A few questions concerning voltage, current and resistance

AI Thread Summary
In the discussion about voltage, current, and resistance in a circuit with a 5V battery and two resistors, participants clarify misconceptions about charge movement and energy transformation. When the circuit is closed, an electromagnetic wave propagates, affecting all charges simultaneously rather than sequentially accelerating them. The current remains constant despite energy loss due to resistance because the power supply continuously provides energy to replace what is lost. The analogy of pushing boxes illustrates how energy is transferred through the circuit, with charges losing energy to resistance while being replenished by the battery. Understanding these concepts is crucial for developing a deeper intuition about electrical circuits and their behavior.
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[ mentor note: post adjusted to conform to homework template and some spellings fixed ]

1. The Problem


I have some few questions that connect voltage, current and resistance.
Imagine that we have a circuit that consists of a battery (5V), 2 resistors. Now once we close the circuit, battery would do work to move charges to positive terminal so they would gain an electric potential energy of 5 joules per coulombs, and it would start decreasing while their kinetic energy increases by the same amount, until they reach the first resistance.

My questions are:

- is that what really happens in the circuit at first?

-What does happen to their kinetic energy, potential energy, after leaving the resistor and why doesn't it affect the current?

-Why is the current constant meanwhile electrons gain acceleration because of electric field? As well as kinetic energy?

- Why do electrons persist to move even after leaving the second resistor, since they supposedly lost all their potential energy? is it because of the kinetic energy?

Homework Equations

are:
[/B]
Electric field work = -Q * v = -ΔPE = +ΔKE
Net force = m*a = q * electric field strength
Resistor's voltage drop = I * r

The Attempt at a Solution


[/B]
My attempts were:
- Yes
- Kinetic energy would be converted to another form of energy , potential energy would drop, current would remain the same because potential energy which was converted to kinetic energy compensated loss in velocity because of resistance and thus current would remain constant, maybe.
- No idea
- I think because of the kinetic energy they still have
 
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Volta said:
Now once we close the circuit, battery would do work to move charges to positive terminal so they would gain an electric potential energy of 5 joules per coulombs, and it would start decreasing while their kinetic energy increases by the same amount, until they reach the first resistance.

This is incorrect. When the circuit is first closed, an EM wave is sent through the circuit. This EM wave travels much faster than the charges are accelerated at, and results in the charges throughout the entire circuit being affected by the battery, not just the charges at the terminal. For the most part, you cannot equate kinetic energy to electric potential energy in a circuit, as the charges are constantly interacting with the voltage source and the charges within the wires, resistors, etc. So the charges in the terminal aren't accelerated, moving faster and faster until they hit the first resistor. They interact with the charges in the wire, which interact with the charges further down the wire, which interact with the charges even further down the wire, which interact with the charges in the resistor, which interact with the charges in the other wire, etc. The battery is acting on all the charges throughout the whole circuit all at once.

Not only that, but the motion of the charges is almost entirely randomized in direction and speed, with only a small net overall that gives rise to current flow. It's kind of like cars on a freeway. You have cars going both directions and they have somewhat random speeds. Current flow is the equivalent of slightly more cars traveling one way than the other. This can take the form of roughly the same number of cars traveling at a higher average speed, or more cars traveling at a slower average speed.

Another analogy would be to imagine you are pushing a series of boxes lined up on the ground, all touching each other, so that when you push the first box all the subsequent boxes move too. It's not correct to say that the first box accelerates until it hits the 2nd box because those two boxes are already touching. Their atoms are thus interacting with each other, and any pushing you do on the first box is almost immediately transmitted through to the rest of the boxes. A similar phenomenon happens in a circuit. You also can't equate a potential difference in energy to their velocity, as friction is constantly acting on all the boxes, much like resistance is constantly acting on the charges. Neither the boxes nor the charges accelerate up to some value determined solely by the available energy.

Note that these two analogies are just that; analogies. They shouldn't be taken too far and if you try to wring too much from them they will collapse.
 
Volta said:
[ mentor note: post adjusted to conform to homework template and some spellings fixed ]

1. The Problem


I have some few questions that connect voltage, current and resistance.
Imagine that we have a circuit that consists of a battery (5V), 2 resistors. Now once we close the circuit, battery would do work to move charges to positive terminal so they would gain an electric potential energy of 5 joules per coulombs, and it would start decreasing while their kinetic energy increases by the same amount, until they reach the first resistance.

My questions are:

- is that what really happens in the circuit at first?

-What does happen to their kinetic energy, potential energy, after leaving the resistor and why doesn't it affect the current?

-Why is the current constant meanwhile electrons gain acceleration because of electric field? As well as kinetic energy?

- Why do electrons persist to move even after leaving the second resistor, since they supposedly lost all their potential energy? is it because of the kinetic energy?

Homework Equations

are:
[/B]
Electric field work = -Q * v = -ΔPE = +ΔKE
Net force = m*a = q * electric field strength
Resistor's voltage drop = I * r

The Attempt at a Solution


[/B]
My attempts were:
- Yes
- Kinetic energy would be converted to another form of energy , potential energy would drop, current would remain the same because potential energy which was converted to kinetic energy compensated loss in velocity because of resistance and thus current would remain constant, maybe.
- No idea
- I think because of the kinetic energy they still have

The problem with your picture here is that you are using the idea of motion of charges in vacuum to the motion of charge carriers in a conductor. Those two scenarios are not compatible with one another.

I am not sure which of the two scenarios that you're trying to work out: charge motion in vacuum, or current in a conductor. If you are simply trying to figure out constant current in vacuum, then it would be easier to deal with that than to use conduction current. After all, we have plenty of examples of the former with particle accelerators and vacuum tubes. If you wish to do the latter, then I strongly suggest that before you come up with your scenario, that you first look at the simplest way that we describe charges in a conductor - the Drude Model.

Zz.
 
Aha, i think that using concept of electric potential energy equating to kinetic energy doesn't work here just like zapper said, i don't really know about drude's model.

Then, how can we define the energy of the charges in a circuit? And how can we relate it her to their energy? I just don't get why do they persist to move if they lose all their energy and what does really happen to a single charge the moment it's about to enter a resistor until it leaves it.

I actually want to have a deeper intuition of how things work to become more intuitive in solving current problems and such, along with reliance on kirchhoff laws.
 
Volta said:
I just don't get why do they persist to move if they lose all their energy

The power supply is supplying a constant voltage. So even if they are constantly losing energy through resistance, the power supply is providing more. It's like pushing boxes across the ground. They are constantly losing energy to frictions, but this lost energy is replaced by the energy you provide by pushing.
 
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Ohh alright
 
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