Calculating E.M.F and Internal Resistance of a Cell in Series Circuit

AI Thread Summary
The discussion revolves around calculating the electromotive force (e.m.f.) and internal resistance of a cell connected in series with a variable resistor. Two sets of voltage and resistance values are provided: 1.20 V at 16.0 Ω and 1.00 V at 8.0 Ω. The user initially calculated currents based on these values but struggled to reconcile the e.m.f. and internal resistance with their findings. They were advised to set up two equations using the formula I = ε / (r + R) to solve for internal resistance. The user confirmed that the voltmeter's position does not affect the calculations in this scenario.
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Homework Statement



A cell of unknown e.m.f. , ε , and internal resistance, r, is connected in series with a
variable resistor. A voltmeter is then connected across the terminals of the resistor.
When the resistor has a value of 16.0 Ω the voltmeter reads 1.20 V. When the
resistance is reduced to 8.0 Ω the voltmeter reading falls to 1.00 V.

Calculate the e.m.f. and internal resistance of the cell


Homework Equations



V=IR
emf = v + vr
emf = v + ir
(r = internal resistance)

The Attempt at a Solution



I (perhaps wrongly) worked out each current for the given voltages and resistances as

1.2/16= 0.075A
1/8 = 0.125A

I then drew a graph of voltage against current, (I think I may be wrong here also) to find that the e.m.f is 1.5v and the internal resistance(gradient with the two points of (current , voltage)) as -0.25.

But I cannot seem to get calculatiosn with these to add up for instance if the emf is 1.5 then v at either point 1.2 or 1 added to current x internal resistance should equate to 1.5.

Anyhelp would be much appreciated.
 
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Currents you have calculated can be written as I = ε /( r + R ). Write two equations for two currents and solve for r.
 
rl.bhat said:
Currents you have calculated can be written as I = ε /( r + R ). Write two equations for two currents and solve for r.


Hi, Thanks alot.

I saw the other thread in which you helped https://www.physicsforums.com/showthread.php?t=352782

I was a little unsure if the currents were correct due to the location of the voltmeter but I have discovered and you have confirmed the location doesn't matter in this case.

Thanks again!
 
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