**Induction Relays:**

#### The **induction type relays** are also called **magnitude relays**.These relays work on the principle of the induction motor or an energy meter.In these relays, a metallic disc is allowed to rotate between the two electromagnets.The coils of the electromagnets are energised with the help of alternating currents.

#### The torque is produced in **Induction relays** due to the interaction of one alternating flux with eddy currents induced in the rotor by another alternating flux.The two fluxes have the same frequency but are displaced in time and space. As the interaction of alternating fluxes is the base of operation of **Induction relays**, these are not used for the d.c. quantities.These are widely used for protective relaying involving only a.c. quantities.

**Types of Induction Relays:**

**1)Shaded pole Relays **

**2)Watt-hour meter Relays **** **

**3)Induction cup Relays **

#### Before studying these types in detail, let us derive the torque equation for the **induction type relays**, which is same for all the **three types of induction relays.**

**Torque Equation of Induction Relay – its Derivation**

#### The induction relay works on the principle of electromagnetic induction similar to the induction motor and energy meter. It consists of two electromagnets and a rotating disc placed between the two electromagnets. The torque produced in these relays is by the interaction of alternating flux with the eddy currents induced in the rotor (disc) by another alternating flux.

#### The two electromagnets when excited produce two alternating fluxes Φ_{1} and Φ_{2}. The two fluxes produced by the electromagnets will be of the same frequency but there exists a phase displacement α (required for the production of torque) between the two fluxes. The mathematical expression for the two alternating fluxes is given by,

*Φ*_{1} = Φ_{1m} sin ωt …(1)

*Φ*

_{1}= Φ_{1m}sin ωt …(1)*Φ*_{2} = Φ_{2m} sin(ωt + α) …(2)

*Φ*

_{2}= Φ_{2m}sin(ωt + α) …(2)#### The two alternating fluxes when links with the disc induce two emfs in the disc which lags their respective fluxes by 90°. The induced emfs leads to the circulation of eddy currents i_{1} and i_{2} in the disc. The interaction of eddy currents produced due to two alternating fluxes results in the production of force or torque in the disc.

#### The above figures show the production of force or torque in the disc of an induction relay. The eddy currents will be in phase (assuming the disc to be non-inductive) with their respective induced voltages. The induced voltages are proportional to the rate of change of fluxes and hence the eddy currents are also proportional to the rate of change of fluxes. Hence we can write,

*i*_{1} = dΦ_{1}/dt

*i*

_{1}= dΦ_{1}/dt*i*_{2} = dΦ_{2}/dt

*i*

_{2}= dΦ_{2}/dt#### Substituting Φ_{1} and Φ_{2} from equations 1 and 2 in the above equations, we get,

#### The force or torque produced on the disc is due to the interaction of Φ_{1} with i_{1} and Φ_{2} with i_{2}.

*∴ F*_{1} ∝ Φ_{1} i_{2}

*∴ F*

_{1}∝ Φ_{1}i_{2}*and F*_{2} ∝ Φ_{2} i_{1}

*and F*

_{2}∝ Φ_{2}i_{1}#### The directions of F_{1} and F_{2} can be found from the Flemings left-hand rule. From the above figure, it can be seen that the two forces are acting in opposite directions. Hence the resultant force acting on the disc is given by,

*F ∝ F*_{2} – F_{1}

*F ∝ F*

_{2}– F_{1}*∴ F ∝ Φ*_{2} i_{1} – Φ_{1} i_{2} …(5)

*∴ F ∝ Φ*

_{2}i_{1}– Φ_{1}i_{2}…(5)#### Substituting the expressions of Φ_{1}, Φ_{2}, i_{1}, and i_{2} from equations 1, 2, 3, and 4 in equation 5, we get,

*F ∝ [Φ*_{2m} sin(ωt + α) Φ_{1m} cos ωt – Φ_{1m} sin ωt Φ_{2m} cos(ωt + α)]

*F ∝ [Φ*

_{2m}sin(ωt + α) Φ_{1m}cos ωt – Φ_{1m}sin ωt Φ_{2m}cos(ωt + α)]*F ∝ Φ*_{1m} Φ_{2m} [sin(ωt + α) cos (ωt) – sin ωt cos(ωt + α)]

*F ∝ Φ*

_{1m}Φ_{2m}[sin(ωt + α) cos (ωt) – sin ωt cos(ωt + α)]*F ∝ Φ*_{1m} Φ_{2m} [sin(ωt + α – ωt)]

*F ∝ Φ*

_{1m}Φ_{2m}[sin(ωt + α – ωt)]*∴ F ∝ Φ*_{1m} Φ_{2m} sin α

*∴ F ∝ Φ*

_{1m}Φ_{2m}sin α#### The above equation gives the resultant force or torque produced in an induction relay. By considering rms values of the alternating fluxes, the above equation is expressed as,

*∴ F ∝ Φ*_{1} Φ_{2} sin α

*∴ F ∝ Φ*

_{1}Φ_{2}sin α