How do voltage and voltage drops over a circuit relate to work done?

Question

I have a few remaining uncertainties when it comes to voltage. I numbered the paragraphs in my thought sequence for easier reference.

1. In a series circuit, I know voltage is the electric potential, and represents the potential for work, but that doesnt mean that any electron travelling on any 10 volt circuit does the same amount of work from start to finish, does it?

2. What i mean is, let's say you have a 10 volt series circuit with a 1 ohm wire. Tracking an electron in the current from start to finish, as work is done on the electron by the voltage, its potential drops, until it reaches 0 at the end of the circuit.

3. But if we increase the resistance of the wire to 5 ohms, it would take 5x longer for the electron to get to the end of the circuit, because the current would be 1/5th, right?

4. But then that means that the electron in the 5 ohm circuit would have done 5x the amount of work (or work done on it) of the 1 ohm circuit over 5x the duration.

5. I know the work done per second is the same, but ultimately, if each electron in the 5 ohm circuit has more work done on it over the course of the circuit, then voltage cant be an expression of total work done from p1 (start of circuit) to p2 (end), right? I think i first thought that because voltage relates to potential for work, and the voltage always drops from full to 0 by the end of the circuit, so i assumed the work must be the same regardless of the circuit resistance.

6. So does this mean that when voltage is divided over the components of the circuit, it's really representing the ratio of where in the circuit the work is done?

7. i.e. in a 12v series circuit with a 3 ohm resistor and a 1 ohm resistor, there would be a voltage drop of 9v over the 3ohm resistor, and 3 volts over the 1 ohm, right? which would mean that 75% of total work done in the circuit was at the 3 ohm resistor?

8. But why is it that the more resistors you have in a series, the less the voltage drops over any one resistor? I know with more resistors the current would be lower. Does voltage drop over a resistor decrease as current decreases?

Answer 1

How do voltage and voltage drops over a circuit relate to work done?

The Volt unit is energy normalized to unit charge; Joule per Coulomb.

Since the Amp unit is Coulomb per second, the product of the voltage across and current through a circuit element is the power associated with the circuit element.

For a DC circuit, voltage and current are constant thus the energy delivered or supplied by a circuit element over some period of time is the product of the voltage across, the current through, and the elapsed time.

Now, the rest of your question contains misconceptions that are too numerous to untangle here. I recommend that you do some more reading and careful thinking about, in particular, basic circuit laws. For example, Ohm's Law immediately gives you the answer to your question "Does voltage drop over a resistor decrease as current decreases?"

Ohm's law states that the current through a conductor between two points is directly proportional to the potential difference across the two points.

Also, I recommend perusing William J Beaty's "Electricity Misconceptions Spread by Textbooks" site.

For a start, take a look at Electric current is a flow of energy? Wrong.

Answer 2

But then that means that the electron in the 5 ohm circuit would have done 5x the amount of work (or work done on it) of the 1 ohm circuit over 5x the duration.

You're confusing work with power here.

Work has nothing to do with duration. If an electron crosses a potential difference of $V$ with any resistance in between, the work done is the same, $eV$. It doesn't matter if it takes a minute or a year for the electron to pass.

That's the definition of potential — it is proportional to the work done in moving a unit charge across two points by any path. The path does not matter as long as the system is an electrostatic one.

I know the work done per second is the same

Again, no. Work done per second is power, and power is $\frac{V^2}{R}$ for this system and is not constant as we vary $R$

Answer 3

This should help.

For a conservative force, such as gravity or an electrostatic field, we can define the negative of the work done by the force as the change in a potential energy. So a change in potential energy is just a simple way to evaluate the work done by a conservative force.

Voltage is the potential energy per unit charge for an electrostatic force. For a circuit, the drop in voltage (a negative value) is the positive work done by the electrostatic field per unit charge.

This is similar to the force of gravity, a conservative force. For an object of mass m close to earth subject to the acceleration of gravity, g, the force is mg. The potential energy is mgh where h is elevation. For an object falling from a high elevation, h1, to a low elevation, h2, due to gravity, the work done by gravity is -mg(h2 - h1), a positive value, and the corresponding change in potential energy is mg(h2 - h1), a negative value. The work from gravity equals the negative of the change in potential energy.

Why do we use potential energy instead of work? It simplifies the evaluation. For example, if an object is moved up, down, and all around it is not easy to calculate the work from gravity as the integral of the force dot product with the differential distance travelled, but it is very easy to calculate the change in potential energy as the difference between the final and initial elevations. Similarly, in a circuit the use of a change in potential energy is simpler than evaluating the work done by the electrostatic force.