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17

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 ...


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 ...


10

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 ...


9

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 ...


9

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 ...


7

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 ...


6

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 ...


6

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 ...


6

Typically batteries involve using the energy stored in some chemical compound (for example, we have batteries of type Lithium-ion). So what happens is we use the potential energy stored in the chemical as an electromotive force to power our device. Now what happens when allt he potential energy stored inside that compound runs out? We have to recharge it. ...


5

One alkaline AA cell has about 11 kJ of energy. For a laptop battery, it is 360 kJ. Chevrolet Equinox Fuel Cell has 58 MJ of energy. One kilogram of TNT carries about 4.184 MJ of energy. Divide the numbers from the previous paragraph by this constant to see that the AA cell, laptop battery, and electric car battery have 2.6 grams, 86 grams, or 14 kilograms ...


5

Crazy Buddy's answer and related comments have made the point that you could indeed use a capacitor to charge a battery, but the amount of energy stored in capacitors is generally less than in batteries so it wouldn't charge the battery very much. However there is a new generation of capacitors called ultracapacitors that are being developed with electric ...


5

I am going to assume what you are thinking about is some sort of power pack that uses AA batteries like this one. There is no physical reason that you can not use AA batteries to charge a cell phone, as long as you have the correct adapters. What might be confusing the issue is that Li-ion batteries have definite charging issues. If overcharged, they can ...


4

The battery has in both cases the same energy content, so it just depends on which method uses more energy per time. This power depends on the resistance $R$ you use to connect both terminals, with a given voltage $U$ derived from Ohm's law: $$P = U^2 /R$$ So, the smaller the resistance, the faster your battery will lose it's stored energy. The copper wire ...


4

An ampere passing through your heart can give you a heart attack. An ampere passing through a wire will not. The human body has a fairly large resistance ($10000\ \mathrm{\Omega}$ perhaps?), so the same voltage that can make a large current pass through a copper wire will not necessarily make any significant current flow through a person.


4

What you are describing is called a series connection. Think of the batteries as pumps, with each pump generating 1.5 PSI. If you connect the outflow of one pump to the inflow of another, then overall the two pumps are going to generate 3.0 PSI. Note that the current capability is not doubled. If the two pumps are each rated for 1 gallon/minute, then the ...


3

Yes, capacitors could function as a battery, but there are some technical problems. A battery generates electricity from a chemical reaction so the voltage and power produced remains approximately constant for the lifetime of the battery. The voltage on a capacitor is proportional to the charge on it so the voltage decreases as you discharge the capacitor. ...


3

From the specifications of your battery, that is 1.5V and 2700mAh, you can compute that there is $14580$ Joules of energy stored in your batteries. The formula $P=U\cdot I$ relates power to voltage and current. You battery specs give voltage and capacity (that is total charge stored). The former is in Volt, the latter in milli-Ampère-hour. The product is ...


3

Apparently Hydrogen/Oxygen are liberated when a Lead-acid battery is charged NOT Always. Just explaining how? Lead-Acid Battery comes under Secondary cells. An LA battery usually has plates of lead & lead oxide (when fully charged) or lead sulfate (when fully discharged) in an electrolyte of 35% sulfuric acid and 65% water solution. Indeed, ...


3

Let's take your question in my thought... For instance, let us assume that we've got a capacitor of capacitance about some $100 \mu F$ and Also, a commonly used $\text{Ni-mH}$ battery of some voltage $1.5 \text V$ with charge capacities about $2000 \text{mA-h}$ Energy of the battery would be $E=1.5\times 2000\times10^{-3}\times3600$ $$=1.08\times10^4 J$$ ...


3

Internal resistance generally increases as you lower the temperature. This is because batteries generate current using a chemical reaction and the reaction generally goes slower at lower temperatures. This article on Wikipedia claims the internal resistance of a manganese alkaline battery increases from 0.1 ohms at 40°C to about 0.9 ohms at -40°C. Car ...


3

Watts (electrical power) = Volts $\cdot$ Amps, so 25W = 12V $\cdot$ 2.1A 150Amp Hour is the total capacity so 150amp $\cdot$ 1hour, 1amp $\cdot$ 150hours, or 2.1amp for 72hours. That's in an ideal world of course, there are heating losses as you charge the battery, the voltage of the solar panel varies with the load and if you entirely empty a 12V lead ...


3

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 ...


3

You asked too many questions to make sense answering individually. 12 V is 12 V, but what distinguishes a car battery from a bunch of AA cells strung together is that it can still maintain this 12 V at high current, like what a starter motor requires to crank a cold engine. Actually even a car battery will exhibit a voltage dip when the large starting ...


2

You have (1.5V)(2.7A)(3600s) = 14580 Watt-seconds = 14580 Joules of energy in the battery. BUT you can only get that much energy out if you extract it the same way the manufacturer did it during their optimized tests, which is typically at a much lower amperage than what you want to drive your motor. Also, remember that there are friction and other losses. ...


2

The mAh hour rating of a battery is how much energy it stores. It's simply the number of mA it can deliver for an hour (in theory) The power of a battery is how much energy it can deliver in a certain time. Since Power is current * voltage, and the voltage of a battery is (almost) fixed, then the maximum power is related to the maximum current. Which ...


2

Why this difference? What is the exact cause for the difference in the discharge times? The difference is due to the difference in the amount of energy stored. Consider, for example, a typical alkaline cell. From this chart, we see that a new alkaline long life AAA cell stores about 5kJ of energy. Now, consider a large electrolytic capacitor of, ...


2

Some complementary remarks: Seen as black boxes, both are simply voltage sources. A battery is designed to supply, as far as possible, a constant voltage whereas a capacitor has no such "dosage capabilities". From a applied standpoint, batteries store a lot of charge, but are slow to charge/discharge. Capacitors are the exact opposite: low storage, fast ...


2

You have a typo that the Amperes output is 4.936, but that is corrected later on. You also slipped a decimal when you did 4.936 * 0.2, which should give 0.9872. The battery is at 130V, so the current out of the battery is $\frac {235}{130}\cdot 4.936=8.922$ and it is really a little higher for the losses in the conversion. Let's say it is 9A out of the ...


2

This is really more of a question about chemistry rather than physics, although ‘expected volume’ might give it a hint of physics… Anyway, you can get the volume of a given gas with a specific pressure (usually 1013 hPa) and temperature (probably about 300 K) from the number of molecules in that gas using either the Ideal Gas Law or the Van der Waals ...



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