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# Mathcad - 02-10 - I br V Z = Since the loads are connected...

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Q 11.52KVAR = P 23.04kW = Q 3 V I line sin θ - ( 29 = P 3 V I line F p = Solving for the three-phase real and reactive power: KVAR kW lagging F p 0.894 = F p cos θ ( 29 = The power factor is the cosine of the impedance angle: I line 30.984 amp = Of course the phase angle of the line current would be 30 degrees behind the phase current I line 3 I br = The line current magnitude is the branch current times the square root of three: θ arg I br ( 29 = arg I br ( 29 26.565 - deg = I br 17.889 amp = I br 16 8j - amp = Current angle Current magnitude (V and Z are complex but V is assumed at angle zero)
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Unformatted text preview: I br V Z = Since the loads are connected in delta, the line voltage appears across the load. The branch current is given by: Z R X j ⋅ + = V 480 volt ⋅ = X 12 ohm ⋅ = R 24 ohm ⋅ = Given values: Three identical impedances, Z = 24 + j12, are connected in delta to a 480 volt three-phase line. Find line current, power factor, real power, and reactive power. Problem 2-10: Solution...
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