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cpautler
Posts: 1
Joined: Sun Feb 19, 2017 7:42 pm

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.

darrellg
Posts: 332
Joined: Fri Oct 17, 2014 10:02 am
Location: Southern California

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.

comforter
Posts: 7
Joined: Tue Mar 14, 2017 10:28 pm

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.

teprojects1
Posts: 51
Joined: Sat Jan 21, 2017 1:28 am
Location: Lahore
Contact:

cpautler wrote: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 you need any help in Engineering Projects | Arduino Projects you can contact me !! Introduction to PIC16F877a

jremington
Support Volunteer
Posts: 2269
Joined: Fri Jun 15, 2007 9:41 pm
Location: Eugene, Or

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.

teprojects1
Posts: 51
Joined: Sat Jan 21, 2017 1:28 am
Location: Lahore
Contact: