# Graph of 1/R against E/V

by Tangent100
Tags: 1 or r, e or v, graph
Emeritus
HW Helper
PF Gold
P: 7,801
 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: 7,801 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: 1,969 Yes, the slope of the curve is the internal resistance.
Emeritus
HW Helper
PF Gold
P: 7,801
 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?
PF Gold
P: 11,942
 Quote by peterspencers 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?
What would be changing?
Given that 'internal resistance' is not Ohmic over the whole range of currents, it is considered to be, in most situations. When it isn't, you don't get a straight line graph.
 P: 58 your right im completely wrong, I took the same exam today, luckily I realised before I answered the question.
 P: 58 I couldnt see that all you needed to do was factor out the 1/R as x

 Related Discussions High Energy, Nuclear, Particle Physics 7 Introductory Physics Homework 5 Classical Physics 1 Introductory Physics Homework 4 Differential Equations 6