Why does closing the switch cause the voltage across the thermistor to decrease?

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SUMMARY

Closing the switch S in the circuit introduces the bulb, which lowers the overall resistance due to the parallel configuration. This results in an increase in current flow through the circuit. Consequently, the voltage across the thermistor decreases because the majority of the current diverts through the bulb, reducing the current available for the thermistor. The thermistor's resistance drops significantly, leading to a further decrease in voltage across it.

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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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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.
 
madmike159 said:
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!
 
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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