Question 7.ASP.19: In the magnetic circuit of Fig. 7.34, the relative permeabil...
In the magnetic circuit of Fig. 7.34, the relative permeability of the core steel is 2000.
Calculate
(a) the maximum flux density
(b) the maximum value of exciting current
(c) the maximum value of the energy stored in the magnetic circuit. What percentage of it resides in the air-gap?
Assume zero coil resistance and no flux leakage. Neglect fringing.
(a) The coil induced emf equals applied voltage
\begin{aligned} V & =E=4.44 f \phi_{\max } N \mathrm{rms} \\ 200 & =4.44 \times 50 \phi_{\max } \times 1600 \\ \text { or } \quad \phi_{\max } & =0.563 \mathrm{mWb} \\ B_{\max } & =\frac{\phi_{\max }}{A_C}=\frac{0.563 \times 10^{-3}}{5 \times 10^{-4}} 1.126 \mathrm{~T} \end{aligned}
(b) Core reluctance
\mathcal{R}_C=\frac{20 \times 10^{-2}}{4 \pi \times 10^{-7} \times 2000 \times 5 \times 10^{-4}}=159 \times 10^3
Air-gap reluctance
\begin{aligned}\mathcal{R}_{\mathrm{g}} & =\frac{0.5 \times 10^{-3}}{4 \pi \times 10^{-7} \times 5 \times 10^{-4}}=159 \times 10^3 \\\mathcal{R}(\text { total }) & =(159+796) \times 10^3=955 \times 10^3\end{aligned}
Now F_{\max }=N i_{\max } ; \phi_{\max }=\frac{N i_{\max }}{R(\text { total })}
Thus
\begin{aligned}i_{\max } & =\frac{\phi_{\max } R(\text { total })}{N}=\frac{0.563 \times 10^{-3} \times 955 \times 10^3}{1600} \\& =0.336 \mathrm{~A}\end{aligned}
Percent energy in air-gap =\frac{\mathcal{R}_g}{R(\text { total })} \times 100=\frac{796 \times 10^3}{955 \times 10^3} \times 100=83.3