How Do You Calculate Internal Resistance and Current Flow in a Circuit?

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Homework Help Overview

The discussion revolves around calculating the internal resistance of a cell and the current flow through a voltmeter connected to it. The original poster presents a scenario involving an electromotive force (EMF) measurement and the effects of connecting a voltmeter with a known resistance.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the application of Kirchhoff's rules to relate the voltages and resistances in the circuit. There are questions about the interpretation of the resistance values and the appropriate use of measured voltages in calculations.

Discussion Status

Some participants have provided alternative equations and clarifications regarding the use of the ideal voltage and the current previously calculated. There is an ongoing exploration of how to solve for the internal resistance with the information given, but no consensus has been reached on the method to proceed.

Contextual Notes

There is some uncertainty regarding the resistance value of the voltmeter and the implications of using it in calculations. Participants are also considering the impact of the internal resistance on the voltage readings.

debwaldy
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Homework Statement


Hi I've tried to solve this question but I keep getting stuck at a certain point, it probably very simple but any tips would be greatly appreciated

The EMF of a cell is measured using a potentiometer and is found to be 1.51V. When a voltmeter, which has a resistance of 3.5k Ohms is connected across the cell, it reads 1.49V. What is the internal resistance of the cell. What current flows through the voltmeter?
If a resistor of resistance 50 Ohms is connected across the cell, what current flows through it?

Homework Equations


Kirchhoff's rules
V=RI


The Attempt at a Solution



I said so far:
(internal resistance)*I = 1.51
1.49= I1*(3.5*10^3) - 1.51
& found I1 current flowing through voltmeter = 8.6*10^-4A but i don't know what to do from here and i think this might be wrong anyway...:frown:
 
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Are you sure its "3.5k" ohms?
Because that would be 3500ohms.
 
If you use kirchhoff rule and consider the two resistances: where v is the ideal voltage (that is under no load of the emf)

V-(R(int)-R(ext))i=0
and V-R(int)i=0.02 solve for R, is that what you got?
 
i don't understand,how do you solve that equation when you dontknow what V is, do i take 1.51V as the value for V in that equation?
 
yes that's the ideal voltage measured under no current. The 1.49 is with the Ir drop due to current across int resistance.
 
and i just use the current value i have already calculated?
ok i think i follow now,thanks for the help
 
debwaldy said:
and i just use the current value i have already calculated?
ok i think i follow now,thanks for the help

I'm a little unclear on your first eqn, if those don't work use the two I posted, remember its 2 unknowns and two equations tho the error from using just the 3500 Ohm resistor to compute i will likely be negligible.
 

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