# Working out EMF of a cell

• DeanBH
Ri and it would be 5.2 amps not 0.06How can you apply Kirchoff's voltage law if you don't know what it is?guessed, that the volts in the battery would have to make up for the difference. but why is the amps 0.06 out of 0.032, if the volts were 5.2 instead of 4.2 in the right part of the circuit it wouldn't fit the rule RI=v/Ri and it would be 5.2 amps not 0.06
I know I need to find the difference of of the volts about the 2 resistors.

it ends up being 1 volt, but how the hell does i(R+r)=emf come into that

DeanBH said:
I know I need to find the difference of of the volts about the 2 resistors.

it ends up being 1 volt, but how the hell does i(R+r)=emf come into that
That equation is valid only for two resistors in series, i.e. where a cell has an internal resistance r and is connected to a load resistance R.

As I said previously, you need to apply Kirchoff's laws here.

I did, the answer is 1

just doesn't make sense.

DeanBH said:
I did, the answer is 1

just doesn't make sense.
I also have emf = 1V. So what doesn't make sense?

Why do i have to count one of those volt readings as negative, and why inst it 1/r it looks like a a parallel

DeanBH said:
Why do i have to count one of those volt readings as negative, and why inst it 1/r it looks like a a parallel
What is Kirchoff's voltage/loop law?

Hootenanny said:
What is Kirchoff's voltage/loop law?

no idea

Hootenanny said:
As I said previously, you need to apply Kirchoff's laws here.
DeanBH said:
I did, the answer is 1
Hootenanny said:
What is Kirchoff's voltage/loop law?
DeanBH said:
no idea
How can you apply Kirchoff's voltage law if you don't know what it is?

guessed, that the volts in the battery would have to make up for the difference. but why is the amps 0.06 out of 0.032, if the volts were 5.2 instead of 4.2 in the right part of the circuit it wouldn't fit the rule RI=v

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