Example problems of magnetic circuits: Difference between revisions
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Now with this, the length and cross sectional area of the core we can solve for reluctance <math> R_c </math> by: <br> |
Now with this, the length and cross sectional area of the core we can solve for reluctance <math> R_c </math> by: <br> |
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<math> R_c = \frac{L}{\mu A} = \frac{1}{1.2566 \times 10^{-6}\times .1} = 7.96 \times 10^{6} |
<math> R_c = \frac{L}{\mu A} = \frac{1}{1.2566 \times 10^{-6}\times .1} = 7.96 \times 10^{6} </math> <br> <br> |
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Similarly to get the reluctance of the gap <br> |
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<math> R_g = \frac {g}{\mu_0 (\sqrt{A} + g)^2} </math> |
Revision as of 17:56, 10 January 2010
Given:
A copper core with susceptibility
length of core L = 1 m
Gap length g = .01 m
cross sectional area A = .1 m
current I = 10A
N = 5 turns
Find: B
Solution:
First we need to find the permeability of copper given by the equation
Which yeilds
Now with this, the length and cross sectional area of the core we can solve for reluctance by:
Similarly to get the reluctance of the gap