hw6soln

hw6soln - 12.6 The load impedance of the balanced Y-Y...

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1 12.6 12.7 440 0 V 68 Y Z j =−Ω Using the per-phase circuit shown above 10 5 11.18 26.56 220 0 19.68 26.56 A 1.18 26.56 120 19.68 146.56 A 120 19.68 93.44 A 220 3 30 38130 V 381 90 V 381 210 V 220 0 V 220 120 V 220 Y an a Y ba ca ab bc ca AN a Y BN bn CN cn Zj V I Z II V V V VI Z VV =+= == = − =− = =− ==− 120 V 440 0 =44 53.13 A 120 =44 66.87 A 120 =44173.13 A a I j = = The load impedance of the balanced Y-Y network is: The line currents are: The line voltages are: The load voltages are: 13.36 22 11 1 - 1 - 1 1 - n n n n n n n n a) b) c) d) 13.43 20 0 10 12 0 12 I V I V = + =+ Redraw the circuit using source transformation: 20 0 V 12 10 12 0 V + V 1 - + V 2 - 1:4 I 1 I 2 1 2 21 - 4 4 I I = = 1 2 4.186 V 16.744 V V V = = The eqns for each circuit are: The properties of the transformer yield the following: Solving these eqns gives:
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2 13.44 Using superposition: Source one (dc) 0 1( ) 2( ) 0 dc dc V i R i = = 1( ) 2( ) 2( ) 1( ) 2( ) 1( ) 1( ) 1( ) 2( ) 2 - cos cos cos ss ss ss ss ss m ss m ss ss m ss i i n vn v vV t v Vt i Rn R i i nn R ω = = = =− = == Using superposition: Source two (sinusoidal) R 0 dc V 1: n R + v ( ss ) - + v 2( ss ) - i 1( ss ) i 2( ss ) cos m 0 1 2 2 cos () A cos A m m VV t it R nR =+ = Sum over currents from both sources: The properties of the transformer yield the following: Now find currents due to sinusoidal source:
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hw6soln - 12.6 The load impedance of the balanced Y-Y...

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