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Ejercicios Circuito 2

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Horiana Gimenez
Horiana Gimenez

1) The document presents calculations to determine currents (I1, I2) and voltages (V1, V2) in electric circuits. It uses mesh current analysis and voltage division to set up and solve systems of equations for the unknown values. 2) A second circuit is analyzed to determine the current I0 flowing through a branch using similar analysis techniques. The current I0 is calculated to be 3.25¡Ï66.04¡ã A. 3) Mesh current analysis is again used to solve for the voltage V0 in another circuit, giving the result V0 = 13.32¡Ï-33.69¡ã V.

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1) Determinar I1,I2,V1,V2 enel circuitode la figura. A = ?1 ?2 = ?2 ?1 A = 1 4 = ?1 ?2 = ? ?2 ?1 Por recorridode mallatenemos: 1) - 12?0¡ã - 2? * I1 + V1 = 0 -2? * I1 + V1 = 12?0¡ã ¢Ù 2) ¨C V2 +4? * I2 + 24?0¡ã = 0 4? * I2 ¨C V2 = - 24?0¡ã ¢Ú 3) A = ?1 ?2 V1 = A * V2 V1 = 1 4 * V2 DespejamosV2 V2 = 4 * V1 ¢Û 4) A =- ?2 ?1 -I2 = A * I1 -I2 = 1 4 * I1 DespejamosI1 I1 = - 4 *I2 ¢Ü Ahorasustituimos ¢Ü en¢Ù. -2? * I1 + V1 = 12?0¡ã -2? (- 4? * I2 ) + V1 = 12?0¡ã 8? * I2 + V1 = 12?0¡ã ¢Ý
¢Û en¢Ú. 4? * I2 - V2 = 24?0¡ã 4? * I2 - ( 4? * V1) = 24?0¡ã 4? * I2 ¨C 4? * V1 = 24?0¡ã ¢Þ De ¢Ý y ¢Þ. I2 = 2 A V1 = -4 V Nuevamentesustituimos De ¢Ý y ¢Þ en¢Û y ¢Ü I2 = 2 A I1 = -4 * I2 I1 = -4 * (2 A) I1 = - 8 A V1 = -4 V V2 = 4 * V1 V2 = 4 * (-4 V) V2 = -16 V 2) En el circuitomostradoenla figura,halle I0 Por recorridode malla 1) ( 2? ¨C j4? + j2? + 2?) * I1 ¨C 2? * I2 ¨C j1? * I2 = 32?0¡ã ( 4? ¨C j2? ) * I1 ¨C ( 2? + j1? ) * I2 = 32?0¡ã ¢Ù
2) ¨C 2? * I1 - j1? * I1 + ( 2? + j2? ¨C j2? + 3? ) * I2 = 0 -( 2? + j1? ) * I1 + 5? * I2 = 0 DespejandoI1 I1 = 5? ??2 2? +?1? ¢Ú Sustituyendotenemos ¢Ú en¢Ù ( 4? ¨C j2? ) * I1 ¨C ( 2? + j1? ) * I2 = 32?0¡ã ( 4? ¨C j2? ) * ( 5? ??2 2? +?1? ) ¨C ( 2? + j1? ) * I2 = 32?0¡ã Factor comunI2 I2 [ 5? ??2 2? +?1? ? (2? + ?1? )] = 32?0¡ã I2 * ( 4? ¨C j9? ) = 32?0¡ã DespejamosI2 I2 = I0 = 32?0¡ã 4???9? I0 = 3.25?66.04¡ã (A)
3) En el circuitomostrado,halle V0 Por recorridode mallaobtenemos: 1) ( 1? ¨C j1? ) * I1 ¨C ( - j1? ) * I2 = 24?0¡ã ( 1? ¨C j1? ) * I1 + j1? * I2 = 24?0¡ã ¢Ù 2) ¨C ( - j1? ) * I1 + ( - j1? + j2? ¨C j2? + j2? + 2? ) * I2 ¨C j1? * I2 ¨C j2? ¨C j2? = 0 j1? * I1 + ( 2? + j1? ¨C j2? ) * I2 = 0 ¢Ú despejamosI1 I1 = ? ( 2???1?) ? ?2 ?1? ¢Û Sustituyendotenemos: ¢Û en¢Ù ( 1? ¨C j1? ) * I1 + j1? * I2 = 24?0¡ã ( 1? ¨C j1? ) * [ ? ( 2???1?) ? ?2 ?1? ]+ j1? * I2 = 24?0¡ã Factor comunI2 [? ( 1??? 1?)? (2??? 1?) ? + j ] * I2 = 24?0¡ã I2 * ( 3? + j2? ) = 24?0¡ã Despejamos I2 I2 = 24?0¡ã 3? + j2? I2 = 6.66?-33.69¡ã (A) V0 = 2? * I2 V0 = 2? * 6.66?-33.69¡ã (A) V0 = 13.32?-33.69¡ã (V)
4) Determine I1,I2,V1 yV2 enel circuitode la figura. A = ?1 ?2 = ?1 ?2 = 1 2 Multiplicamos encruz V2 = 2 * V1 ¢Ù Tenemos A = ? ?2 ?1 = 1 2 I1 = - 2 * I2 ZR = A2 * Z2 ZR = ( 1 2 ) 2 * ( 4? + j4? ) ZR = 1? + j1? Zp = [ 1 ? ?1? + 1 1+?1? ] ?1 Zp = 1? - j1?
V1 = 10?30¡ã?( 1? ?j1? ) 2? +1?? ?1? (V) V1 = 10?30¡ã?( 1? ?j1? ) 3??1? (V) V1 = ?. ????.??¡ã (V) I1 = ?1 ?? I1 = 4.47?3.43¡ã (V) 1? ? j1? I1 = ?. ?????.??¡ã (A) I2 = - ?1 2 I2 = - 3.16?48.43¡ã (A) 2 I2 = ?. ???? ???.??¡ã (A) V2 = 2 * V1 V2 = 2 * 4.47?3.43¡ã (V) V2 = ?. ????.??¡ã (V)

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Ejercicios Circuito 2

  • 1. 1) Determinar I1,I2,V1,V2 enel circuitode la figura. A = ?1 ?2 = ?2 ?1 A = 1 4 = ?1 ?2 = ? ?2 ?1 Por recorridode mallatenemos: 1) - 12?0¡ã - 2? * I1 + V1 = 0 -2? * I1 + V1 = 12?0¡ã ¢Ù 2) ¨C V2 +4? * I2 + 24?0¡ã = 0 4? * I2 ¨C V2 = - 24?0¡ã ¢Ú 3) A = ?1 ?2 V1 = A * V2 V1 = 1 4 * V2 DespejamosV2 V2 = 4 * V1 ¢Û 4) A =- ?2 ?1 -I2 = A * I1 -I2 = 1 4 * I1 DespejamosI1 I1 = - 4 *I2 ¢Ü Ahorasustituimos ¢Ü en¢Ù. -2? * I1 + V1 = 12?0¡ã -2? (- 4? * I2 ) + V1 = 12?0¡ã 8? * I2 + V1 = 12?0¡ã ¢Ý
  • 2. ¢Û en¢Ú. 4? * I2 - V2 = 24?0¡ã 4? * I2 - ( 4? * V1) = 24?0¡ã 4? * I2 ¨C 4? * V1 = 24?0¡ã ¢Þ De ¢Ý y ¢Þ. I2 = 2 A V1 = -4 V Nuevamentesustituimos De ¢Ý y ¢Þ en¢Û y ¢Ü I2 = 2 A I1 = -4 * I2 I1 = -4 * (2 A) I1 = - 8 A V1 = -4 V V2 = 4 * V1 V2 = 4 * (-4 V) V2 = -16 V 2) En el circuitomostradoenla figura,halle I0 Por recorridode malla 1) ( 2? ¨C j4? + j2? + 2?) * I1 ¨C 2? * I2 ¨C j1? * I2 = 32?0¡ã ( 4? ¨C j2? ) * I1 ¨C ( 2? + j1? ) * I2 = 32?0¡ã ¢Ù
  • 3. 2) ¨C 2? * I1 - j1? * I1 + ( 2? + j2? ¨C j2? + 3? ) * I2 = 0 -( 2? + j1? ) * I1 + 5? * I2 = 0 DespejandoI1 I1 = 5? ??2 2? +?1? ¢Ú Sustituyendotenemos ¢Ú en¢Ù ( 4? ¨C j2? ) * I1 ¨C ( 2? + j1? ) * I2 = 32?0¡ã ( 4? ¨C j2? ) * ( 5? ??2 2? +?1? ) ¨C ( 2? + j1? ) * I2 = 32?0¡ã Factor comunI2 I2 [ 5? ??2 2? +?1? ? (2? + ?1? )] = 32?0¡ã I2 * ( 4? ¨C j9? ) = 32?0¡ã DespejamosI2 I2 = I0 = 32?0¡ã 4???9? I0 = 3.25?66.04¡ã (A)
  • 4. 3) En el circuitomostrado,halle V0 Por recorridode mallaobtenemos: 1) ( 1? ¨C j1? ) * I1 ¨C ( - j1? ) * I2 = 24?0¡ã ( 1? ¨C j1? ) * I1 + j1? * I2 = 24?0¡ã ¢Ù 2) ¨C ( - j1? ) * I1 + ( - j1? + j2? ¨C j2? + j2? + 2? ) * I2 ¨C j1? * I2 ¨C j2? ¨C j2? = 0 j1? * I1 + ( 2? + j1? ¨C j2? ) * I2 = 0 ¢Ú despejamosI1 I1 = ? ( 2???1?) ? ?2 ?1? ¢Û Sustituyendotenemos: ¢Û en¢Ù ( 1? ¨C j1? ) * I1 + j1? * I2 = 24?0¡ã ( 1? ¨C j1? ) * [ ? ( 2???1?) ? ?2 ?1? ]+ j1? * I2 = 24?0¡ã Factor comunI2 [? ( 1??? 1?)? (2??? 1?) ? + j ] * I2 = 24?0¡ã I2 * ( 3? + j2? ) = 24?0¡ã Despejamos I2 I2 = 24?0¡ã 3? + j2? I2 = 6.66?-33.69¡ã (A) V0 = 2? * I2 V0 = 2? * 6.66?-33.69¡ã (A) V0 = 13.32?-33.69¡ã (V)
  • 5. 4) Determine I1,I2,V1 yV2 enel circuitode la figura. A = ?1 ?2 = ?1 ?2 = 1 2 Multiplicamos encruz V2 = 2 * V1 ¢Ù Tenemos A = ? ?2 ?1 = 1 2 I1 = - 2 * I2 ZR = A2 * Z2 ZR = ( 1 2 ) 2 * ( 4? + j4? ) ZR = 1? + j1? Zp = [ 1 ? ?1? + 1 1+?1? ] ?1 Zp = 1? - j1?
  • 6. V1 = 10?30¡ã?( 1? ?j1? ) 2? +1?? ?1? (V) V1 = 10?30¡ã?( 1? ?j1? ) 3??1? (V) V1 = ?. ????.??¡ã (V) I1 = ?1 ?? I1 = 4.47?3.43¡ã (V) 1? ? j1? I1 = ?. ?????.??¡ã (A) I2 = - ?1 2 I2 = - 3.16?48.43¡ã (A) 2 I2 = ?. ???? ???.??¡ã (A) V2 = 2 * V1 V2 = 2 * 4.47?3.43¡ã (V) V2 = ?. ????.??¡ã (V)