As Internal resistance and emf question

If a cell is connected with an external resistor of 6V and a switch and a voltmeter is connected across the terminals of the cell which reads 2.6 V when the switch is open and 2.4 V when the switch is closed, how do you find the emf and the internal resistance?

so far i tried doing:

V = IR
hence 2.4/6 = 0.4 A

V = Ir
2.6/0.4 = 6.5 V

substituting the above into:

V = E - Ir
2.6 = E - (0.4 x 6.5)
E = 2.6 + (0.4 x 6.5)
therefore E = 5.2 V

but something seems very wrong here..

any help would be much appreciated =)

The Attempt at a Solution

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Hi there.
Well your first problem is misunderstanding of Emf. When the voltmeter is connected across the supply with no other components, this means that there is no current being drawn from the supply. Therefore, there is no resistance, so the reading on the voltmeter when the switch is closed is the Emf. After you understand that I think the problem should fix itself.
You have correctly identified the current in the circuit, so you can then use the equation
E=Ir+V to work out the internal resistance.
SO: 2.6=(0.4 x r)+2.4
0.2= 0.4r
r=0.5 (ohms)

E = 2.6
r = 0.5

Hope this helps and good luck with the exam :) xx