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## Q. 7.6

The core of Fig. 7.11 has a relative permeability of 4000. The central limb is required to carry a flux of 0.01 Wb. Find the current needed for the exciting coil.

## Verified Solution

The analogous electrical circuit of Fig. 7.11 is drawn in Fig. 7.12, wherein

$R_{L}$ = reluctance of the left limb (A to B)

$R_{C}$ = reluctance of the central limb (A to B)

$R_{R}$ = reluctance of the right limb (A to B minus air-gap)

$R_{G}$ = reluctance of the air-gap

Various reluctances are calculated below:

$I_{L}$ = 2 × (20 + 4) + (25 + 4) = 77 cm

$I_{C}$ = 25 + 4 = 29 cm

$I_{R}$ = 2 × (20 + 4) + (25 + 4) – 0.02 = 76.98 cm

$I_{G}$ = 0.02 cm

$R_{L} = \frac{77 × 10^{–2}}{4π × 10^{–7} × 4000 × 16 × 10^{–4}}$ = 0.0957 × $10^{6}$

$R_{C} = \frac{29 × 10^{–2}}{4π × 10^{–7} × 4000 × 16 × 10^{–4}}$ = 0.0361 × $10^{6}$

$R_{R} = \frac{76.98 × 10^{–2}}{4π × 10^{–7} × 4000 × 16 × 10^{–4}}$ = 0.0957 × $10^{6}$

$R_{G} = \frac{0.02 × 10^{–2}}{4π × 10^{–7} × 16 × 10^{–4}}$ = 0.0995 × $10^{6}$

$F_{AB}$ = 0.01 × $R_{C}$ = 0.01 × 0.0361 × $10^{6}$ = 361 AT

$\phi _{R} = \frac{F_{AB} }{R_{R} + R_{G} } = \frac{361}{(0.0957 + 0.0997) × 10^{6} }$

= 1.8861 × $10^{-3}$ Wb

$\phi _{L} = \phi _{C} + \phi _{R}$ = (10 + 1.8861) × $10^{-3}$ = 11.8861 × $10^{-3}$

Going round the left loop

F = $\phi _{L} R_{L} + F_{AB}$

= 11.8861  ×  $10^{-3}$  ×  0.0957  ×  $10^{6}$ + 361
= 1137 + 361 = 1498 AT

∴     Exciting coil current     I = $\frac{1498}{501 }$  = 3 A  