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R or r from the gradient of V by I graph...? watch

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    I have a problem understanding something which my teacher was unable to explain from my recent exam. When given a graph of Current (x-axis) by Voltage (y-axis), to find the resistance if I'm not mistaken you simply find out the gradient and that's all since it is in the form of V/I which is equal to R. However even though this is the case, the teacher stated the internal resistance is apparently also the gradient as a result of using the equation E-Ir=V (in the form of y=mx+c). How can the internal resistance and resistance both be the gradient??? Can anyone clear up my confusion?
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    (Original post by TotenhamsHokage)
    I have a problem understanding something which my teacher was unable to explain from my recent exam. When given a graph of Current (x-axis) by Voltage (y-axis), to find the resistance if I'm not mistaken you simply find out the gradient and that's all since it is in the form of V/I which is equal to R. However even though this is the case, the teacher stated the internal resistance is apparently also the gradient as a result of using the equation E-Ir=V (in the form of y=mx+c). How can the internal resistance and resistance both be the gradient??? Can anyone clear up my confusion?
    By dimensional analysis the gradient of any plot of voltage(y-axis) and current(x-axis) will be in Ohms and both internal resistance and resistance are measured in Ohms.

    However, the graphs you are referring to are subtly different. Here the voltage is the potential difference across a cell/battery (e.m.f source) and NOT the potential difference across a component. We know from Ohm's Law that V=IR for a component hence R, the resistance, is the gradient. However, since we have an e.m.f source the equation is V=E-Ir so the gradient here is -r. It is all about what we aree measuring.
 
 
 
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