The discussion revolves around finding the electromotive force (emf) and the internal resistance of a battery when connected to two different resistors, R1 and R2, which produce currents I1 and I2, respectively. The problem is situated within the context of introductory physics, specifically dealing with circuit analysis and Ohm's law.
Several participants have engaged in rewriting the problem statement for clarity and have provided equations that relate the currents to the emf and internal resistance. There is an ongoing verification of the equations and calculations presented, with some participants expressing confidence in the correctness of their approaches.
Participants are working under the constraints of typical homework guidelines, which may limit the extent of assistance provided. There is a focus on ensuring that the equations are correctly formatted and interpreted, with some emphasis on the proper application of Ohm's law.
Here's what I didBvU said:Can I rephrase the problem statement as :
With a resistor R1 connected to a battery, the current is I1.
With a resistor R2 connected to a battery, the current is I2.
Find emf and the internal resistance.
Your equation requires brackets: ## I_i = emf / (R_i + r_{internal})##
So you have two equations with two unknowns: ## emf## and ##r_{internal}##. Can you write them down and post ?
And: the mentor will move this thread to introductory physics homework.
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V=IR and not I/R.prishila said:efm=I1/(R1+r)=I2/(R2+r)
You're right. i corrected itcnh1995 said:V=IR and not I/R.
Check the 3rd equation.prishila said:I1/I2=(R2+r)/(R1+r)
I1R1+I1r=R2I2+I2r
r(I1-I2)=R2I2+I1R1
Correctedcnh1995 said:Check the 3rd equation.
Looks correct!prishila said:Here's what I did
efm=I1*(R1+r)=I2*(R2+r)
I1*(R1+r)=I2*(R2+r)
I1/I2=(R2+r)/(R1+r)
I1R1+I1r=R2I2+I2r
r(I1-I2)=R2I2-I1R1
r=(R2I2-I1R1)/(I1-I2)
I replaced R with what I found and it results
efm=I1I2(R2-R1)/(I1-I2)
Did I find efm and r correctly?