Exploring Electrical Potential Energy Across Series Circuits

[Jacob Gawel]

Assuming the resistance of a wire in a series circuit, consisting only of 1 component (e.g. filament lamp) and a battery, is negligible; does each Coulomb of charge commit all of its electrical potential energy, supplied by the battery's potential difference, as work done across the component (thus electrical potential energy gained at battery = electrical potential energy lost across component)?

If so, how is voltage divided evenly across 2 identical components in series?
i.e. how do the circuit electrons give up only half their electrical potential energy at component 1, to 'save' some electrical potential energy for component 2?

 

[Drakkith]

does each Coulomb of charge commit all of its electrical potential energy, supplied by the battery's potential difference, as work done across the component (thus electrical potential energy gained at battery = electrical potential energy lost across component)?

Ideally, yes. In real circuits some is lost in the wire and in the battery.

If so, how is voltage divided evenly across 2 identical components in series?
i.e. how do the circuit electrons give up only half their electrical potential energy at component 1, to 'save' some electrical potential energy for component 2?

Remember that electrons do not exist by themselves in a circuit, but as a collection of a great many electrons. The battery does work on the entire circuit, including all of the electrons, all at once. So if you look at electron A, you cannot say that it has 10 volts of electric potential energy, of which it gives half to resistor 1 and half to resistor 2. Instead you have trillions of electrons whose collective behavior delivers the energy to each component in the circuit.

In fact, the motion of each electron is completely random, and the battery only causes a small net motion in a certain direction. This slight imbalance between directions is what gives rise to the movement of charge through the circuit. If you could look at a single slice of the circuit, you would see electrons crossing it from both directions, with one direction having slightly more electrons over time.

 

[jim hardy]

If so, how is voltage divided evenly across 2 identical components in series?
i.e. how do the circuit electrons give up only half their electrical potential energy at component 1, to 'save' some electrical potential energy for component 2?

When starting out it's difficult to work these conundrums in your head.
Use your everyday experience on them.
As a kid, did you ever slide down a not very steep, slow sliding board made for little guys and feel it warm the seat of your pants ? Halfway down you've converted half your potential energy into heating your backside.

Any help ?

old jim

 

[davenn]

If so, how is voltage divided evenly across 2 identical components in series?

lets look at a couple of examples

The left hand part just shows a single resistor across the 10V power supply
Using Ohms Law, we can work out the total current flowing in the circuit

I = V / R ( Current = Voltage / Resistance)
I = 10V / 30 Ohms
I = 0.33 Amps

in the second circuit if we want to find the voltage drop across each resistor
We first do that same step again work out the total current flowing in the circuit

I = V / R
I = 10V / 60 Ohms ( 30 Ohms + 30 Ohms)
I = 0.166

Now that we know the total current flowing, we can now use Ohms Law rearranged to find the voltage drop across each resistor
NOTE ... because the resistors are the same value, the voltage drop will be the same across each of them ... therefor we only need to work out the voltage drop across one of the resistors

V = I x R
V = 0.166 x 30
V = 4.98V

Close enough to 5V drop across each resistor

Now you could do an example where the two resistors are different values and work out the voltage drop across
each resistor.
NOTE .. the addition of the 2 voltage drops will always equal the total voltage across the circuit.Dave