# Determining EMF and Internal Resistance

1. Nov 8, 2009

### joseppi

Basically I have been given this question:

A battery of unknown EMF and internal resistance is connected in series with an ammeter and a resistance box. The current was 2.0A when R= 4.0 Ohms and 1.5A when R= 6.0 Ohms. Calculate the EMF and internal resistance.

I feel you may have to arrange this somehow although I am probably wrong:

Internal resistance = v1-v2/I2-I1

I have had no attempt at a solution as I am utterly confused by this question.

x

2. Nov 8, 2009

### rl.bhat

Hi joseppi, welcome to PF.
You have to write two equations for two resistances.
Use the equation
I = E/(R + r). Here r is the internal resistance.
Then solve for r.

3. Nov 8, 2009

### joseppi

Hi, sorry for me not having understood your instruction but how can I use the below equation for the answer if I have neither the: r or the E value?

I = E/(R + r)

x

4. Nov 8, 2009

### Seannation

E in this case is the emf, which you can work out with Ohm's Law.

5. Nov 8, 2009

### rl.bhat

You can write two equations for two resistances. With two equations you can find two unknowns.
I1 = E/(R1 + r)-------(1)
I2 = E/(R2 + r)-------(2)
Substitute the values given in the problem. Try to eliminate one unknown to find the other unknown.

6. Nov 8, 2009

### joseppi

2A = ?E/ (4.0 Ohms + ?r)

1.5A = ?E/ (6.0 Ohms + ?r)

I'll be honest I am quite bad with Maths even though I take AS level physics. I have no clue how to eliminate the r or E unknown or even really how to move it around ;/ You couldn't explain how and why you do this? It would be a great help, thank you.
x

7. Nov 8, 2009

### rl.bhat

From the first equation
E = 2A(4.0 Ohm + r)----(1)
Similarly from the second equation write E = ........---(2)
Equate eq 1 and 2, and solve for r.