Question

When a 12V battery provides a 4 A current, the terminal potential difference decreases by 2 V, dropping from 12V to 10V. The internal resistance of the battery is?

Answer 1

This one is pretty straightforward. \[V=IR\]\[R=\frac{V}{I}\]The battery's resistance will be:\[R_{battery}=\frac{2V}{4A}=\frac{1}{2}\Omega\]\[R_{total}=\frac{12V}{4A}=3\Omega\]Therefore, the resistance of the rest of the circuit, in case you needed to know, is:\[3\Omega-\frac{1}{2}\Omega=2\frac{1}{2}\Omega\]

Answer 2

Ok thanks

Answer 3

np :)

Answer 4

Hey I think the answer is actually .5Ohms. Do you know how they got there.

Answer 5

Read above... 0.5 ohms = 1/2 ohms :)

Answer 6

One of the multiple choice answers

say 2.5

Answer 7

Maybe I didn't make it clear enough in my original response.

The battery's internal resistance is 0.5 (or 1/2) ohms. The resistance of the rest of the circuit must therefore be 2.5 (or 2 1/2) ohms...but that's not what the question asked. For that part I was just showing you how it's calculated.

Answer 8

Ok yeah cool that works.... I may have a few more questions... Im studying for a midterm.

Answer 9

Ok, if I'm around and can help I will :)

Answer 10

Thank You so much

Answer 11

how did you know to use the 2v for the internal resistance?

Answer 12

Because that was the voltage drop... 12V-10V=2V