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

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SUMMARY

The discussion focuses on calculating the electromotive force (e.m.f.) and internal resistance of a cell in a series circuit using a variable resistor. The voltmeter readings at resistances of 16.0 Ω and 8.0 Ω were 1.20 V and 1.00 V, respectively. The calculated e.m.f. is 1.5 V, while the internal resistance is determined to be -0.25 Ω. The user confirmed the correctness of their current calculations and the placement of the voltmeter does not affect the results.

PREREQUISITES
  • Understanding of Ohm's Law (V = IR)
  • Familiarity with the concept of electromotive force (e.m.f.)
  • Knowledge of internal resistance in electrical circuits
  • Ability to solve simultaneous equations
NEXT STEPS
  • Learn how to derive e.m.f. and internal resistance using Kirchhoff's laws
  • Study the impact of internal resistance on battery performance
  • Explore graphical methods for analyzing voltage-current relationships
  • Investigate the effects of varying load resistance on circuit behavior
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Students in physics or electrical engineering, educators teaching circuit analysis, and anyone interested in understanding battery characteristics and circuit behavior.

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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 a lot.

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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