Deathlotus
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circuit EMF 2V, has a rheostat and wire in series. The wire is 3.4 ohms, and rheostat goes from 0-1500ohms

Asks to calculate the maximum and minimum p.d. that the wire can have.
I understand that maximum p.d. is 2, but i figured minimum would be 3.4/1503.4, however the MS says that it has to be 2x(3.4/1503.4). Why does it have to be doubled?
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uberteknik
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(Original post by Deathlotus)
circuit EMF 2V, has a rheostat and wire in series. The wire is 3.4 ohms, and rheostat goes from 0-1500ohms

Asks to calculate the maximum and minimum p.d. that the wire can have.
I understand that maximum p.d. is 2, but i figured minimum would be 3.4/1503.4, however the MS says that it has to be 2x(3.4/1503.4). Why does it have to be doubled?
You have correctly stated the ratio between the resistances as

\frac{R_{wire}}{R_{wire} \rm \ + \ R_{rheostat}}

This must now be multiplied by the source voltage to get the actual p.d. developed across the wire:

V_{emf}(\frac{R_{wire}}{R_{wire} \rm \ + \ R_{rheostat}})

Hence:

2 \rm \ x \ (\frac{3.4}{3.4 + 1500})
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