# The Effect of Temperature on Resistivity

• I-need-help
It means that the 12 V is constant, regardless of the current through it. So, when the switch is first closed, the initial current of 14 mA is the highest current the thermistor will see, and therefore the highest power dissipation. As the thermistor heats up, its resistance drops, allowing more current to flow. However, the 12 V source remains constant, so the power dissipation will not increase. In summary, the thermistor will dissipate more power initially when the switch is first closed, as the initial current is the highest it will see due to the constant voltage source.

#### I-need-help

1) A thermistor is connected through a switch to a 12V power supply of negligible internal resistance. When the switch is closed, the initial current in the circuiy is 14mA. A few minutes later it has risen to a constant 20mA.

Does the thermistor dissipate more power initially, or after it has been left for some minutes? Justify your answer.

My attempt:

Well the question tells you that the current increases, which must mean that the temperature of the circuit components increase, resulting in a decrease in resistance - as this is a thermistor.

I think that the thermistor would therefore dissipate more power after it has been left for some minutes, because if P = I x V , then as the current rises over time, and the voltage supply remains constant, the overall Power should rise...

However I'm not sure...because it could also be initially, as the increase in the current will eventually slow down, and level off...so surely, in proportion to the time the circuit has been left on for...more power will have been dissipated right at the start (initially) than after a while, where the current rise has leveled off more... I have no idea!