Voltage & Current & Push & Force & Potential

  • #1
Barclay
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1

Homework Statement


I’m trying to understand what voltage is. Getting confused. I've written my confusion as questions below and answered the questions.

(i) What is voltage?

(ii) What is voltage passing through a lightbulb? (iii) Is POTENTIAL just another word for FORCE?

(iv) The battery connected to a three 5V bulbs must be a 15 V battery?

Homework Equations

The Attempt at a Solution

(i) Voltage a FORCE that can give PUSH to coulombs to make them flow in a wire. That force is found in the power source e.g. battery(ii) It a measure of how much energy can be transferred per coulomb to, say a light bulb, when a voltmeter is connected parallel to the light bulb OR is it a measure of the FORCE between the two points of the voltmeter(iii) POTENTIAL just another word for FORCE. At the battery the potential is greatest(iv) The battery connected to a three 5V bulbs must be a 15 V battery if it is to deliver 5 V energy per coulomb per bulbThanks for your help
 
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  • #2
Barclay said:
Voltage a FORCE that can give PUSH
No. Voltage can be thought of as electrical analog of force (refer Force-Voltage analogy) but voltage itself is not a force.
Barclay said:
It a measure of how much energy can be transferred per coulomb to, say a light bulb, when a voltmeter is connected parallel to the light bulb
Yes.
Barclay said:
OR is it a measure of the FORCE between the two points of the voltmeter
No.
Barclay said:
POTENTIAL just another word for FORCE.
No.
Circuits work on potential "difference".
Barclay said:
The battery connected to a three 5V bulbs must be a 15 V battery if it is to deliver 5 V **joule** energy per coulomb per bulb
Yes, if the bulbs are in series.
 
  • #3
To designate a bulb as a "5 V" bulb is meaningless. Bulbs are rated by power and voltage. For example, a 100 Watt - 120 Volt bulb will consume 100 Watts of power if connected to a 120 Volt power source. Also, note that any three identical bulbs (not necessarily "5 V" ones) connected to a 15 V battery will have "5 V joule energy per coulomb per bulb."

On edit: The three identical bulbs must be connected in series for this to be true.
 
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  • #4
cnh1995 said:
No. Voltage can be thought of as electrical analog of force (refer Force-Voltage analogy) but voltage itself is not a force.

Circuits work on potential "difference".

If voltage is not a type of force in the battery then what is pushing the electrons along the wire? I keep reading things about the electrons receiving a PUSH down a river on a incline. A PUSH is a force. What is the ELECTROMOTIVE FORCE?

What is voltage then? Is it just a MADE UP CONCEPT to say how much joules of energy each coulomb of electrons gain as they pass through the battery?

I think I'm having a EUREKA moment; please read this:
Voltage is the difference in charge between two points. So at the +ve end of the battery the coulombs of electrons have no charge. But by the time they have passed through the battery they have gained charge from the chemicals in the battery. The charge gives the electrons energy so they have joules of energy. They have a certain number of joules per coulomb of energy (which is the voltage).
Current is the rate at which charge is flowing (number of coulombs per second).

When the coulombs of electrons pass through a bulb they lose some energy (joules per coulomb) so there is a potential difference between the entry and exit of the bulb. The PD is a measure of how much joules per coulomb of energy has been used up by the bulb

PLEASE TELL ME WHAT YOU THINK
 
  • #5
I've amended my EUREKA moment shown above. there were probably too many errors to sort out so no one replied so here's my revised version:

Voltage / potential difference is the difference in the energy available to be released by the coulombs of electrons between two points (joules per coulomb).

So at the -ve end of the battery the coulombs of electrons arriving at the terminal hold no energy per coulomb (J/C). But by the time they have passed through the battery they have gained some J/C from the chemicals in the battery.
They now have a certain number of joules per coulomb (which is the voltage).

Current is the rate at which the electrons are flowing (number of coulombs per second).

When the coulombs of electrons pass through a bulb they transfer some energy (joules per coulomb) so there is a potential difference between the entry and exit of the bulb. The PD is a measure of how much joules per coulomb of energy has been used up by the bulb

Please tell me what you think
 
  • #6
Barclay said:
Voltage / potential difference is the difference in the energy available to be released by the coulombs of electrons between two points (joules per coulomb).
That's OK. Have you ever wondered why you pay so much money to the electric company? They send you electrons which you send back to them keeping none! The deal is that they send you electrons at high potential energy and you send them back to the company after you use some of that energy for, say, to light your light bulbs.
Barclay said:
So at the -ve end of the battery the coulombs of electrons arriving at the terminal hold no energy per coulomb (J/C). But by the time they have passed through the battery they have gained some J/C from the chemicals in the battery.
They now have a certain number of joules per coulomb (which is the voltage).
Something like that.
Barclay said:
Current is the rate at which the electrons are flowing (number of coulombs per second).
You nailed that one.
Barclay said:
When the coulombs of electrons pass through a bulb they transfer some energy (joules per coulomb) so there is a potential difference between the entry and exit of the bulb. The PD is a measure of how much joules per coulomb of energy has been used up by the bulb
It is better to think of it as the battery setting up the potential difference. The job of a battery in a circuit is to maintain a constant potential difference between its terminals. If it is connected to a bulb, there will be electrons at the high energy end of the bulb and electrons at the low energy of the bulb. The higher energy electrons will "fall" losing their potential energy without gaining kinetic energy. The lost energy appears as heat in the filament which gets so hot that starts glowing. The power (joules per second or Watts) dissipated in the bulb is the product of current times voltage.

I think that you have considerably enhanced your understanding. Eureka moments are the best because you "get it" better when you explain it to yourself. :smile:
 
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