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

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SUMMARY

The discussion focuses on calculating the internal resistance of a cell and the current flow through a voltmeter connected to it. The electromotive force (EMF) of the cell is measured at 1.51V, while the voltmeter reads 1.49V with a resistance of 3.5k Ohms. Using Kirchhoff's rules and the equation V = RI, participants derive the current flowing through the voltmeter as 8.6 x 10^-4 A. The conversation emphasizes the importance of understanding the relationship between voltage, internal resistance, and current flow in circuit analysis.

PREREQUISITES
  • Understanding of Kirchhoff's rules
  • Familiarity with Ohm's Law (V = RI)
  • Basic knowledge of electrical components, specifically resistors and voltmeters
  • Ability to solve simultaneous equations
NEXT STEPS
  • Study the application of Kirchhoff's rules in complex circuits
  • Learn how to calculate internal resistance in various circuit configurations
  • Explore the effects of load on voltage readings in electrical circuits
  • Investigate the use of potentiometers for measuring EMF
USEFUL FOR

Students studying electrical engineering, hobbyists working with circuits, and anyone interested in understanding the principles of current flow and resistance in electrical systems.

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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