# Mathcad - 05-12 - a 0.433 = a V prated V srated = First...

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polar form arg I 0 ( 29 69.897 - deg = I 0 5.962A = rectangular form I 0 2.049 5.599j - A = I 0 I fe I m + = I m 5.599j - A = I m V prated jX mls = I fe 2.049A = I fe V prated R fels = b. The primary side is the low side and the core parameters are given on the low side. Using the cantilever circuit we can find the exciting current directly Z eq 0.026 0.038j + = Z eq R p j X p + ( 29 R s j X s + ( 29 a 2 + = a. The equivalent winding impedance referred to the low side is found by referring everything to the primary side in this case:
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Unformatted text preview: a 0.433 = a V prated V srated = First find the turns ratio Solution X mls 37.15 Ω ⋅ = X s 0.10 Ω ⋅ = R s 0.07 Ω ⋅ = R fels 101.5 Ω ⋅ = X p 0.0189 Ω ⋅ = R p 0.0132 Ω ⋅ = V srated 480 volt ⋅ = V prated 208 V ⋅ = S rated 75 kVA ⋅ = Problem 5-12: Solution kVAR kW ≡ kVA kW ≡...
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## This note was uploaded on 03/26/2012 for the course ECET 231 taught by Professor Piller during the Spring '12 term at Purdue University.

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