Question 7.11: An attracted armature relay shown in Fig. 7.29 (a) has an ai...

Note: AT vs \phi plot will be the same result as i vs λ plot as iN vs λ/N is AT vs \phi (a) As the armature moves very fast, flux has no time to change. It remain 0.012 Wb. There is no induced emf in the excitation coil, so current changes instantaneously and AT reduces from 1200 to 750 corresponding to the magnetization curve of closed position.

Thus

Δ W _{e} = 0
Δ W _{ƒ} = area OAF – area OCF
= – area OAC, a reduction

By energy balance

 Δ W _{e}  =  Δ W _{ƒ}  +  Δ W _{m}

0 = – area OAC + ΔW _{m}

or             Δ W _{m} = area OAC \frac{1}{2} = area EDAC

= \frac{1}{2} (1200 – 750) × 0.012 = 2.7 J

The mechanical energy output is provided by conversion of the field energy.

(b) As the armature move very slowly, the flux increase while excitation current remains constant. Therefore the magnetic circuit state corresponds to point B.

Δ W _{c}  =  Δ  \phi  .  F_{0} = area FABG

Δ W _{ƒ} = area OBG – area OAF \frac{1}{2} = area FABG

= \frac{1}{2} Δ W _{e}

So                  Δ W _{m}  =  Δ W _{e}  –  Δ W _{ƒ}  =  \frac{1}{2}  Δ W _{e}

= \frac{1}{2} area FABG

\phi_{G}  =  \frac{0.012}{750} × 1200 = 0.0192 Wb

Flux corresponding to point G is \phi_{G}  =  \frac{0.012}{750} × 1200 = 0.0192 wb

Hence      Δ W_{m}  =  \frac{1}{2} (0.0192 – 0.012) × 1200

= 4.32 J

Half the electrical energy input is converted into mechanical energy and half is stored in the field.

(c) We know that

F_{ƒ}  =  – \frac{1}{2}  \phi² ∂R / ∂x

We need to find R(x). From Fig. 7.29

R_{A} (relay open) = \frac{F}{\phi}  =  \frac{1200}{0.012} = 100000

R_{B} (relay open) =   \frac{1200}{0.092} = 62500

Thus

R(total) = R(core) + R(air-gap)

= R_{C}  +  \frac{(10000  –  62500)}{2 × 0.05} · 2x

= R_{C} + 0.75 × 10^{6} x

Relay open

F_{0} = – \frac{1}{2} × (0.012)² × ∂R/∂x

= \frac{1}{2} × (0.012)² × 0.75 × 106 = –54 N

The minus sign means that the force tends to reduce the air-gap. Relay closed

F_{C} = – \frac{1}{2} (0.0192)² × 0.75 × 10^{6} = –138.24 N