Experts say "current depends on voltage". So, if the voltage is high, current would be high. Agreed; (I = V/R)

If the voltage is low, the current would also be low. Agreed -> I = V/R

But why then do two different batteries available with the same voltage (say 2 V) not deliver the same current?

If the voltage is same for two different batteries, current should also be the same, right?

Which factor plays the role in delivering a different current for these batteries?

If two different batteries (with the same voltage) delivers different currents, how can we say that they are both 2 V batteries?

Why do the batteries not obey the rule? Do they have another principle?

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    \$\begingroup\$ Current is limited by voltage drop on internal series resistance Rs or (ESR) and voltage on charged electrolyte. Geometry and aging factor affects Rs. \$\endgroup\$ Nov 29, 2012 at 23:56
  • \$\begingroup\$ Given ohm's law, you would understand it better if you read the current as "up to x amps". \$\endgroup\$
    – gbarry
    Jun 1, 2014 at 21:45

5 Answers 5


Current depends on Voltage". So, if the voltage is high, current would be high. Agreed; (I=V/R)

True, if you're asking about resistance.

But, you're asking about a (non-ideal) voltage source - a battery.

The voltage to current relationship of a battery depends on the chemistry, temperature, etc. Cells and batteries are not resistors.

Now, it is the case that a first approximation of a battery is an ideal voltage source in series with an ideal resistor, as another answer points out, but it must be kept in mind that this effective resistance depends on a number of factors. See, for example this.

enter image description here

If two different batteries(with same voltage) delivers different currents, how can we say that they are both 2V batteries?

You've answered your question yourself. They have the same open-circuit voltage.

Clearly, two identical cells connected in parallel provide the same open-circuit voltage as one of the cells but two cells in parallel can provide twice the short-circuit current of the single cell.


Battery is not an ideal voltage source**. For the purposes of your question, a battery can be modeled as an ideal voltage source \$V_{ideal}\$ with a series resistance \$R_{series}\$. While voltages of the batteries may be the same, the series resistances may be different. Let's model two of your batteries as identical ideal voltage sources, but with different series resistances.

\$V_{batt}=V_{ideal}-I_{load}R_{series}\$ is the observed battery voltage. When there is no load current, the voltages of both of our batteries are the same. Open circuit voltages.

Let's gradually increase \$I_{load}\$. The voltage of the battery with higher \$R_{series}\$ will be falling more, and it will fall out of spec at a smaller current than that of a battery with lower \$R_{series}\$.

** Battery is a nonlinear function of everything.tm

  • \$\begingroup\$ Here Videal could be named VIN (input voltage on the charger side) \$\endgroup\$
    – baddy
    Dec 30, 2020 at 13:46

The basic concept you need to know is that a battery tries to keep its voltage constant, it is what they are designed for - it is basically a voltage source.

When a load is connected to a battery, it is the nature of this potential (voltage) to deliver current to this load. The load is basically a way for charge to travel from high to low potential (one terminal of the battery to the other).

Since accumulation of charge is what creates the potential in the first place, if they find a way to go from one terminal to the other externally to the battery, voltage would decrease, so it is up to the battery's chemistry to internally force them through and elevate them back to the source terminal so that voltage can be maintained.

If the current being allowed to flow is too much for the battery to handle, voltage cannot be maintained and will drop. Hence the higher current capacity the battery has, the better it is at maintaining that voltage from dropping without getting internally damaged.

You see now that batteries can have the same voltage, but their current is a capacity rating, not something they are outputting all the time.

Resistance is then a measure of how much the load is, well, loading this voltage source (battery), so low resistance means that it will let a lot of charges travel through it per second for a given voltage. High resistance means that few charges will be let through for a given voltage (the battery's voltage in this case), so the battery can relax and not work very hard to keep that voltage up there.


The critical parameter to batteries, capacitors, and any power source is the (effective) Series Resistance, Rs or ESR. Generally the bigger the capacity [Ahr] rating, the lower the ESR [Ω].

Here the batteries are graded by ESR. enter image description here

In your case, if you are using the same grade and size of battery the differnce will lie in the aging factor and remaining charge capacity which both affect ESR. So you can estimate each batteries current load capacity based on the ESR which must be tested somehow.


The voltage a battery carries depends on the concentration of the chemical compounds (acid or whatever) and the current delivering capacity depends on the amount (quantity) of the chemical compound. Higher the chemical concentration higher the voltage. And more the quantity of the chemical compound, higher the current delivering capacity.

A simple example of this would be a motorcycle battery and a car battery.

Both carry 12v but the car battery can deliver upto 40-60 amps and the bike battery upto 7-9 amps.

And the sizes of both batteries you know!


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