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Zadatak sa takmi čenj a iz OET

Zadatak sa takmi čenj a iz OET. I 1 = 6 - j8 A I 2 = 11 + j2 A Z 1 = 10 + j27,5 Ω Z 3 = 6 + j12 Ω Z 4 = 2 – j9 Ω Z 1 = ? u( t ) = ? odrediti kakakter kola ?. Z 2. Z 1. i 1. u. Z 4. Z 3. i 3. i 2. U = U 1 + U 3.

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Zadatak sa takmi čenj a iz OET

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  1. Zadatak sa takmičenja iz OET

  2. I1 = 6 - j8 A I2 = 11 + j2 A Z1 = 10 + j27,5 Ω Z3 = 6 + j12 Ω Z4 = 2 – j9 Ω Z1 = ? u( t ) = ? odrediti kakakter kola ? Z2 Z1 i1 u Z4 Z3 i3 i2 U=U1+ U3 U1= Z1*I1=(10 + j27,5 ) ( 6 - j8 ) = 60 – j80 + j165 +220= U1= 280 + j85 = 292,6 ej16,8 I1=I2+ I3 => I3=I1 - I2 = 6 - j8 - 11 - j2 = -5 – j10 A U3= Z3*I3=(6 + j12 ) (-5 – j10 ) = -30 – j60 – j60 + 120 = U3= 90 – j120 = 150 e-j53,1

  3. U=U1+ U3 Z2 Z1 i1 U= 280 + j85 + 90 – j120 Z4 u Z3 i3 i2 U= 370 – j35 = 371,65 e-j5,4 Zred=Z2+ Z4 Z2=Zred - Z4=6-j12- 2 +j9= 4 – j3

  4. Z2 Z1 i1 I1 = 6 - j8 = 10 e-j53,1 A Z4 u Z3 i3 U= 370 – j35 = 371,65 e-j5,4 i2 Fazna razlika kola: ϕ = -5,4 – ( -53,1 ) = 47,7 90 < ϕ > 0 redno RL kolo , realno induktivno kolo -53,10 -5,40 47,70 U I1 Struja i1 kasni u odnosu na napon u, a to znaci da je ovo kolo ima pretezno induktivni karakter ( RL kolo )

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