Might be okay actually. As current increases, voltage decreases, therefore as voltage decreases resistance decreases. Considering the internal resistance of the cell to be r, V = E - Ir so as current increases, Ir becomes larger so the terminal pd (E - Ir) decreases.
Is this okay or is it a bit long winded/have you got any better explanation?
Might be okay actually. As current increases, voltage decreases, therefore as voltage decreases resistance decreases. Considering the internal resistance of the cell to be r, V = E - Ir so as current increases, Ir becomes larger so the terminal pd (E - Ir) decreases.
Is this okay or is it a bit long winded/have you got any better explanation?
Yeah, looks ok to me! You can also think of it in terms of the resistances. If the variable resistor is the same as the internal resistance then the emf is divided equally between the two, so the terminal p.d=EMF-V(int res) as you said. When the variable resistor is far larger than the internal resistance it gets much more of the emf so V(int res) is smaller and so the terminal voltage approaches the EMF. Hope that makes sense.
Yeah, looks ok to me! You can also think of it in terms of the resistances. If the variable resistor is the same as the internal resistance then the emf is divided equally between the two, so the terminal p.d=EMF-V(int res) as you said. When the variable resistor is far larger than the internal resistance it gets much more of the emf so V(int res) is smaller and so the terminal voltage approaches the EMF. Hope that makes sense.
That's a great explanation, I actually thought of that since I realised it's a 'potential divider' circuit but couldn't finish my argument. Thanks