Question

Monitoring of temperatures in electronic devices is often accomplished using

thermistors, circuit elements whose resistance varies rapidly with

temperature. The actual circuits are more complicated than the one at right,

but it illustrates the basic principle. As the temperature of the thermistor rises,

its resistance drops; a small rise in temperature leads to a large drop in

resistance. At 25 °C, the thermistor has a resistance of 3.0 k Ω. As the

temperature rises to 60 °C, the resistance drops to 200 Ω. This changes the

voltage measured by the meter—an electrical quantity that can be easily

monitored.

V

10 V

2.0 kΩ

Thermistor

1) What is the reading on the voltmeter at 25 °C?

A. 1 V

B. 2 V

C. 4 V

D. 6 V

2) What is the reading on the voltmeter at 60 °C?

A. 1 V

B. 2 V

C. 4 V

D. 6 V

You have a voltmeter that will trigger an alarm if the measured voltage rises above 5.0 V. Sketch a circuit using a 10 V power

supply, the above thermistor, one or more resistors, and this voltmeter that will trigger an alarm if the temperature of the

thermistor rises above 60 °C.

Answer 1

At 25^o C

Rt = 3 x 10^3 ohm

R = 2 x 10^3 ohm

Req = R + Rt

Req = (3+2) x 10^3 ohm

Req = 5 x10^3 ohm

V = I* Req

I = V/Req

I = 10/( 5 x 10^3) A

Reading of Voltmeter At 25^o C, V = 10/( 5 x10^3) x (3 x 10^3) V

Reading of Voltmeter At 25^o C, V = 6V

b)

At 60^o C

Rt = 200 ohm

R = 2 x 10^3 ohm

Req = R + Rt

Req = 2 x 10^3 + 200 ohm

Req = 2200 ohm

V = I* Req

I = V/Req

I = 10/2200 A

Reading of Voltmeter At 60^o C, V = 10/(2200 x 200 V)

Reading of Voltmeter At 60^o C, V = 1V