# Mathcad - 05-20 - arg I s 29 18-deg = I s 1 10 3 × A = I s...

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arg V p ( 29 2.409deg = V p 4.305kV = V p 4.301 10 3 × 180.963j + V = V p V sref I sref Z eqp + = I sref I s a = V sref 4.16 10 3 × V = V sref a V s = a V prated V srated = The primary voltage, referred to the secondary side at full load, is calculated by adding the drop across the equivalent impedance (caused by the load current referred to the high side) to the secondary full load voltage after it has been referred to the high side V s V srated = The transformer delivers rated voltage so:
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Unformatted text preview: arg I s ( 29 18-deg = I s 1 10 3 × A = I s 951.057 309.017j-A = I s I L e j θ ⋅ ⋅ = The secondary load current is given in complex form by: Solution angle is negative so lagging F p θ 18-deg ⋅ = I L 1000 A ⋅ = Z eqp 2.72 j 7.48 ⋅ + ( ) Ω ⋅ = S rated 125 kVA ⋅ = V srated 120 volt ⋅ = V prated 4160 V ⋅ = Problem 5-20: kVAR kW ≡ kVA kW ≡...
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