anode and cathode due to the differing charge ...


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