Today I got a question from Garry on the Ohmify community.
“When I measure the voltage of my 9V battery using my multimeter I get a reading of 9.06 when it is not connected to any circuit. When I hook the battery up to a simple circuit (i.e. 325 Ohm resistor in series with a red LED) and measure the voltage drop across the battery I get a reading of 8.65 V – why the difference?”
Great observation! That’s the difference between an ideal battery that we often use to make calculations in a circuit and a real battery.
The reason for this is that a real battery has an internal resistance. The size of the internal resistance depends on the battery type.
So the voltage you can squeeze out of a battery actually depends on what you connect to it!
In the image above, the battery has an internal resistance of 1 ohm. If you connect that to something of 500 ohms, you can figure out the battery voltage by using the voltage divider formula:
V = 9V * 500 ohm / (1 ohm + 500 ohm) = 8.98 V
If you connect it to something of 50 ohms, you get:
V = 9V * 50 ohm / (1 ohm + 50 ohm) = 8.82V
So there you have it! Because of the internal resistance of the battery, the voltage from the battery will change depending on the load you connect to it.
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Keep On Soldering!
Oyvind @ build-electronic-circuits.com