Question 5.9: For the core shown in Fig. 5.29, it is required to produce a......
For the core shown in Fig. 5.29, it is required to produce a flux of 2 mWb in the limb CD. The entire core has a rectangular cross section of 2cm × 2cm. The magnetizing coil has 800 turns. The relative permeability of the material is 1200. Calculate the amount of magnetizing current required.
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; μr = 1200
Total flux, \phi=\phi_1+\phi_2
N = 800, \phi_2=2 \times 10^{-3} \mathrm{~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. 5.30.
The voltage drop across CD is II_2 R_2 .
The voltage drop across BE is equal to the voltage drop cross CD.
Therefore,
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
\phi_1=\phi_2 \frac{S_2}{S_1}S_2 is the reluctance of path BCDE
S_1 is the reluctance of path BE
AT required for portion BAFE (=26 \mathrm{~cm})=\phi \times \mathrm{S}_3
= \frac{7.2 \times 10^{-3} \times 26 \times 10^{-2}}{4 \pi \times 10^{-7} \times 1200 \times 4 \times 10^{-4}}=3105
AT required for portion BE = \phi_1 \times S_1.
\begin{aligned} & =\frac{5.2 \times 10^{-3} \times 10 \times 10^{-2}}{4 \pi \times 10^{-7} \times 1200 \times 4 \times 10^{-4}} \\ & =862 \end{aligned}In the electric circuit, we see that by applying KCL
E-I R-I_1 R_1=0or, \mathrm{E}=\mathrm{IR}+\mathrm{I}_1 \mathrm{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
\mathrm{I}=\frac{3967}{\mathrm{~N}}=\frac{3967}{800}=4.95 \mathrm{~A}