A student is provided with a 12 V battery of negligible internal resistance and four resistors with the following resistances: 100 ohms, 30 ohms, 20 ohms, and 10 ohms. The student also has plenty of wire of negligble resistance available to make connections as desired.

Draw a circuit diagram in which each resistor has a nonzero current flowing through it, but which the current from the battery is as small as possible. Then, draw one in which the current from the battery is as large as possible without short circuiting the battery.

For the first one, the one with that would produce the smallest current, I drew all the resistors in series because current remains constant, I=V/R=12/(100+30+20+10)=.075 A, and that was the smallest I could find...

For the second, the one that would produce the largest current overall. I drew all the resistors in parallel, found the total R to be 5.17 (I know this is right) and then each current on each resistor to be .12 A, .4 A, .6 A, and 1.2A (I know these are right as well), making the overall current 2.32 A.

My question is this: Am I correct in assuming that parallel combinations create the largest current while series combinations create the smallest current? I don't need to be told how to do it, just a simple yes or no would be helpful really. No math involved, as I'm sure my numbers are right.

Thank you for your time and reading this. I appreciate it immensely.:rofl: