Question 3.8: For the core shown in Fig. 3.20, it is required to produce a...

Length CD = BE = AF = 10 cm
Length BC = ED = AB = EF = 8 cm
Length BCDE = 8 + 10 + 8 = 26 cm
Length BAFE = 8 + 10 + 8 = 26 cm
Length BE = 10cm; μ_{r} = 1200
Total flux, Φ= Φ_{1}+ Φ_{2}
N = 800, Φ_{2} = 2 × 10^{−3}  Wb, current, I = ?

Let us draw the equivalent electrical circuit of the given magnetic circuit. The equivalent electric circuit will be as shown in Fig. 3.21.
The voltage drop across CD is I_{2}  R_{2}.

The voltage drop across BE is equal to the voltage drop cross CD.
Therefore,

I_{1}  R_{1} = I_{2}  R_{2}

or,                              I_{1}= I_{2}  \frac{R_{2}}{R_{1}}

For the magnetic circuit, from the analogy of the above equivalent electric circuit, we can write

Φ_{1} = Φ_{2}  \frac{S_{2}}{S_{1}}

S_{2} is the reluctance of path BCDE
S_{1} is the reluctance of path BE

S_{2}= \frac{1}{μ_{0}  μ_{r}  A}= \frac{26  ×  10^{-2}}{4π  ×  10^{-7}  ×  1200  ×  4  ×  10^{-4}}

S_{1}= \frac{10  ×  10^{-2}  }{μ_{0}  μ_{r}  A}= \frac{10  ×  10^{-2}}{4π  ×  10^{-7}  ×  1200 ×  4  ×  10^{-4}}

Φ_{1} = Φ_{2}  \frac{S_{2}}{S_{1}}  = 2  ×  10^{-3} \frac{26}{10}  =  5.2  ×  10^{-3}  Wb.

Φ =Φ_{1} + Φ_{2}= 2  ×  10^{-3}  +  5.2  ×  10^{-3} = 7.2  ×  10^{−3}  Wb.

AT required for portion BAFE (=26cm) = Φ × S_{3}

AT required for portion BE = Φ_{1}  ×  S_{1}

= 862

In the electric circuit, we see that by applying KCL

E − IR  –  I_{1}  R_{1} = 0

or,                                        E = IR  + I_{1}  R_{1}

Similarly, for the magnetic circuit
Total AT = AT required for portion BAFF + AT required for the portion BE

= 3105 + 862
= 3967.

The number of turns of the exciting coil is 800.

AT = NI = 3967

I=\frac{3967}{N}=\frac{3967}{800}=4.95  A.