Fig. 35.1 shows a fire alarm circuit that uses a thermistor and an electromagnetic relay. The thermistor is placed in a room. At normal room temperature (20...

(a) At 20 degrees C, the thermistor has a resistance of 5000 ohms. The total resistance in the sensor circuit is approximately 5000 ohms (the relay coil resistance is comparatively small). Using Ohm's law:

\(I = \dfrac{V}{R} = \dfrac{9.0}{5000} = 0.0018\) A = 1.8 mA

(b) The relay coil requires a minimum current of 25 mA to close the contacts. At 20 degrees C, the current through the relay coil is only 1.8 mA, which is far below the 25 mA threshold. The coil does not generate enough magnetic force to attract the armature and close the contacts, so the alarm circuit remains open and the bell does not sound.

(c) At 60 degrees C, the thermistor resistance drops to 200 ohms. The current is now:

\(I = \dfrac{V}{R} = \dfrac{9.0}{200} = 0.045\) A = 45 mA

This exceeds the 25 mA threshold required to operate the relay.

(d) The 45 mA current flowing through the relay coil creates a magnetic field that magnetises the soft iron core inside the coil. The magnetised core attracts the soft iron armature, which pivots and closes the relay contacts in the alarm circuit. This completes the 240 V alarm circuit, allowing current to flow through the bell, which then sounds the alarm.

(e) There are two key advantages of using a relay:

• The alarm bell requires a 240 V mains supply, which would be dangerous if connected directly to the exposed thermistor wiring in the room. A fire or fault could expose someone to lethal mains voltage through the sensor wires.
• The relay provides electrical isolation between the low-voltage sensor circuit (9 V) and the high-voltage alarm circuit (240 V). The small, safe sensor circuit controls the powerful alarm circuit without any direct electrical connection between them.