Physicist: Chemical batteries use a pair of chemical reactions to move charges from one terminal to the other with a fixed voltage, usually 1.5 volts for most batteries you can buy in the store (although there are other kinds of batteries). The chemicals in a battery litterally strip charge away from one terminal and deposite charge on the other. In general, the more surface area the chemicals have to deposit charge onto, and take charge away from, the higher the current the battery can produce.
The best way to represent the way a real battery works is to replace the battery in a circuit with an ideal voltage source (which is what we usually think of batteries as being) and an imaginary resistor called the battery’s “internal resistance“. The internal resistance can be used to describe why an AA battery is incapable of generating an arbitrary amount of power; the more current that the battery creates, the more the voltage across the internal resistor drops according to Ohm’s_law (V=IR). You can picture this as being a little like pushing a cart; if the cart isn’t moving you can really put your shoulder into it, but as the cart moves faster it becomes harder and harder to apply force to it.
A car battery only produces 12 volts, which is the same as 8 ordinary batteries in series. That voltage is so low that you can put your dry hands on the terminals of a car battery and feel nothing (please don’t trust me enough to try it; I don’t even trust me enough to try it). And yet the internal resistance is so low that if you connected the terminals with a normal wire, the current in the wire would be so high that the wire would melt or explode.
You can model the way a battery dies by increasing the internal resistance. A nearly dead battery still provides 1.5 volts, but has a very high internal resistance so that drawing even a trickle of current zeros out the voltage gain.
The voltage across a capacitor on the other hand is always proportional to the charge presently stored in the capacitor (this is the defiition of capacitance). You can think of a battery as being like a water pump, always providing the same pressure, and a capacitor as being like a water ballon, the pressure increasing the more water is in the ballon. The amount of energy in a capacitor is much easier to measure because of this (if you can measure the voltage across it, you can know the energy immediately). But, because the voltage supplied by a capacitor changes dramatically as it drains, special adaptive circuits are needed to step down the voltage to a fixed, consistent level in order to power a device. Alternatively, the device can be made to work over a wide range of voltages, but that tends to be more difficult.
It’s only fairly recently that capacitors have become small and powerful enough to store energy on par with chemical batteries. A few decades ago electrical engineers would prank the new guy by asking them to go into the backroom for a 1 Farad capacitor, which at the time was ludicrous. The poor sap would be back there forever (electrical engineers think they’re so funny). However, that joke has run its course, because today you can buy an over-the-counter, several-thousand Farad capacitor, that’s small enough to fit in your hand (and they pack a punch)!
So, as a general rule of thumb, batteries have a fixed voltage but:
big or new batteries tend to have a low internal resistance, so they can deliver a high current
small or old batteries tend to have a high internal resistance, so they can’t deliver much current
dude that doesn’t help
It doesn’t really answer the question. Do try to explain in layman terms and make it as simple as possible.
Pls I did get the explanation pls explain in the language I will understand
You can absolutely touch both terminals of a car battery at the same time. Your skin’s resistance is high enough that you won’t feel the minimal amounts of current that will pass through you. I’ve done this many, many times safely.
Ok, but I don’t see any mention to how current is affected. Does the current stay the same and only the voltage changes due to internal resistance? Does a dead battery still put out the same current?
Anonymous should try standing barefoot in a puddle and wetting both hands while touching the car battery terminals for science.
Crafty – The current decreases. Voltage stays the same. Resistance increases. A dead battery produces no current.
Equation -> V=IR
Where V -> Voltage (volt)
Where I -> Current (amp)
Where R -> Resistance (Ohm)
In layman’s terms…
Over the life span of the battery, the voltage basically remains the same. However, the internal resistance increases. Which means, the current must decrease, in order to balance the equation.
This is proven with your car battery. Your battery powers the engine starter. When there isn’t enough “juice”, the engine wont turn over. It is because you do not have enough “cranking amps”. That is why, in colder climates, you will see batteries with higher “cranking amps”.
Any battery which is used commercially is basically an electrochemical cell. It converts the energy released during a chemical reaction to electric current. While the cell runs, the chemicals within the cell are used up, this in the words of “Physicist” and “Engineer” (Colourful names btw) increase the internal resistance of the cell. In theory the cell is supposed to have the same voltage, but IT MAY NOT (There’s a equation called the “Nernst Equation” which governs the voltage of these cells at varying concentrations of the reactants). As for the current, it is not possible for the current to stay constant, in fact even when you start using cells the current isn’t constant.
And to the main issue initially raised, take for example a AAA and a AA battery. Both of them put out 1.5 volts. However their capacities are different (which “Physicist” has explained brilliantly with the balloon example). Thats the basic difference between cells of different sizes with similar voltages.
The relation between the voltage or the current with the battery life is very vague. The battery life is dependent on how long the chemicals last and how they can be replenished (for a rechargeable only). There’s no doubt that the current reduces with battery use.
Please I need to know, do battery powered devices turn off due voltage starvation or current starvation? or both?
how do we apply the use of battries in our everyday life?
sir i have a problem please guide me.. how much distance is affective in between battery and ups.. battery cabnet is 5 yard away between ups and battery..is this work proper or not ?? if yes what type of wire is required?/
I had a checklist in my project and then one said ” identify why different batteries have different voltages and capacities?” and I used a different source to find the capacity part but I still don’t know how voltage is affected. does it depend on the chemicals used? the mass of chemicals like capacity does? does it depend on how many cells?
for my earlier comment maybe the voltage is different in different batteries because the composition of metals and/or the electrolyte?
—–[A-M]—————-
~ [VM] L
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1. Get 1 AC classic lightbulb.
2. Get a voltmeter and an ampmeter and add them to the circuit (correctly, of course).
3. Connect the bulb to AC power.
The voltage (V) will measure the same amount to the AC source and the light bulb.
The current (A) will be the same across the whole circuit.
The bulb will emit light, therefore consume something.
So if eveything is the same (both V and A), what is actually consumed ?
I want to build my own eletricmotorcycle.
does it become a 96v18ah battery if i connect two 48v9ah batteries?
This is what I found on alibaba website
http://www.ebike-bmsbattery.com.cn/product/690588130-215068276/e_bike_48v_lithium_ion_battery_ups_battery_48v_lithium_ion_battery.html
Lifepo4 48v9ah E Bike Battery,e-bike battery,ebike battery
Item No. Voltage Capacity battery type Lifespan
BMS
Option
Charger
Option
Dimensions Weight
G-BP4809A 48v 9ah lifepo4
>2000
cycles
15Amps 3Amps 69*148*360mm 4.9kgs
48v 9ah electric bike li ion battery Characteristics
a. Very Security: No fire,no explosion,no leakage ;
b. Portable handle and hidden charging port;
c. Aluminium alloy Back rack(Silver and Black for option);
d. Well die-casting slide board and lock, for easy installation and theft-prevention.
c. 30A Fuse and smart BMS are included for protection
Pictures and detaild data sheet
Model G-BP4809A
Norminal voltage 48V
Rated capacity 9Ah @0.5 C3A, 20°C
Operating current 15Amps(can be customized)
Peak current 30Amps(can be customized)
Charge voltage 58.4V
Standard charge current 3Amps
Cycle life >2000times
Dimensions 69*148*360mm
Weight 4.9kg
Battery box material aluminium alloy
Built in battery cells 3.2V 9AH polymer lifepo4 cells
Assembly 16 cells in series
Operating temperature – 20 °C to 60 °C
Charge temperature 0 °C to 45 °C
it weights about 12 pounds.
What do you think? No one can answer my question.
Hi Jason~
If you connect two batteries in series you will double your voltage (just like putting two D-cell batteries in a flashlight and getting 3 V nominal). The current will actually depend on the internal resistance of each battery, but in general can be considered to be the same as a single battery.
If you connect two batteries in parallel, you will double your available current, with the same voltage as if using a single battery.
Unfortunately, you can’t get away with doubling voltage and available current simultaneously.
I completely respect your initiative in building your own electric bike, but will suggest that if you are asking these kinds of questions, you may not have the necessary experience to do this safely. Batteries, even the ones you have listed, can have a great deal of stored energy and if they are miss-handled can be very dangerous. Also, from a mechanical perspective, electric vehicles can produce a tremendous amount of torque and thus result in incredible acceleration. Please be careful and if you don’t know something, be sure to consult with someone who does!
Hello Everyone, what will be the effect on batteries which are connected in series with different Ah ratings ? I.e 150Ah, 200Ah ?
Luqman Bangash,
“Ah” means no more than charge, in essence, and could be expressed as “C” (Coulombs) times the factor of 3600 units because the time in International System is in seconds.
Let’s think about. The current in series are the same for the mesh. It means that if you draw 200 A from the batteries in series you’ll be drawing the same 200 A from the 150 Ah battery. This means you’ll be drawing more than the battery can give you. This will short the battery lifespan (because you’re using “more than it can gives”) and will make the battery discharge quicker.
“Here’s one”,
If I understood well, you’re measuring the voltage at the light bulb terminals and knowing the current is the same because it’s a series circuit. So if V is equal to A, in values, the power consumed is the square of V or A.
“The voltage (V) will measure the same amount to the AC source and the light bulb.”
If you’re measuring exactly at the light bulb terminals, you’ll have different voltages (slightly difference) from the AC source. This means that you have a voltage drop according to the Ohm’s law. Measuring the voltage drop across the wire and the current, you can calculate the wire resistance, whose will be a low value because the voltage drop is usually a lot lower than the current (R=V/I).
Planning to start a battery manufacturing unit of all types… Some serious technical help needed …
Think about a Dam and a heart.
Think about a Dam first.
Voltage is water level at surface.
Current is water quantity per time(don’t confuse with speed( ft/sec)) at the bottom canal at Dam.
Even water level is high(12V), Current will be low due to waste build up inside Canal after several years later without maintenance. Low current can not crank up the turbine and generator. Usually regular maintenance is required inside canal and turbine blade in hydraulic Dam and Nuclear Power Plant.
Think about heart(car engine).
Voltage is the total supplying blood quantity into heart muscle. Current is the actual final blood quantity into heart muscle.
Current is just like blood in artery at outside of your heart.
If the waste is deposited in artery blood vessel then Resistance is high, the waste(used chemical) in heart blood vessel is high risk to your heart(engine).
V(Voltage)=I (Current) x R(Resistance)
Car battery can be 12V after 3~5 years but actual current will be low and resistance will be high(chemical used).
Then car battery voltage is still 12V but current(cranking ampere) is low then battery is no more useful.
In electrochemistry, the galvanic cell will operate until the substance which is getting oxidised runs out. The voltage remains more or less the same. Potential difference merely provides an electron pressure required for the electrones to flow from the anode to the cathode. Basically the electrones will flow from the anode to the cathode(conventional current to the other direction) as long as the chemical reaction takes place.
Please guys i am very confused about current in a circuit.on one hand we say that the battey have specific data about voltage and current.for a reachargable aa battery it may be 1.5 v ,1200mah.but when we attach a battery to a circuit say it has a 10k Resister then it should draw the currrent according to ohms law with the applied voltage.then please tell me what was the meaning of giving the rating 1200 mah on it.
I DID NOT GET THE STRAIGHT FORWARD
ANSWER
A battery stores “charge” using chemical reactants.
Charge is measured in coulombs and denoted Q
Current is the rate of flow of charge, measured in Coulombs per second. denoted I.
Voltage/Potential is the pressure which pushes charge through a conductor. denoted V.
Resistance is the “friction” of the circuit, measured in Ohms, denoted R.
Ah means Amp hours. 1 Ah means that the device can sustain a current of 1 A for 1 hour.
Since Current measures the rate of flow of charge, then if charge flows at 1 Amp for 1 second, then 1 Coulomb has been transferred. Q = It, where t denotes time in seconds.
For 1 Ah, that is 1 Amp for 3600 seconds so Q = 1 A x 3600 s = 3600 coulombs.
Therefore, the life expectancy of a battery is determined by the current which it is used at,
and the total charge that it has available.
Resistance is like dynamic friction, the faster that you travel the more effort is required to maintain your speed. So when resistance is low, a high voltage will result in a high current.
V = IR. So if resistance is 1 ohm, and the voltage is 2 Volts. The current will be 2 Amps.
The voltage that a battery “produces” is actually the potential difference between the cathode and anode. So by connecting them in series the potential adds up.
Batteries in parallel however will result in no extra potential.
This voltage divided by the Resistance will be the current ( I = V / R)
A Battery works by storing, generally, two mixtures separated by a bridge. The bridge lets positive ions move across. The two mixtures will have opposing reactions, one will absorb electrons, the other release them. However, this reaction is only feasible when the charge “density” is low so that an equilibrium is reached and the reactions stop.
When the terminals are connected (by a conductor), however, the difference in electron density will cause an electrostatic potential accelerating the electrons (which are the charge carriers) around the circuit.
The volume of chemicals in the two terminals will determine the total “charge carriers” which can ultimately be released. But the equilibrium of the chemical reactions keeps the potential difference low.
That is how different sized batteries can have the same voltage rating.
A capacitor does not have any balancing chemical reactions. So all the charge simply builds up an enormous potential difference. (And the two terminals are separated to prevent the charge from leaking across the gap).
In both cases it is the total charge exchange which determines the capacity of the device.
High current will expend the charge faster, low current will expend it slower.
High voltage will produce a higher current than low voltage with the same resistance.
Lifetime = Q / I = Q / (V / R) = Q x R / V
Q is coulombs, (or A s), R is ohms, V is voltage. In hours:
Lifetime (hours) = Charge ( Amp hours ) x Resistance( Ohms) / Voltage( Volts)
In simpler terms, V = IR. V = Voltage, I = Current and R = resistance. Let say, V = 100A X 1Ohms, here V = 100V, the battery has 100Volts with 100Amps pushing the voltage. Now lets say that V = 100Ohms X 1A, here the battery still has 100Volts but the current(force that pushes voltage) is only 1 Amp. Thus meaning that the battery is not good.
how to check lamp :-13 volt,2 Amp.,
but i used plc analog input mAmp.
please send electricl dig.
Demonstrate how the voltage across a battery is affected when current is drawn from it.
does anyone know what the potential difference rating of a battery tells you ? if you do could you please tell me , thanks
What happens if battery of the .mobile gives less voltage than mobile needed the power (my mobile gets off at 88% every time when its gets off it is in 3.2volt????
Plz reply….
Which AA will make my LED flashlight shine brighter – eneloop Pro, standard eneloop 2000 mAh, or other NIMH AA? Does one have a lower internal resistance than the others?
I have a project that I am hoping you might help me with.
I have an outdoor lighting project in a remote area, thus I use lights, solar panel, and battery with a solar panel control box.
What I want to know is, do I have the right mix of components to keep the battery charged to use the lights all night long?
Battery: 12V, 35 Amp Duracell
Lights: LED drawing 1.5 amps each (2 of them), and 18Watts
Solar Panel: Pmax, 20W; Vmp, 18.2V, Imp, 1.1A; Voc, 22V, Isc, 1.21A
What I have found thus far is that that after about a week, the battery is no longer able to provide power to the lights. I am using the largest Amp battery that I can fit into the space/housing.
Questions: Obviously, I could reduce the power needs of the lights, or cut back to one light. What is the brightest light that I could use for about 8 hours each night (sun down to sun up) and still be able to charge back up?
Do I need a larger solar panel if I get 6 hours of light a day? 4 hours of light a day?