# How do I calculate a battery ohm

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Calculate voltage drop with known internal battery resistance:

When loaded with higher currents, the battery voltage drops by a certain value. This is due to the internal resistance of the cells.
If the internal resistance of the battery type is known, the voltage drop can be calculated using Ohm's law:
U = R x I;
U > Voltage drop in the entire battery [volts]; the voltage is reduced by this amount
R. > Internal resistance of the entire battery block [Ohm]; 1 mOhm corresponds to 0.001 Ohm
I. > assumed current load [ampere];

Example 1: The internal resistance of a NiMH cell is given as 4 mOhm. A battery block consists of 30 cells. 4 blocks have been installed.
The internal resistance of a block (36V) is 30 x 4 mOhm = 120 mOhm. Since four identical blocks are used, the total resistance is 120 mOhm / 4 = 30 mOhm.
With an assumed current load of 120 A (realistic value when accelerating) the voltage drop is:
0.03 ohms x 120 A = 3.6V. The battery voltage breaks down to 36 V - 3.6 V = 32.4V a.

### Calculate power loss if the internal resistance of the battery is known

When exposed to higher currents, power loss in the form of heat is released in the battery cells. This is due to the internal resistance of the cells.
If the internal resistance of the battery type is known, the power loss can be calculated using Ohm's law:
P = R x I²;
P. > Power dissipation [watt]; it heats the battery cells
R. > Internal resistance of the entire battery block [Ohm]; 1 mOhm corresponds to 0.001 Ohm
I. > assumed current load [ampere];

Example 2: The internal resistance of the battery pack and the current load from Example 1 are assumed (0.03 Ohm and 120 A).
P = 0.03 ohms x 120 A x 120 A = 432 W.. The battery is therefore additionally heated with a total of 432W when accelerating (120 A).
This energy is taken from the cells, but is not available to the motor!

### Calculate the internal resistance of the battery

The internal resistance of the battery can be calculated using the formula R = Delta U / Delta I. Procedure: The battery is slightly loaded with a consumer. The current flowing now and the voltage on the battery are noted (e.g. 12V and 10A). Now the current is increased and the voltage on the battery is noted again (e.g. 11.8V and 100A). Delta "U" is now 0.2V and Delta "I" is 90A. This results in an internal resistance of 0.2V / 90A = 0.00222 Ohm or 2.22 mOhm