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Electricity

Why is the Electromotive force always less than the terminal potential difference ? I cant seem to get my head around his concept.
Reply 1
Original post by abdilighter
Why is the Electromotive force always less than the terminal potential difference ? I cant seem to get my head around his concept.

It's been a while but I'll give it a go:
The reverse holds actually, the battery has an internal resistance which causes the e.m.f to drop so the e.m.f actually exceeds the potential difference.

However, this is not always the case, for a battery being recharged, the terminal potential difference will actually exceed the e.m.f.

A quick google gave a couple of links you might like to look at:
Regarding the internal resistance.
Regarding the recharging battery.
(edited 8 years ago)
Original post by abdilighter
Why is the Electromotive force always less than the terminal potential difference ? I cant seem to get my head around his concept.


emf=V+Ir
Reply 3
Original post by abdilighter
Why is the Electromotive force always less than the terminal potential difference ? I cant seem to get my head around his concept.


TPD = EMF - Lost Volts, so the EMF is always less as it takes into consideration the lost volts which are deducted as it results in energy dissipated as heat etc
Reply 4
Original post by derpz
TPD = EMF - Lost Volts, so the EMF is always less as it takes into consideration the lost volts which are deducted as it results in energy dissipated as heat etc

Thank you !
Reply 5
Original post by abdilighter
Why is the Electromotive force always less than the terminal potential difference ? I cant seem to get my head around his concept.


Work has to be done to get the charges to leave the battery, hence some of the energy is lost due to this 'internal resistance'.
Original post by abdilighter
Why is the Electromotive force always less than the terminal potential difference ? I cant seem to get my head around his concept.


Original post by joostan
It's been a while but I'll give it a go:
The battery has an internal resistance which causes the e.m.f to drop.

However, this is not always the case, for a battery being recharged, the terminal potential difference will actually exceed the e.m.f.

A quick google gave a couple of links you might like to look at:
Regarding the internal resistance.
Regarding the recharging battery.


Original post by thefatone
emf=V+Ir


Original post by derpz
TPD = EMF - Lost Volts, so the EMF is always less as it takes into consideration the lost volts which are deducted as it results in energy dissipated as heat etc


Original post by abdilighter
Thank you !


Original post by Kyx
Work has to be done to get the charges to leave the battery, hence some of the energy is lost due to this 'internal resistance'.



isn't V = emf - Ir.
So emf is never less than V
Reply 7
Original post by Xenon17
isn't V = emf - Ir.
So emf is never less than V


Yes.
Emf is always greater.
Reply 8
Original post by Xenon17
isn't V = emf - Ir.
So emf is never less than V


Yeah, just remember that equation and you should be fine!
Original post by Kyx
Yes.
Emf is always greater.


Original post by derpz
Yeah, just remember that equation and you should be fine!



Yh but

Original post by abdilighter
Why is the Electromotive force always less than the terminal potential difference ? I cant seem to get my head around his concept.


thats where the OP has gone wrong
Original post by Xenon17
isn't V = emf - Ir.
So emf is never less than V


Yes, that's what I said, although I think I misread the question, so my answer makes less sense.
Thanks for this, will edit.
Reply 11
Original post by Xenon17
Yh but



thats where the OP has gone wrong


Yes
The OP went wrong saying it is less: it is more :smile:
Original post by Xenon17
isn't V = emf - Ir.
So emf is never less than V


just rearranged it.. :smile:
Reply 13
Original post by thefatone
just rearranged it.. :smile:


v = emf - Ir

adding Ir to both sides gives: v + Ir = emf - Ir + Ir

v + Ir = emf

Or

emf = v + Ir

:wink:

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