# Graph of 1/R against E/V

by Tangent100
Tags: 1 or r, e or v, graph
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 Quote by Tangent100 Emf was calculated without the resistors in circuit. Then the circuit switch was placed on and the current had to pass through resistors. The voltage was then measured with a volt meter attached in parallel, and so it decreased as more energy was lost through the resistors, depending on what resistors were in the circuit at that instance, so R1,R2,R3,R1+R2,R1+R3,... and so on.
OK.

Now it's making more sense.

I'll try to get back to this soon. In the meantime, someone else may jump in.

In your circuit, the battery itself is usually modeled as an ideal voltage source providing emf, E, in series with a resistor, having (usually small) resistance, r.

Thus the equation:
E = IR + Ir
The R is whatever you used, R1, R2 , ... , R2 + R3, ...
 P: 1 Yh i have my test tomorrow on the 7th May. AQA Physics Unit 03X task 3.
 Emeritus Sci Advisor HW Helper PF Gold P: 7,819 Take ## \ E=IR+Ir \ ## and divide by ##\ IR\ .##
P: 10
 Quote by SammyS Take ## \ E=IR+Ir \ ## and divide by ##\ IR\ .##
Oh my... that would mean the gradient is internal resistance, could someone check if what I done here is correct?

 P: 1,988 Yes, the slope of the curve is the internal resistance.
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 Quote by Tangent100 Oh my... that would mean the gradient is internal resistance, could someone check if what I done here is correct? [ Img]http://s24.postimg.org/ro4owqox1/2014_05_07_11_15_16.png[/img]
Yes, that's it.

It strikes me as kind of a strange way to combine two (or is it three?) very common quantities.

Substituting in other orders can make it very difficult to see this result.
 P: 58 No. I don't believe you can deduce the internal resistance from that gradient. gradient = EMF/V by 1/R (IR + Ir) / IR = 1 + r/R (1 + r/R) x R = R+r given the gradient is a constant, however the sum of the two resisitances (R+r) the total resistance of the circuit is not constant. I don't see how this can be accurate?