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With internal resistance experiment, why does I increase when V decreases? watch

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    For a normal wire at a constant temperature, I know that the graph would have a positive correlation, as when V increases, I increases too because V=IR.

    However, I see that for internal resistance experiments, they vary the value of R with an external variable resistor, and plot a graph of V against I, where as V increases, I decreases.

    Why is that? Can you explain that to me please? I have a feeling it is very simple but i cannot understand why when you vary the resistance, the current would increase when the voltage decreases.


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    So the cell produces a fixed e.m.f right? So e.m.f = V(across internal resistant) + V(across variable resistor)

    therefore e.m.f. = IR(internal) + IR(variable)
    e.m.f. is fixed so as you increase R(variable), I has to fall to maintain constant e.m.f, but by a smaller amount (because I is the same for both terms, so as I drops, so does the value of both terms.
    Therefore, looking at just the variable resistor now, V = IR. R goes up, I goes down, but by less, therefore V goes up.
    Therefore there is a negative correlation between I and V on the variable resistor.

    I may not have explained this very well, so if you still have any questions, just say and I'll draw some graphs and send some pics

    as you decrease resistance, current increases, which increases lost volts (as lost volts = IR) and decreases terminal p.d. (which is what you measure). The emf doesn't change and is equal to terminal p.d. + lost volts.
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Updated: January 28, 2017
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