hw6_solution

# hw6_solution - EE321 Spring 2008, HW #6 Problem 25 The...

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EE321 Spring 2008, HW #6 Problem 25 The magnetic geometry is all the same, so the reluctances do not change. Applying a current i to the two new terminals means i/2 goes to each of the two parallel coils, so the MEC is the same except the two mmf's are now each Ni/4 (adding up to a total mmf of Ni/2 in the equivalent circuit). For the aligned case, Φ Ni 2 R a = Ni 2R a = Since the input current i is only linked by the equivalent of N/2 turns (i/2 flows into each set of N/2 turns), the flux linkage equation is as follows: λ N 2 Φ = N 2 4R a i = L aligned N 2 4R a = Therefore Similarly, L unaligned N 2 4R u = The aligned and unaligned inductance are each a quarter of their values for the series case, so the LB for the parallel case is also one quarter of its previous value. For the series case, T 1 2 L B I 2 = Now, T 1 2 L B 4 i parallel 2 = So the required current to get the same torque is 2I.

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Problem 26 For the series case the resistance of the two coils added. r
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## This note was uploaded on 09/22/2011 for the course ECE 321 taught by Professor Staff during the Spring '08 term at Purdue University-West Lafayette.

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hw6_solution - EE321 Spring 2008, HW #6 Problem 25 The...

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