# E.M.F and internal resistance

**1. The problem statement, all variables and given/known data**

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

**2. Relevant equations**

V=IR

emf = v + vr

emf = v + ir

(r = internal resistance)

**3. 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.

http://ift.tt/NL3rIE

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