I understand that a voltage source drives the current to flow from the terminal with lower potential to the terminal with higher potential. However, if the electrons are gaining potential energy then how can it also be supplying energy to the circuit?
[Physics] How does the voltage source provide energy in a circuit
batterieselectric-circuitsenergypowervoltage
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A current source is a device that will (ideally) provide a constant pre-specified current to the connected load no matter what the load's resistance is, so that it will adjust the voltage across its terminals to get the current right (instead of the other way around as in a(n ideal) voltage source).
Of course, this is only ever approximately possible, and the quality of that approximation will depend (as with voltage sources!) on how much is known or expected about the load resistance and how fancy the device is -- see for example wikipedia.
I was just wondering what happens in a circuit in terms of different types of energy transformations.
You can think of this just like gravity. See below.
If you apply a voltage to a circuit then electrons start moving (very slowly).
Voltage is a name for electrical potential difference.
- There is a gravitational potential energy difference between the floor and a higher shelf, because a book wants to fall down, if it could.
- There is an electrical potential energy difference between two points in a circuit, if the charge wants to move to another point.
But the electrons are also now moving once the voltage is applied as there is a current so would they have kinetic energy as well?
Yes.
- A book falling from a high shelf is trading the gravitational potential energy into kinetic energy.
- A charge moving because a voltage is suddenly applied, is trading the electrical potential energy into kinetic energy.
Some information I have found online says that electrons collide with atoms in a bulb and this is why the filament heats up. But this would imply that kinetic energy is being transformed into heat [...]
Yes.
- Drop a steel ball into a fluid, and the fluid resistance because of viscosity (because of "bumbing" into many water particles) with pull out some of the kinetic energy, slowing it down. This kinetic energy loss is indeed converted into heat.
- A charge "bumbing" into atoms similarly are slowed down, and the loss in kinetic energy is converted into heat.
[...] and this can't be correct because surely any change in kinetic energy would alter the current flowing?
It does alter the current flowing. Without the resistance, the current would be much, much higher.
- The steel ball falling through the water is slowed down and soon reaches a constant speed (terminal speed).
- The charge moving through a resisting material is slowed down and soon reaches a new equilibrium speed, where the "push" that makes it move (the voltage) balances out the resistance that holds it back.
The end-result is indeed a slower flow, i.e. a lower current, than without this resistor present (in that particular part of the circuit).
Also, in a series circuit, if you measure the potential difference between any 2 points after all of the loads then you always get zero. I was just wondering why it is zero because don't the electrons keep moving even after passing through all the load in order to get back to the power source so surely they cannot do this without some form of energy?
Remember Newton's 1st law. If charges exit the last resistor, they are no longer slowed further down, no. But the speed they came out with will stay. If nothing prevents them along the wire alone, they will not speed up, no, but also not slow down. So they continue.
And should one be stopped by whatever reason, the next charge will come and push it forward (like-charges repel). This is the case when the wire is not a completely perfect conductor but has a little resistance.
Best Answer
Actually, they are not.
There indeed is a reduction in potential energy, transformed into kinetic energy.
It all comes down to the answer to your first line:
This is correct. But. Two types of charges exist. A positive and a negative. What would happen if you suddenly replaced all negative charge-carriers with positive charge-carriers?
The flow will be opposite. The battery is creating a high negative net charge at one terminal, which is the reason that electrons are repelled and want to move away from this end towards the other end. Had they been positive, then they would be attracted to this terminal instead. The current flow would be opposite. In some circuits (most common ones with metallic wiring) the current flow is made from electrons. In others (semiconductors, ionic solutions, etc.) it is made from positive charges (or a mix). So it is not an irrelevant consideration.
So, which terminal is the "high-potential" one and which is the "low-potential" one? You are thinking, that a charge flowing from a low to a high potential must gain energy - but clearly, electrons moving one way or positive charges moving the other way, gives the same final result - energy will be spent as they move through the circuit in the same manner. They are two equivalent ways of thinking of the circuit.
Which do we call high and which low potential? People in the past have made the decision to always name it as if the charge-carrier was positive. The terminal (or point in the circuit) of higher potential is thus the terminal positive charges will move away from - and a "lower potential" point is a point that attracts positive charges. Regardless of what the actual sign of the charge is. This is a convention, so we didn't have to think in charge-carriers all the time.
So it is correct to think that the "high potential" battery terminal actually is a "low potential" point for the electrons. Because they loose energy on their way to that point. Like a ball falling from high potential on a shelf to low potential on the flow. We just don't say this conventionally. But keep it in mind, and then you always understand what potentials mean.