- #1
fadecomic
- 10
- 0
Hi,
I am having difficulty understanding why potential must drop completely over a circuit from the high side of the source to the low side of the source. I've seen this statement in several books now with no further explanation other than "it must".
Consider a simple dc circuit consisting of a battery and a resistor. In this case, charges leaving the source are raised in PE by an amount V per charge. A number of charges, dependent upon the resistance of the resistor, enter the circuit. The resistor reduces the PE energy of each charge as heat. That much is clear to me. But it's not just any amount of PE that is lost in the resistor, but exactly as much PE/charge as was imparted by the source must be dissipated by the resistor. Why? Why can't the charges return to the low side of the source with some of the PE they were given? The classic answer seems to be that this violates the conservation of energy. Why? The energy is imparted to the low side, which one might assume would convert the PE back to chemical PE.
Can someone illuminate this problem for me? I have a feeling I'm just looking at it the wrong way.
EDIT: I should add that it seems like the water analogy breaks down here, too. If you consider a pipe containing a pump and a constriction, the equivalent argument is that not only does the pump decrease the pressure across its length, it reduces it to the unpumped pressure! Why again?
I am having difficulty understanding why potential must drop completely over a circuit from the high side of the source to the low side of the source. I've seen this statement in several books now with no further explanation other than "it must".
Consider a simple dc circuit consisting of a battery and a resistor. In this case, charges leaving the source are raised in PE by an amount V per charge. A number of charges, dependent upon the resistance of the resistor, enter the circuit. The resistor reduces the PE energy of each charge as heat. That much is clear to me. But it's not just any amount of PE that is lost in the resistor, but exactly as much PE/charge as was imparted by the source must be dissipated by the resistor. Why? Why can't the charges return to the low side of the source with some of the PE they were given? The classic answer seems to be that this violates the conservation of energy. Why? The energy is imparted to the low side, which one might assume would convert the PE back to chemical PE.
Can someone illuminate this problem for me? I have a feeling I'm just looking at it the wrong way.
EDIT: I should add that it seems like the water analogy breaks down here, too. If you consider a pipe containing a pump and a constriction, the equivalent argument is that not only does the pump decrease the pressure across its length, it reduces it to the unpumped pressure! Why again?