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## Q. 10.2

An inductor constructed with an EI assembly EI42 (corresponding to IEC YEI 1-14) with M530-50A material, as described in Example 10.1, was constructed with g = 0.5 mm, core length $l_{c}$ =8.4 cm, core cross-sectional area $A_{g}$ = 2.072 $cm^{2}$ and with N = 365 turns. Calculate the inductance L and the effective inductance L$_{\mathrm{eff} }$ as a function of current for the range 0.25 to 4 A.

## Verified Solution

For low values of current before the onset of saturation, the inductance is a function of the gap and is:

$L_{max}= \frac{N^{2}}{R_{g}} = \frac{\mu _{0}N^{2}A_{g}}{g} = \frac{(4\pi\times10^{-7})(365)^{2}(2.072\times10^{-4})}{0.5\times10^{-3}} \times10^{3}= 69.4 mH$

L is found from Equation 10.7:

$L= \frac{N^{2}}{R_{g}}\frac{1}{\frac{\mu_{\mathrm{eff}}}{\mu_{r}} }$  (10.7)

$\mu _{\mathrm{eff}}= \frac{l_{c}}{g} = \frac{84}{0.5}= 168$

And for i = 1.5 A:

a = H$_{0}$ = 100 A/m

b = $\frac{Ni}{g}+ H_{0}\mu _{\mathrm{eff}} -H_{m}= \frac{(365)(1.5)}{0.5\times10^{-3}} + (100)(168)-1.3\times10^{6}= -18.82\times10^{4 } A/m$

c = $-H_{m}\mu _{\mathrm{eff} }= -(1.3\times10^{6})(168)= -2.184\times10^{8} A/m$

μ$_{r}=\frac{-b+ \sqrt{b^{2}-4ac} }{2a} = \frac{(18.82\times10^{4})+\sqrt{(-18.82\times10^{4})^{2}-(4)(100)(-2.184\times10^{8})} }{(2)(100)} = 2693$

L  = $\frac{N^{2}}{R_{g}} \frac{1}{1+ \frac{\mu_{\mathrm{eff} }}{\mu_{r}} } = (69.4\times10^{-3})\frac{1}{1+ \frac{168}{2693} } = 65.3 mH$

Leff is found from Equation 10.15:

$L_{\mathrm{eff} }= L(i)+ i\frac{dL(i)}{di} = L(i)+ i\frac{dL}{d\mu_{r}}\frac{d\mu_{r}}{di}$             (10.15)

$\frac{dL}{d\mu_{r}}= L\frac{\mu_{\mathrm{eff} }/\mu^{2}_{r}}{1+ \mu_{\mathrm{eff} }/\mu_{r}} = (0.0653)\frac{168/(2693)^{2}}{1+ 168/2693} = 1.424\times10^{-6}H$

$\frac{d\mu_r}{di} = \left[-\frac{1}{2a}+ \frac{b}{2a\sqrt{b^{2}-4ac} } \right] \frac{N}{g}$

= $\left[ -\frac{1}{(2)(100)}+ \frac{-18.82\times10^{4}}{(2)(100)\sqrt{(18.82\times10^{4})^{2}-(4)(100)(-2.184\times10^{8})} } \right]\times \frac{365}{0.5\times 10^{-3}}$

= $-5.61 \times10^{3} A^{-1}$

$L_{\mathrm{eff} }= L(i)+ i\frac{dL}{du_{r}}\frac{d\mu_{r}}{di} = 0.0653+ (1.5)(1.424\times10^{-6})(-5.61\times10^{3})= 53.3 mH$

This process is repeated for the range of current and the plots of L and L$_{\mathrm{eff} }$ are shown in Figure 10.4. 