Finding Internal Resistance with a Graph: How to Derive r+δr from the Gradient?

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Homework Help Overview

The discussion revolves around analyzing a circuit involving an oscilloscope, a signal generator with internal resistance, and a variable resistor. The original poster is attempting to derive the internal resistance from a graph of 1/V_R versus 1/R, having measured corresponding values of R and V_R. There is confusion regarding the expected shape of the graph and how to interpret the results.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the relationship between the variables and the expected linearity of the graph. There are attempts to clarify the nature of the graph for V_R versus R, with some suggesting it resembles a logarithmic or hyperbolic shape. Questions arise about the implications of these shapes and how they relate to the internal resistance.

Discussion Status

The conversation is ongoing, with participants providing insights and corrections regarding the graph's shape. Some guidance has been offered on the relationship between the variables, but there is still uncertainty about the interpretation of the graph and the derivation of internal resistance.

Contextual Notes

There is mention of a plateau in the graph at approximately 300 ohms, which may indicate a specific behavior in the circuit. Participants are also navigating misunderstandings about the mathematical relationships involved in the problem.

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Homework Statement


Homework Equations


The Attempt at a Solution



I got a simple circuit with an oscilloscope, a SG (with internal resistance), and a variable resistor.
r and R are in series.
for V_R=V_0*(R/R+r)
I measured a series of R and V_R values from the oscilloscope.
Now, I'm supposed to plot those values on a graph 1/V_R vs 1/R.
This should give me a straight line, and I am supposed to derive r+δr from the gradient.
But how? and from where? The graph of V_R vs R gave an expontential (I had a plateau at approx. 300 ohms)..I'm really confused
 
Last edited:
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1/V_R = 1/V_0 (R+r)/R = 1/V_0 (1+r/R)
If you set y=1/V_R and x=1/r, this equation is just y=1/V_0 (1+r*x), a straight line, and you can get r.

The graph of V_R vs R gave an expontential (I had a plateau at approx. 300 ohms)
It is not an exponential function.
 
Argh! You are right about the shape! It resemble a log graph actually!
Many thanks for your help.
 
safat said:
Argh! You are right about the shape! It resemble a log graph actually!
No, it follows the shape y=a - b/x, as you can see in the original equation.
 
edit. as I misunderstood your answer.

TO recap:
the graph for 1/V_R vs 1/R should give me a straight line
and the graph for V_R vs R and hyperbola?
Right?
 
ok, last question, and then I promise I will stop bothering you!
Can you explain me why (in a broad sense) the graph is an hyperbola? Which kind of relationship there is between V_R and R to lead to an hyperbola?
 
You can derive V_R=V_0*R/(R+r) from the usual circuit laws. This is equivalent to V_R=V_0 - V_0*(r/R+r)
 

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