I've attached a picture of the circuit below for reference, but the question is asking what the potential is at point D.
V = IR[/B]
The Attempt at a Solution
I was told that the answer was 9V, but I'm a little confused because I thought if I looked at the battery on the left side that by the time the current starting at this battery reached point C, there would be 0 voltage and therefore 0 current. Going with this, I thought that if I kept going to point D it would still be 0 since there would be no voltage drop at resistor 5 if there was no longer any current. Therefore I thought I could find current I1 by taking 12V/3ohms = 4A
Looking at the battery on the right side I thought that by the same logic, the current starting at the right battery would reach 0 A by the time it reached point C. Likewise I thought current I2 would = 18V/3 ohms = 6A.
However when starting at point A and going around the left side path to point D, I would get 0V. But if I take the right side path using current 2 of 6A, I'm getting 18V - (6A*2ohms) = 6V.
I'm not sure what I'm doing wrong here; I know I'm definitely missing some conceptual bridge here.
66.9 KB Views: 162