# simple Temperature monitoring circuit

#### Bigcountry

Joined Jul 4, 2008
76
I have problems with this problem I can't seem to figure it out I have attached the diagram so hopefully someone can help me. here is what the problem says:

This is simple temperature monitoring circuit. As the temperature of the liquid increases in the tank changes, so the resistance of the thermistor. This will change current through it, and also the current through the 10 ohm resistor Rx. The voltmeter which is calibrated directly in degrees centigrade will measure this. If the circuit is to be accurate we must have a constant voltage across A and B. Therefore we use a zener diode. At 0 degrees C the voltage drop across Rx is .125 volt. What is the resistance of the thermistor ?

the instructor gives a hint has This si circuit with 10V across A and B and Rx and T in a series if Rx drops .125V what is the drop across T ? Then find I and Rt

that hint really messed me up. so I am totally lost. thanks for any useful help.

sorry about the bad drawing it the was best I could on such short notice. I got to figure this out by Monday.

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#### beenthere

Joined Apr 20, 2004
15,819
Rx is in series with the thermistor. The drop across Rx subtracted from the source gives the drop across the thermistor. Ohm's law gives the current - I = E/R.

#### Bigcountry

Joined Jul 4, 2008
76
Rx is in series with the thermistor. The drop across Rx subtracted from the source gives the drop across the thermistor. Ohm's law gives the current - I = E/R.

so the resistor for the thermistor is 9.87 ohms

#### hgmjr

Joined Jan 28, 2005
9,027
so the resistor for the thermistor is 9.87 ohms
The answer you have gotten is not correct.

Can you post your calculations so that we can find where you are having problems?

hgmjr

#### Bigcountry

Joined Jul 4, 2008
76
The answer you have gotten is not correct.

Can you post your calculations so that we can find where you are having problems?

hgmjr
ok this is my first step. 10-.125 = 9.875

then I got really discouraged..... I think this is where I am making my mistake. but then again I am doubting myself. that hint threw me off. the source is actually the 14-17 volts right ? I think I need to take some baby steps now to figure it out. thanks for your help up this far. also thanks for not giving me the answer I would like to figure this one out.

Last edited:

#### beenthere

Joined Apr 20, 2004
15,819
Hint - all elements in a series circuit have the same current. The zener keeps the voltage for the resistor and the thermistor at 10.

#### Bigcountry

Joined Jul 4, 2008
76
Hint - all elements in a series circuit have the same current. The zener keeps the voltage for the resistor and the thermistor at 10.

so the current is 1 amp ?

#### mik3

Joined Feb 4, 2008
4,843
If you assume that the thermometer has an infinite resistance and draws no current then the current through Rx is

0.125/10=0.0125 Amps

where
0.125 is the voltage across Rx
and 10 is the resistance of Rx

Now, the voltage across Rx and T is constant and it is 10 Volts (the problem says that) thus the voltage across T is

10-0.125=9.875 Volts

Also, because Rx and Tare in series the have the same current through them.

Thus to find the resistance of T

R=V/I

R=9.875/0.0125=790 Ohms

Just simple logic!

#### Bigcountry

Joined Jul 4, 2008
76
If you assume that the thermometer has an infinite resistance and draws no current then the current through Rx is

0.125/10=0.0125 Amps

where
0.125 is the voltage across Rx
and 10 is the resistance of Rx

Now, the voltage across Rx and T is constant and it is 10 Volts (the problem says that) thus the voltage across T is

10-0.125=9.875 Volts

Also, because Rx and Tare in series the have the same current through them.

Thus to find the resistance of T

R=V/I

R=9.875/0.0125=790 Ohms

Just simple logic!
I knew i was overlooking the simplest part..thank you