# Homework Help: Internal resistance-gradient

1. Mar 6, 2013

### safat

1. The problem statement, all variables and given/known data
2. Relevant equations
3. 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: Mar 6, 2013
2. Mar 6, 2013

### Staff: Mentor

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.

It is not an exponential function.

3. Mar 7, 2013

### safat

Argh! You are right about the shape! It resemble a log graph actually!
Many thanks for your help.

4. Mar 7, 2013

### Staff: Mentor

No, it follows the shape y=a - b/x, as you can see in the original equation.

5. Mar 7, 2013

### safat

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?

6. Mar 7, 2013

Right.

7. Mar 7, 2013

### safat

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?

8. Mar 8, 2013

### Staff: Mentor

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)