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# Mathcad - 05-02 - I lsec 62.354 44.504j − amp = I lsec V...

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CapRating 10.681 KV amp = CapRating Q := To correct to unity power factor, a capacitor must offset the inductive reactive power of the load: Z in 2.295 10 3 × 1.638j 10 3 × + Ω = Z in a 2 Z lsec := To find the input impedance, multiply the load impedance by turns ratio squared: reactive Q 10.681 KV amp = P 14.965 kW = S 1.497 10 4 × 1.068j 10 4 × + watt = Q Im S ( ) := P Re S ( ) := S V s1 I lsec := S = P + jQ = VI* I lpri 2.078 1.483j amp = I lpri I lsec a := so the primary side current is the secondary current divided by "a" a 30 = a V prated V s1 := the turns ratio is:
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Unformatted text preview: I lsec 62.354 44.504j − amp = I lsec V s1 Z lsec ( ) := since the voltage is 240: Z lsec 2.55 j 1.82 ⋅ + ( ) ohm ⋅ := V prated 7200 volt ⋅ := V s1 240 volt ⋅ := Problem 5-2 Solution: The load on the above transformer is 2.55+j1.82 ohms at the 240 volt terminals. Find P and Q to the load, the input impedance of the transformer and load, and the KVAR rating of a capacitor to correct the load to unity power factor....
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