According to my physics book, the total e.m.f. around a circuit = the sum of the p.d.s accross each component. What about internal resistance? I thought the e.m.f. was the total energy per coloumb including the energy lost from internal resistance.

Question

anonymous · Answer

Emf of cell= (current)(extl. resistance)  + (current)(internal resistance)

anonymous · Answer

http://farside.ph.utexas.edu/teaching/302l/lectures/node57.html See here for detailed explanation

anonymous · Answer

On that page, e.m.f is described as the 'pure voltage source'. I understand that, I just don't understand how the total e.m.f. around a circuit = the sum of the p.d.s accross each component, where does that take into account internal resistance?

anonymous · Answer

It does take into account the internal resistance of the cell.  See the equation I wrote above in first post

anonymous · Answer

I mean this equation 'total e.m.f. around a circuit = the sum of the p.d.s accross each component'

anonymous · Answer

ɛ = ΣIR <- no internal resistance?
ɛ = IR + Ir

anonymous · Answer

r--Internal  resistance of celll

anonymous · Answer

Yes, so why isn't r involved in the first equation?

anonymous · Answer

Your source for 1st equation would be appreciated. I always use this equation:
ɛ = IR + Ir
where
ɛ ->EMF of cell
R->External resistnance in the circuit
r-> Internal resistnace of cell
I-> current in the circuit

anonymous · Answer

Same, which is why ɛ = ΣIR doesn't make sense to me. It's given in my revision guide. Here's a picture of it http://i.imgur.com/nTTe2.jpg

anonymous · Answer

ΣIR = IR + Ir

anonymous · Answer

So the 'internal resistor' of the battery is counted as a component?

anonymous · Answer

Yes. Basically there is no internal resistor.  The internal resistance of the cell is due to the resistance faced by electrons while traveliing in electrolyte of the cell.

anonymous · Answer

Okay, thanks for your help.

anonymous · Answer

Welcome! :)