Question 5.4: A magnetic circuit is having its winding on its central limb......
A magnetic circuit is having its winding on its central limb. The cross-sectional area of the central limb is 10 cm² whereas the cross-sectional area of the outer limbs is 5 cm². The effective length of the central limb is 16 cm and that of the outer limbs is 25 cm. Calculate the current required to flow through the winding which has 1000 turns to produce a flux of 1.2 mWb in the central limb. Assume that for a flux density of 1.2 Wb/m² , the magnetizing force required is 750 AT/m. Draw the equivalent electric circuit.
The details of the magnetic circuit are shown in Fig. 5.23.
The equivalent electric circuit is drawn as shown in Fig. 5.24.
This is an example of a parallel circuit. As current in an electric circuit gets divided into two parallel branches, the flux produced in the central limb will get divided into the two outer limbs.
We will calculate the MMF required for the central limb as also for any of the outer limbs which will maintain the desired flux in the core. For a flux density of 1.2 Wb/m² , the value of H has been given. Let us calculate the flux density in the central limb first.
Flux density in the central limb, \mathrm{B}_{\mathrm{c}}=\frac{\phi_{\mathrm{c}}}{\mathrm{A}_{\mathrm{c}}}=\frac{1.2 \times 10^{-3}}{10 \times 10^{-4}}=1.2 \mathrm{~Wb} / \mathrm{m}^2
The Flux density in the outer limb will be the same as that in the central limb since half the flux is available in each of the outer limbs and their cross-sectional area is half of that of the central limb.
\mathrm{B}_{\mathrm{o}}=\frac{\phi_{\mathrm{o}}}{\mathrm{A}_{\mathrm{o}}}=\frac{0.6 \times 10^{-3}}{5 \times 10^{-4}}=1.2 \mathrm{~Wb} / \mathrm{m}^2The corresponding H i.e., AT/m for flux density of 1.2 Wb/m² has been given as 750. The total MMF required = MMF required for the central limb + MMF required for one outer limb (and not for both the limbs).
Since \mathrm{H}=\frac{\mathrm{MMF}}{l}, \mathrm{MMF}=\mathrm{H} \times l
Total MMF required =\frac{750 \times 16}{100}+\frac{750 \times 25}{100}=307.5 [considering length in m]
\begin{aligned} \mathrm{AT} & =307.5=\mathrm{NI} \\ \mathrm{N} & =1000 \\ \mathrm{I} & =\frac{307.5}{1000}=0.3075 \mathrm{~A} \end{aligned}
