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

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Thread starter 3 years ago
#1
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.

Thanks

0
3 years ago
#2
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
0
3 years ago
#3
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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