Current Electricity: Calculating Power at 0 & 100 Degrees C

[fran008]

I am revising the chapter on current electricity and i came across this question which i got stuck in the last part. It shows a circuit with a thermistor having a resistane of 60k ohms at 100 deg C and 100k ohm at 0 deg C. The termistor is in series witha 12 k ohm resistor There is a voltmeter but we are told that we can consider it behaving ideally. Now the last questions were: depermine the power dissipated in the thermistor at
a 100 deg C
b 0 deg C

Checking my answer for a , it was worked out rigght :). but d problem is b, i got a wrong ans when I used the same method ie

I found R total by adding the 2 resistances. Then using ratios I found voltage across the thermistor which is 10.7, then using I=V/R i found I and then used P=IV. What is wrong pls? My book says that the ans is 4microwatts

 

[mezarashi]

You are most likely using the wrong value for voltage or current at some stage. If you could write out your calculations.

One method in doing the problem is to find the total resistance, then find the total current. To find the power dissipating through any resistor, then use $$P_1 = I^2 R_1$$.

 

[quicknote]

I would like to first clarify some concepts and assumptions in the given scenario. The question mentions a thermistor, which is a type of resistor whose resistance changes with temperature. In this case, the resistance of the thermistor is given as 60k ohms at 100 degrees Celsius and 100k ohms at 0 degrees Celsius. The question also states that the thermistor is in series with a 12k ohm resistor and a voltmeter is present in the circuit. It is also mentioned that the voltmeter can be considered to behave ideally, which means it has infinite resistance and will not affect the circuit.

Now, let's look at the approach used to calculate the power dissipated in the thermistor at 0 degrees Celsius. It is correct to first find the total resistance in the circuit by adding the 12k ohm resistor and the thermistor's resistance at 0 degrees Celsius (100k ohms). This gives a total resistance of 112k ohms. However, the next step of finding the voltage across the thermistor by using the ratio method is incorrect.

The ratio method is used to find the voltage drop across a specific resistor in a circuit. In this case, the thermistor is not the only resistor in the circuit, so it is not appropriate to use the ratio method to find its voltage drop. Instead, you can use the voltage divider rule, which states that the voltage drop across a resistor is proportional to its resistance in comparison to the total resistance in the circuit. Using this rule, the voltage drop across the thermistor at 0 degrees Celsius can be calculated as:

V = (Rthermistor / Rtotal) * Vtotal

V = (100k ohms / 112k ohms) * 10.7 volts

V = 9.6 volts

Now, using Ohm's law (I = V/R), the current through the thermistor can be calculated as:

I = 9.6 volts / 100k ohms

I = 0.096 milliamps

Finally, the power dissipated in the thermistor can be calculated as:

P = I * V

P = 0.096 milliamps * 9.6 volts

P = 0.9216 milliwatts

The given answer of 4 microwatts is incorrect and seems to be a typing error.

In summary, the mistake in