are designed for sustained overcurrent protection, typically set between 105-120% of the full load current with a time delay. They are not intended for momentary overcurrents.

The overcurrent setting depends on the following:

• The electrical system's maximum continuous load capacity determines the relay settings to ensure the system operates without tripping under normal conditions.
• The relay settings are influenced by the insulation class and its capacity to withstand elevated temperatures.
• The amount of power consumption and heat generated during normal operation is evaluated to set the relay accurately.
• The relay takes into account the cooling mechanism in place to ensure that heat dissipation during operation is factored into the thermal protection settings.

Advantages of Thermal relays over Magnetic relays:

• Thermal relays provide a time delay, preventing tripping from momentary overcurrents that might not cause actual damage. Magnetic relays respond much faster.
• Thermal relays operate based on the heat generated by an overcurrent, offering a more accurate reflection of the actual thermal stress on the system. Magnetic relays respond to the magnitude of the current, irrespective of heat generation.
• Thermal relays are more economical because they use bimetals instead of more expensive magnetic solenoid coils.
• Thermal relays are effective for sustained overcurrents, providing protection that is independent of other factors like the system voltage or magnetic field fluctuations.

$$Resistance \space of \space solenoid \space = \space 35Ω$$

$$Supply \space voltage \space = 220V, \space 50Hz$$

$$Initial \space current \space = \space 2A$$

$$Current \space in \space "Full-in" \space position \space = \space 0.7A$$

When plunger is out (initial stage),

$$Impedance \space (Z) \space = \space {{220} \over 2} \space = \space 110Ω$$

$$Inductive \space reactance \space (X) \space = \space \sqrt{Z^2 - R^2}$$

$$X \space = \space \sqrt{110^2 - 35^2} \space = \space 104.28Ω$$

$$Also, \space X \space = \space 2 \pi fl$$

$$l \space = \space {{X} \over 2 \pi f} \space = \space {{104.28} \over 2 \times 3.14 \times 50} $$

$$l \space = \space 0.33H$$

Inductance of solenoid when plunger is out = 0.33H

When "Full-in",

$$Impedance \space = \space {{V} \over I} \space = \space {{220}\over 0.7} \space = \space 314.286Ω$$

$$∴ \space X \space = \space \sqrt{314.286^2 - 35^2} \space = \space 312.33Ω$$

$$l \space = \space {{312.33} \over 2 \times 3.14 \times 50 } \space = \space 0.994H $$

$$Also, L \space = \space {{N \phi} \over I_{peak}}$$

$$∴ \space N \phi \space = \space L \space I_{peak}$$

$$∴ \space N \phi \space = \space 0.994 \times \sqrt{2} \times 0.7$$

$$∴ \space N \phi \space = \space 0.984 \space weber-turn$$