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