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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
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