The equivalent resistance you would have to place in series with an ideal battery (which of course does not exist) of the same voltage to get the same behavior (voltage drop with load) as the real battery has. It is a mathematical modeling technique to help in circuit analysis.
The voltage of the battery, and the resistance of the circuit (including the resistance of the wire and the internal resistance of the battery).
As it ages, its Kirchhoff virtual 'internal resistance' increases. Pictured this way, the practical effect is that the open-circuit voltage of a failing cell still measures good, but the voltage quickly sags when you try to draw any current from it
The internal battery is a battery in the game pack that stores energy to keep track of things like the passage of time when the game is not in use.
The voltmeter would read 12 volts. An ammeter connected to to battery would only read 4 amps (12 volts divided by 3 ohms =4)
It's all relative. A battery has internal resistance. Batteries are not "ideal" voltage sources. If there is a "heavy" load on the battery (low resistance), the voltage you measure on the outside of the battery will be lower. A car battery is normally thought of as a 12 volt battery. When there is no load (disconnected) you can often measure in excess of 13 volts if it's fully charged. During normal cranking of the engine, the voltage can drop below 11 volts.
No, it is desirable for a battery to have a low internal resistance.
The value of internal resistance of 1.5 volt battery is 0.5 ohms.
The battery internal resistance chart provides data on how much resistance the battery has to the flow of electricity within it. This information can help determine the battery's efficiency and performance.
Yes
Yes, increasing temperature can cause the internal resistance of a battery to increase. This is because higher temperatures can lead to material degradation and changes in chemical reactions within the battery, which can ultimately result in higher internal resistance.
Internal resistance is approximately equal to 94.667
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.
The voltage of the battery, and the resistance of the circuit (including the resistance of the wire and the internal resistance of the battery).
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.
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.
If Rin = Rout, then the voltage at the output of the device goes down to half of the value that the circuit has without the external resistance. Scroll down to related links and look at "Interconnection of two audio units".
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