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

AI Thread Summary
To calculate the internal resistance of a cell and the current flow through a voltmeter, the EMF is given as 1.51V, while the voltmeter reads 1.49V with a resistance of 3.5k Ohms. The internal resistance can be determined using Kirchhoff's rules and the relationship V = RI. The current through the voltmeter was initially calculated as 8.6 x 10^-4 A, but further clarification on using the ideal voltage and the internal resistance is needed. The discussion emphasizes the importance of setting up two equations with two unknowns to solve for the internal resistance accurately. Understanding the voltage drop due to internal resistance is crucial for solving the problem effectively.
debwaldy
Messages
34
Reaction score
0

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:
 
Physics news on Phys.org
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.
 
Back
Top