You have to imagine the internal resistance as being in parallel with any load you connect. You get the maximum possible current when the load is zero. In this case, just apply Ohm's Law. That is, divide the voltage by the internal resistance.
That will depend on the internal resistance of the battery. I = E / R Where I is the current, E is the open circuit battery voltage, and R is the internal resistance of the battery.
Internal resistance is approximately equal to 94.667
There is internal resistance in a battery because a battery is not an ideal voltage source. It may be close, but it is not ideal. As a result, analytically, there will be some series resistance, resistance which places a limit on the maximum current that the battery can provide. While no battery is ideal, most are sufficiently ideal to not require any consideration of the internal resistance. If your circuit is dependent on the internal resistance of a battery, then it is probably not well suited for that application.
When a voltage source, such as a battery or a generator, is on open circuit -in other words, when it is not supplying a load- the voltage appearing across its terminals is called its 'open circuit voltage' and corresponds numerically to its electromotive force.However, when the voltage source supplies current to a load, that current also passes through the voltage source itself. This causes an internal voltage drop, which is the product of this current and the voltage source's internal resistance. This voltage drop acts in the opposite direction to the electromotive force and reduces the source's terminal voltage. This internal voltage drop will increase, of course, if either the load current increases or the internal resistance increases.So, in order to keep that the source's internal voltage drop is as low as possible, its internal resistance must be as low as possible. In the case of a battery, the internal resistance is due to the ionic resistance of the electrolyte/plates, whereas in a generator it is due to the resistance of the windings.
I think you are asking why is it necessary for an automobile battery to have low internal resistance. That is because the internal resistance of a battery limits the amount of current it will produce. Imagine connecting a wire from the plus side of the batter to the minus side. Then, the amount of current the battery produces is the voltage of the battery divided by its internal resistance. So, if you take a flashlight battery and connect a wire between the two terminals, it might get a little warm. If you take a car battery cannot a wire between the two terminals, the wire will probably melt! (Don't try this!) If you take a lithium ion battery from your computer and short the two terminals, the battery will catch on fire. (Computer companies build extra resistance into the batteries to help to prevent that. So, why do cars need low internal resistance batteries? They need lots of current to start those large gasoline engines turning. You car battery does a tremendous amount of work to start your engine (and remember, energy and work are the same in physics).
That will depend on the internal resistance of the battery. I = E / R Where I is the current, E is the open circuit battery voltage, and R is the internal resistance of the battery.
Internal resistance is approximately equal to 94.667
The voltage of the battery, and the resistance of the circuit (including the resistance of the wire and the internal resistance of the battery).
Battery maximum current is limited by the internal resistance of the battery. As the current is increased towards this maximum, you will notice the output voltage appear to shink towards zero. What this means is the voltage the battery is capable of supplying is being dropped almost completely across the internal resistance, so no real power is available to use.This internal resistance is dependent on the chemical and physical makeup of the battery.
It causes the battery's voltage to drop when a current is drawn from it.
A battery is rated to supply a certain number of volts. However, it actually supplies less, because they are "lost" as the current has to get out of the battery in the first place.(The battery has internal resistance)The amount of lost volts depends on the current being drawn:The less resistance a circuit has, the more current is drawn, because it's easier to flow.Example:If the circuit has little resistance, it draws a large current and the battery's internal resistance causes more lost volts.If the circuit has high resistance, it draws a small current and there are fewer lost volts.This is why when you short-circuit a battery (give it hardly any resistance to go through) it heats up and may explode. A large current is drawn and all the volts are used by the battery's internal resistance.
Doesn't work like that. Current drain is dependent on the (internal resistance of the battery and the) resistance/power requirements of what's connected to the battery. If shorted out, the current - unless the battery is fused or otherwise protected - can go into tens of amps.
There is internal resistance in a battery because a battery is not an ideal voltage source. It may be close, but it is not ideal. As a result, analytically, there will be some series resistance, resistance which places a limit on the maximum current that the battery can provide. While no battery is ideal, most are sufficiently ideal to not require any consideration of the internal resistance. If your circuit is dependent on the internal resistance of a battery, then it is probably not well suited for that application.
its internal resistance V/R= A its internal resistance V/R= A
Enormous initial current, rechargeable, cheap, low internal resistance, availability
That will depend on the sum of the load resistance and the internal resistance of the battery (this is true for all power sources, not just 6 volt batteries). Small compact batteries tend to have higher internal resistance and therefore are more limited in the current they can deliver to a given load than larger batteries.
internal resistance is always infinite in ideal current source .the internal resistance is in shunt with current source