I have been working on this problem:

enter image description here

The way I approached this problem is given (I used Kirchhoff's laws on Electrical equivalent of magnetic circuit) :

enter image description here

I have two doubts here:

  1. My working does not give the right answer, can someone tell the mistake in my solution?

  2. As per the book's solution, "Total ampere turns is equal to that required for path E and that required for two paths C or D".

Can someone explain this?

PS: If my solution is not clear due to handwriting or other issues, please contact me by commenting, I shall explain you my method.

  • \$\begingroup\$ No, I don't see figure 6.35. Put yourself in the reader's shoes. \$\endgroup\$ – Andy aka Feb 6 at 10:45
  • \$\begingroup\$ @Andyaka It's a typo I guess, because that's what is given. \$\endgroup\$ – programmer Feb 6 at 10:46
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    \$\begingroup\$ If you want this answered by someone (given that I have now unsuccessfully tried to decode your handwriting), I suggest you present the question in your own "typed" words with the relevant information and use properly formatted equations in mathjax. For instance, to make this example formula: \$L = \dfrac{N\cdot I}{\Phi}\$ requires you to type this \$L = \dfrac{N\cdot I}{\Phi}\$. You might also explain how the text in the first picture can possibly suggest that the MMF in C and D are the same. I disagree with that but, realistically, I've spent enough time on this so now. \$\endgroup\$ – Andy aka Feb 6 at 11:04
  • \$\begingroup\$ I will try to write in MathJax, by the way do you agree with the book's claim that "MMF in C and D "are same? @Andyaka \$\endgroup\$ – programmer Feb 6 at 11:11
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    \$\begingroup\$ I'm guessing it actually says the m.m.f. ACROSS paths C and D is the same. \$\endgroup\$ – Finbarr Feb 6 at 13:13

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