Calculating Internal Resistance of a Voltage Source from Terminal Voltage Drop

  • Thread starter Thread starter nautica
  • Start date Start date
  • Tags Tags
    Emf Voltage
AI Thread Summary
The discussion revolves around calculating the internal resistance of a voltage source based on a terminal voltage drop and current increase. The internal resistance is derived from the equation E = V - Ir, leading to the conclusion that r = -0.4 Ohms when applying the given values. However, the instructor argues that this calculation is incorrect due to the lack of information about the load resistance, indicating that multiple unknowns prevent a definitive solution. Additionally, it is clarified that the electromotive force (emf) cannot be determined without knowing the potential difference across the terminals. The conversation highlights the importance of considering all variables in such calculations.
nautica
Okay here is the question:

a) what is the internal resistance of a voltage source if its terminal voltage drops by 2 V when the current supplied increases by 5 A.

b) Can the emf of the voltage source be found with the information supplied?

I know that E = V - Ir, and I am pretty sure that a simple algebraic equation can be put together. So what formulat am I missing??

Thanks
Nautica
 
Physics news on Phys.org
You know that E = V - Ir so ΔE= ΔV- ΔI r. You are told that ΔE= 2volts when &Dekta;I= -5 amps. Of course, ΔV= 0 since the voltage source is fixed.
Thatis: 2= -5r. Can you solve for r now? :smile:.

The answer to b is NO. In effect V is the "constant of integration" when you "integrate" dE= -rdI. There is no way to find that from the information given.
 
Last edited by a moderator:
I am in College Physics, we can not use calculus for an answer. So is there a reason algebraically.

thanks
nautica
 
If u know the potential diff b/w two terminals, then u can find the emf.

With this information u cannot find
 
So ther is no way to find emf without Potential difference?

thanks
nautica
 
Okay For a) here is what I got

V=E-Ir

or dV=dE - dIr

2V = -5r

so r = -.4Ohms

BUT, my instructor said this was wrong b/c there I did not no the Rload and there are too many unknowns for this to be solved.

I can see how part B could not be solved but this looked so simple for part a)

He said this was the other equation I should have considered, but I do not see why it was needed.

I = E / (Rload + r)

Thanks
nautica
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top