Recent content by blakmamba619

i got it, thx.

[Wrong, because not all of that current goes through the headlight.] current is the same in all resistors in series right? but your potential difference is not. So V = I(R+r) is that correct?

i just don't understand the wording of the problem, i am usually good at physics lol (i received an A in motion and mechanics) Can someone describe what the problem means when it states, it takes 35 A from the battery? is it possible to draw more current than what's provided? does

that doesn't make sense, they want the voltage drop across the 6.00 ohm resistor. But i tried your method, and i get I = 37.07 (too high??) then V = IR = 37.07*6= 222.42 tooo high. Am i doing the calculations right? or is this method wrong?

naw, i tried that, i get too high of a potential across the 5.00 ohms. i got 12.003 v too high

1.An automobile battery has an emf of 12.6 V and an internal resistance of 0.0620 . The headlights together present equivalent resistance 6.00 (assumed constant). What is the potential difference across the headlight bulbs when they are the only load on the battery? What is the potential...