Series and Parallel Magnetic Circuits

Electrical Circuit Analysis > Magnetic Circuits

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As one might expect, the close analogies between electric and magnetic circuits will eventually lead to series-parallel magnetic circuits similar in many respects to those encountered in previous chapters. In fact, the electric circuit analogy will prove helpful in defining the procedure to follow toward a solution.

Example 1: Determine the current I required to establish a flux of $1.5 \times 10^{-4}$ Wb in the section of the core indicated in [Fig. 1].

Fig. 1: For Example 1.

Solution:

Fig. 2: Equivalent magnetic circuit of Fig. 1.

The equivalent magnetic circuit and the electric circuit analogy appear in [Fig. 2]. We have

$$B_2 = {\Phi_2 \over A}$$

From B-H curve of [Fig. 3],

$$H_{bcde} = 40 \text{At/m}$$

Applying Ampere's circuital law around loop 2 of [Fig. 2] and,

$$\begin{split} H_{be}l_{be} - H_{bcde}l_{bcde} &= 0 \\ H_{be}(0.05 m) - (40 \text{At/m})(0.2 m) &= 0 \\ H_{be}(0.05 m) &= (40 \text{At/m})(0.2 m) \\ H_{be}&= {(40 \text{At/m})(0.2 m) \over (0.05 m)} \\ H_{be}&= {(8) \over (0.05 m)} = 160 \text{At/m} \\ \end{split} $$

From B-H curve of [Fig. 3],

$$B_1 = 0.97 T $$

and

$$\begin{split} \Phi_1 &= B_1A = (0.97 T)(6 \times 10^{-4} \, m^2)\\ &=5.82 \times 10^{-4} \text{Wb}\\ \end{split} $$

Table 1

The table reveals that we must now turn our attention to section efab:

$$\begin{split} \Phi_T &= \Phi_1 + \Phi_2 \\ &= 5.82 \times 10^{-4} Wb + 1.5 \times 10^{-4} Wb \\ &= 7.32 \times 10^{-4} Wb \\ \end{split} $$

and

$$\begin{split} B &= {\Phi_T \over A}\\ &= {7.32 \times 10^{-4} Wb \over 6 \times 10^{-4}} \\ &=1.22 T\\ \end{split} $$

From B-H curve of [Fig. 3],

$$H_{efab} = 400 At$$

Applying Ampere's circuital law,

$$ \begin{split} +NI - H_{efab}l_{efab} - H_{be}l_{be} &= 0\\ NI &= (400 At/m)(0.2 m) + (160 At/m)(0.05 m)\\ (50 t)I &= 80 At + 8 At\\ I &= 1.76 A\\ \end{split} $$

Example 2: Calculate the magnetic flux $\Phi$ for the magnetic circuit of [Fig. 3].

Fig. 3: For example 2.

Solution: By Amperes circuital law,

$$NI = H_{abcda}l_{abcda}$$

$$\begin{split} H_{abcda} &= {NI \over l_{abcda}}\\ &= {(60t)(5A) \over 0.3 } = 1000 \text{At/m}\\ \end{split} $$

and from B-H curve of [Fig. 3],

$$ B_{abcda} = 0.39 T$$

Hence

$$ \begin{split} \Phi &= BA = (0.39 T)(2 \times 10^{-4} m^2 \\ &=0.78 \times 10^{-4} \text{Wb} \\ \end{split} $$

Example 3: Find the magnetic flux $\Phi$ for the series magnetic circuit of [Fig. 4] for the specified impressed mmf.

Fig. 4: For example 3.

Solution: Assuming that the total impressed mmf= NI is across the air gap,

$$\begin{split} NI &= H_gl_g \\ H_g &= {NI \over l_g}\\ &= {400At \over 0.001} = 4 \times 10^5 \text{ At/m} \end{split} $$

And

$$\begin{split} B_g &= \mu_o H_g = (4\pi \times 10^{-7} )( 4 \times 10^5)\\ &=0.503 T \\ \end{split} $$

The flux

$$\begin{split} \Phi_g &=\Phi_{core} = B_g A\\ &=(0.503 T)(0.003 m^2 )\\ &=1.51 \times 10^{-3}\text{Wb}\\ \end{split} $$

Table 2.

$$H_{core}l_{core} = (1500\text{ At/m})(0.16 m)= 240 \text{At}$$

Applying Ampere's circuital law results in

$$\begin{split} NI &= H_{core}l_{core} + H_gl_g \\ & = 240 \text{At} + 400 \text{At} NI &= 640 At > 400 At\\ \end{split} $$

Since we neglected the reluctance of all the magnetic paths but the air gap, the calculated value is greater than the specified value. We must therefore reduce this value by including the effect of these reluctances. Since approximately

$$640 At - 400 At)/640 At = 240 At/640 At = 37.5% $$

of our calculated value is above the desired value, let us reduce $\Phi$ by 30% and see how close we come to the impressed mmf of $400 At$:

$$\begin{split} \Phi &= (1 - 0.3)(1.51\times 10^{-3} \text{Wb}\\ & = 1.057 \times 10^{-3} \text{Wb}\\ \end{split} $$

Using of reduced flux

$$\begin{split} B &= {\Phi \over A} \\ &={1.057 \times 10^{-3} \text{Wb} \over 0.003m^2} \\ &= 0.352 T \\ H_gl_g &=(7.96 \times 10^5 )B_gl_g\\ &=(7.96 \times 10^5 )(0.352 T)(0.001 m)\\ &=280.19 At \end{split} $$

From the B-H curves,

$$\begin{split} H_core &=850 At/m\\ &=H_{core}l_{core} = (850 At/m)(0.16 m)= 136 At \end{split} $$

Applying Ampere's circuital law yields

$$\begin{split} NI &= H_{core}l_{core} + H_gl_g\\ &= 136 At + 280.19 At \\ NI &= 416.19 At > 400 At\\ \text{(but within 5% and therefore acceptable)} \end{split} $$

The solution is, therefore,

$$ \Phi = 1.057 \times 10^3 \text{Wb}$$

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