1. The problem statement, all variables and given/known data You have a 9V battery with an internal resistance of 1.5 ohms. If a 3 ohms load is connected to the battery, how much current will flow through it? When the 3 ohms load is connected, what potential difference will an ideal voltmeter read across the battery's terminals? 2. Relevant equations V=IR 3. The attempt at a solution Okay, for the first problem, I used the above equation for the battery and for the resistance load and set them equal to each other. So the equation for the battery from above is I=V/R, where V= 9V and R=1.5 ohms. The equation for the resistance load is V=IR, where R=3 ohms. Taking both equations, I got V/R=IR and solved for I, the current. The current came out to be 2 amps. What I don't understand from this problem is why I had to set those equations equal to each other to get the current for the resistance load. Why can't I do I=V/R, where V = 9V and R = 1.5 ohms. If I did that, is that the current through the battery itself? I don't get why the current for the resistance load is the same as for the battery. For the second problem, I used V=IR again where I took the current from above, 2 A, and plugged it into that equation. V=(2A)(3ohms)=6 V. That is the potential difference an ideal voltmeter will read across the battery's terminals when the 3 ohms load is connected. I don't understand why I am using the 3 ohms load and its current to find the potential difference across a battery's terminals? I don't understand why the battery's own current and its own resistance, 1.5 ohms, was not used to solve this problem. Wouldn't the 6V be the potential difference across the resistance load itself? I'm just having a hard time understanding the story behind the equations and what I did in these problems. Can someone please explain?