- #1
JFS321
- 75
- 6
Hi all, I'll continue on my quest to get my misconceptions straight. I appreciate the help so far.
Let me explain my thinking about the Voltage Law and please identify where I might be going wrong.
I've been reading as much as possible on voltage drop and how it relates to work. Let's take an example battery of 10V. At the end of the circuit, voltage drops to zero of course.
A) Is this voltage drop equivalent to the work done by the circuit...or is work ONLY done by the battery to raise 0V back up to 10V?
B) I realize that charges flowing "downhill" much like a bowling ball dropped off a mountain eventually return to zero volts with zero potential energy. But, I'm fuzzy on what intuitively causes this change in a battery loop. Is the voltage drop in this case due to the inherent loss voltage as current encounters wire resistance, or is this the loss simply because current is reaching the negative terminal and thus no longer traveling toward a potential difference?
C) Also, similar to A, I am confused on whether or not work is only considered "work" if it opposes the current flow. Meaning, resistors/bulbs/etc. do work because they cause a voltage drop by opposing the direction of current.
THANKS to all...I can't wait to get this straight!
Let me explain my thinking about the Voltage Law and please identify where I might be going wrong.
I've been reading as much as possible on voltage drop and how it relates to work. Let's take an example battery of 10V. At the end of the circuit, voltage drops to zero of course.
A) Is this voltage drop equivalent to the work done by the circuit...or is work ONLY done by the battery to raise 0V back up to 10V?
B) I realize that charges flowing "downhill" much like a bowling ball dropped off a mountain eventually return to zero volts with zero potential energy. But, I'm fuzzy on what intuitively causes this change in a battery loop. Is the voltage drop in this case due to the inherent loss voltage as current encounters wire resistance, or is this the loss simply because current is reaching the negative terminal and thus no longer traveling toward a potential difference?
C) Also, similar to A, I am confused on whether or not work is only considered "work" if it opposes the current flow. Meaning, resistors/bulbs/etc. do work because they cause a voltage drop by opposing the direction of current.
THANKS to all...I can't wait to get this straight!