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Graph of 1/R against E/V 
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#19
May614, 05:15 PM

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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 + IrThe R is whatever you used, R_{1}, R_{2} , ... , R_{2} + R_{3}, ... 


#20
May614, 05:22 PM

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Yh i have my test tomorrow on the 7th May. AQA Physics Unit 03X task 3.



#21
May614, 06:52 PM

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Take ## \ E=IR+Ir \ ## and divide by ##\ IR\ .##



#22
May714, 05:24 AM

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#23
May714, 08:27 AM

P: 1,970

Yes, the slope of the curve is the internal resistance.



#24
May714, 02:50 PM

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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. 


#25
May1314, 03:10 AM

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? 


#26
May1314, 05:04 AM

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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. 


#27
May1314, 12:48 PM

P: 58

your right im completely wrong, I took the same exam today, luckily I realised before I answered the question.



#28
May1314, 12:49 PM

P: 58

I couldnt see that all you needed to do was factor out the 1/R as x



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