How exactly are charge carriers driven around a complete circuit??

Watch
BTAnonymous
Badges: 22
Rep:
? You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#1
Report Thread starter 3 years ago
#1
Everywhere I look says that the battery's potential difference drives these charge carriers thus a current is produced. But HOW exactly does p.d drive these charge carriers around a circuit? It's not enough for me to say that p.d 'moves' these electrons out of the battery!
0
reply
Fox Corner
Badges: 21
Rep:
? You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#2
Report 3 years ago
#2
Hi! I've moved this to the Physics forum for you - hopefully someone will help you out here
0
reply
mphysical
Badges: 14
Rep:
? You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#3
Report 3 years ago
#3
(Original post by BTAnonymous)
Everywhere I look says that the battery's potential difference drives these charge carriers thus a current is produced. But HOW exactly does p.d drive these charge carriers around a circuit? It's not enough for me to say that p.d 'moves' these electrons out of the battery!
Inside the battery a chemical reaction produces a flow of ions to the + ve.
There is now a p.d. between the +ve and -ve terminals of the battery.
The loosely bound outer electrons on the copper atoms are attracted to the +ve terminal and are dislodged from their shell by the pd.
The movement of electrons in the circuit is the current.
0
reply
BTAnonymous
Badges: 22
Rep:
? You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#4
Report Thread starter 3 years ago
#4
(Original post by mphysical)
Inside the battery a chemical reaction produces a flow of ions to the + ve.
There is now a p.d. between the +ve and -ve terminals of the battery.
The loosely bound outer electrons on the copper atoms are attracted to the +ve terminal and are dislodged from their shell by the pd.
The movement of electrons in the circuit is the current.
ok so is it right to say (from conventional current) that electrons have high potential at the positive terminal but as they flow towards to negative their potential decreases?
0
reply
the bear
Badges: 20
Rep:
? You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#5
Report 3 years ago
#5
:iiam:
0
reply
BTAnonymous
Badges: 22
Rep:
? You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#6
Report Thread starter 3 years ago
#6
(Original post by the bear)
:iiam:
I'll never be happy then .-.
0
reply
mphysical
Badges: 14
Rep:
? You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#7
Report 3 years ago
#7
(Original post by BTAnonymous)
ok so is it right to say (from conventional current) that electrons have high potential at the positive terminal but as they flow towards to negative their potential decreases?
Electrons don't have any potential (not sure what you mean). They have a negative charge so they are attracted to the +ve.
0
reply
BTAnonymous
Badges: 22
Rep:
? You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#8
Report Thread starter 3 years ago
#8
(Original post by mphysical)
Electrons don't have any potential (not sure what you mean). They have a negative charge so they are attracted to the +ve.
but they have the potential to move from the -ve to the +ve?

this is what I'm confused about. the analogy I use is think of a pond and a reservoir which is elevated above the pond. they are connected by a pipe. when water flows up the pipe, work is done against gravity to do so and this energy is stored as gpe (much like the energy of the electrons on the -ve plate). when water flows down the pipe, this energy is dissipated and it's gpe decreases, much lie how electrons deposit their energy across components in a circuit. therefore the gpe is lowest at he pond.
0
reply
mphysical
Badges: 14
Rep:
? You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#9
Report 3 years ago
#9
(Original post by BTAnonymous)
but they have the potential to move from the -ve to the +ve?

this is what I'm confused about. the analogy I use is think of a pond and a reservoir which is elevated above the pond. they are connected by a pipe. when water flows up the pipe, work is done against gravity to do so and this energy is stored as gpe (much like the energy of the electrons on the -ve plate). when water flows down the pipe, this energy is dissipated and it's gpe decreases, much lie how electrons deposit their energy across components in a circuit. therefore the gpe is lowest at he pond.
Using this analogy is fine for water = current. But not really the movement of electrons. Conventional current flows the opposite direction to the electron flow.
The movement of electrons creates the current but the individual electron movement is not a simple flow from one end to the other
1
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Back
to top
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

Would you give consent for uni's to contact your parent/trusted person in a mental health crisis?

Yes - my parent/carer (118)
33.71%
Yes - a trusted person (95)
27.14%
No (93)
26.57%
I'm not sure (44)
12.57%

Watched Threads

View All
Latest
My Feed