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Voltages in Potential Divider

  • Thread starter nokia8650
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  • #1
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See attachment.

Question: "The switch S is now closed. Explain, without calculation, why the voltage across the
thermistor will fall"

I am able to deduce that by turning the switch on, the bulb becomes part of the circuit, effectively loweing the resistance of the parrallel "section" and hence the resisitance of the overall circuit. This will cause a greater current to flow. Why, however, will the voltage across the thermister fall? Thanks
 

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  • #2
madmike159
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I can't see the picture so I don't know what the circuit looks like. Try uploading it to imageshack and putting in a link or something like that.
 
  • #3
cristo
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I can't see the picture so I don't know what the circuit looks like.
Sometimes it takes a while for a mentor to spot the attachment to approve it-- an hour's not bad going!
 
  • #4
madmike159
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Wellif theresistanceof the thermistor was high then most of the current would go round via the bulb. If the resistance in the thermistor droped low enough then the bulb should go out.
 

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