For the circuit of Figure B.16, where [latex]i=2 \mathrm{~A}, \phi=1 \times 10^{-3} \mathrm{~Wb}[/latex], cross-sectional area [latex]=5[/latex] in [latex]^{2}[/latex], and the mean flux path length [latex]=2[/latex] in, the total reluctance [latex]\mathcal{R}[/latex] of the magnetic circuit in [latex]\left(\mathrm{A} \cdot t \cdot \mathrm{in}^{2}\right) / \mathrm{Wb}[/latex] is a. [latex]1 \times 10^{5}[/latex] b. [latex]2 \times 10^{5}[/latex] c. [latex]1.5 \times 10^{5}[/latex] d. [latex]3.5 \times 10^{4}[/latex] e. [latex]2 \times 10^{5}[/latex]

From Chapter 16, the relationship between magnetomotive force and flux is [latex]\mathcal{F}=\phi \mathcal{R}[/latex] Also, the magnetomotive force is related to the current [latex]i[/latex] by [latex]\mathcal{F}=N i=100 \times 2=200 \mathrm{~A} \cdot \mathrm{t}[/latex] which means that the reluctance is [latex]\mathcal{R}=\frac{\mathcal{F}}{\phi}=\frac{200}{1 \times 10^{-3}}=200,000 \frac{\mathrm{A} \cdot \mathrm{t} \cdot \mathrm{in}^{2}}{\mathrm{~Wb}}[/latex] Therefore, the answer is b.

Question B.27: For the circuit of Figure B.16, where i=2 A,ϕ=1×10^−3 Wb, cr...

Principles and Applications of Electrical Engineering

For the circuit of Figure B.16, where

i=2 \mathrm{~A}, \phi=1 \times 10^{-3} \mathrm{~Wb}

, cross-sectional area

=5

^{2}

, and the mean flux path length

=2

in, the total reluctance

\mathcal{R}

of the magnetic circuit in

\left(\mathrm{A} \cdot t \cdot \mathrm{in}^{2}\right) / \mathrm{Wb}

1 \times 10^{5}

2 \times 10^{5}

1.5 \times 10^{5}

3.5 \times 10^{4}

2 \times 10^{5}

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