My son understands currents better than me, but he’s also stumped on this. He’s doing a science experiment testing the current of a 9V battery with different resistors in the circuit. Now from what we can tell, the current of a 9v battery with no resistance should be around 500mA.
When he used a 10ohm resistor, he got around 300mA.
But when you put the numbers into an ohm’s law calculator, it comes up with 900mA.
We don’t understand how this can be true. How can it have a number higher than the usual current of 500mA?
Hoping one of you smarty-pants out there can help us with this.
In an ideal world, the current from a battery with a dead short would be infinite. In the real world, the battery has its own internal resistance that limits its current output. In addition, if you are measuring the current with a multimeter in series, the meter itself will be adding its own series resistance. Though these resistances are low practically (2ohms or less, typically), they can have a relatively large effect when the test resistance is low like it is in this case.
Correct. I actually figured it out when I installed my newly purchased AMP Research Power step. I thought it was a simple installation, good thing a fellow forumer here helped me understand about the definition of current and how it is affected when applied with a certain load of an resistor. It ended as a complicated installation, but all worth it.
cpautler:
My son understands currents better than me, but he’s also stumped on this. He’s doing a science experiment testing the current of a 9V battery with different resistors in the circuit. Now from what we can tell, the current of a 9v battery with no resistance should be around 500mA.
When he used a 10ohm resistor, he got around 300mA.
But when you put the numbers into an ohm’s law calculator, it comes up with 900mA.
We don’t understand how this can be true. How can it have a number higher than the usual current of 500mA?
Hoping one of you smarty-pants out there can help us with this.
Hey dear you need to understand the Ohm’s law first.
V=IR
I=V/R
R=V/I
If I=300 mA when external R=10 Ohms, the battery has internal resistance of about 20 Ohms.
Assuming open circuit voltage = 9V, the short circuit current would be about 450 mA.
jremington:
If I=300 mA when external R=10 Ohms, the battery has internal resistance of about 20 Ohms.
Assuming open circuit voltage = 9V, the short circuit current would be about 450 mA.
Thanks buddy exactly what I want to convey, but the fact is to just understand ohms law to get clear each point 