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Nodal Analysis

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The document discusses Kirchoff's Current Law and how it can be used to analyze electrical circuits. Specifically: - Kirchoff's Current Law states that the sum of all branch currents entering a node must equal zero. - An example circuit is given and Kirchoff's Current Law is applied to obtain the equation IA + IB + IC = 0. - Kirchoff's Voltage Law and matrix methods are then used to solve the example circuit for the unknown currents I1, I2, and IC.

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By Pn Zabidah Binti Haron Jabatan Kejuruteraan Elektrik, PSA EE602/ZAB/JUN2012
EE602/ZAB/JUN2012
Kirchoff’s Current Law states that the the branch currents incident at a node must sum to zero. i2+i3-i1-i4 = 0 Note : nodes" (points where elements or branches connect) in an electrical circuit in terms of the branch currents. EE602/ZAB/JUN2012
Z1 IA V IB Z2 IC V1 + _ Z3 EE602/ZAB/JUN2012 I
      By using Kirchoff’s Current Law IA+IB+IC=0 The equation IA:( V-V1)/Z1 IB=-I Ic=V/Z3  V is Node Voltage  V1 is supply voltage EE602/ZAB/JUN2012
4Ω IA IB 3Ω IC 4+j4 Ω 20>0ºV + _ I I 1 + _ 2 Find the value of I 1 , I 2 and Ic EE602/ZAB/JUN2012 15>0ºV
Using Kirchoff Voltage Law (KVL) : Loop 1 : -(20>0º) + 4 I 1 + (4 + j4)(I 1- I 2 ) = 0 4 I 1 + 4 I 1 + j4 I 1 - 4 I 2 - j4 I 2 = 20>0º (4 + 4 + j4 ) I 1 + (- 4 - j4 ) I 2 (8+ j4 ) I 1 + (- 4 - j4 ) I 2 = 20>0º = 20>0º ----- ( 1 )   EE602/ZAB/JUN2012
Loop 2 : +(15>0º) + (4 + j4)(I 2- I 1 ) + 3 I 2= 0 4 I 2 + j4 I 2 - 4 I 1 - j 4 I 1 + 3 I 2 = -(15 >0º ) (4 + j4 + 3 )I 2 +( - 4 - j 4 )I 1 = -(15 >0º ) (7 + j4)I 2 +( - 4 - j 4 )I 1 = -(15 >0º ) (- 4 - j 4 )I 1 (7 + j4)I 2 = -(15 >0º ) -----( 2 ) Apply to matrix equation 8 + j4 -4-j4 -4-j4 7 + j4 I1 I2 = EE602/ZAB/JUN2012 20>0º -(15 >0º )
= 8 + j4 -4-j4 -4-j4 7 + j4 = (8+j4)(7+j4) – (-4-j4)(-4-j4) = 40 + j 28 = 48.83 < 34.99 º 1 = 20>0º -(15 >0º ) -4-j4 7 + j4 = (20>0º)(7+j4) - (-4-j4)(-(15 >0º )) = 80 + j20 =82.46< 14.04 º  EE602/ZAB/JUN2012
2 = 8 + j4 -4-j4 20>0º -(15 >0º ) = (8 + j4)(-(15 >0º )) - (-4-j4) (20>0º ) = -40 + j20 = 44.72 < 153.43º EE602/ZAB/JUN2012
I1 = 1 /  =(80 + j20)  / (40 + j 28) = 1.58 - j0.60 Ampere I2= 2 /  =(-40 + j20)  / (40 + j 28) = -0.44+j 0.81 Ampere Ic= I1 – I2 = (1.58 - j0.60 ) –(-0.44+j 0.81 ) = 2.01-j1.42 Ampere EE602/ZAB/JUN2012
5Ω IA IB 4Ω IC 5+j3 Ω 100>0ºV + _ I I 1 + _ 2 Find the value of I 1 , I 2 and Ic EE602/ZAB/JUN2012 50>0ºV
  The principles of Mesh Analysis is the algebraic sum of the voltage supply and the voltage drop across the load, around a closed path is equal zero. One of the method to solve the electrical circuits problems. EE602/ZAB/JUN2012

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Nodal Analysis

  • 1. By Pn Zabidah Binti Haron Jabatan Kejuruteraan Elektrik, PSA EE602/ZAB/JUN2012
  • 3. Kirchoff’s Current Law states that the the branch currents incident at a node must sum to zero. i2+i3-i1-i4 = 0 Note : nodes" (points where elements or branches connect) in an electrical circuit in terms of the branch currents. EE602/ZAB/JUN2012
  • 5.       By using Kirchoff’s Current Law IA+IB+IC=0 The equation IA:( V-V1)/Z1 IB=-I Ic=V/Z3  V is Node Voltage  V1 is supply voltage EE602/ZAB/JUN2012
  • 6. 4Ω IA IB 3Ω IC 4+j4 Ω 20>0ºV + _ I I 1 + _ 2 Find the value of I 1 , I 2 and Ic EE602/ZAB/JUN2012 15>0ºV
  • 7. Using Kirchoff Voltage Law (KVL) : Loop 1 : -(20>0º) + 4 I 1 + (4 + j4)(I 1- I 2 ) = 0 4 I 1 + 4 I 1 + j4 I 1 - 4 I 2 - j4 I 2 = 20>0º (4 + 4 + j4 ) I 1 + (- 4 - j4 ) I 2 (8+ j4 ) I 1 + (- 4 - j4 ) I 2 = 20>0º = 20>0º ----- ( 1 )   EE602/ZAB/JUN2012
  • 8. Loop 2 : +(15>0º) + (4 + j4)(I 2- I 1 ) + 3 I 2= 0 4 I 2 + j4 I 2 - 4 I 1 - j 4 I 1 + 3 I 2 = -(15 >0º ) (4 + j4 + 3 )I 2 +( - 4 - j 4 )I 1 = -(15 >0º ) (7 + j4)I 2 +( - 4 - j 4 )I 1 = -(15 >0º ) (- 4 - j 4 )I 1 (7 + j4)I 2 = -(15 >0º ) -----( 2 ) Apply to matrix equation 8 + j4 -4-j4 -4-j4 7 + j4 I1 I2 = EE602/ZAB/JUN2012 20>0º -(15 >0º )
  • 9. = 8 + j4 -4-j4 -4-j4 7 + j4 = (8+j4)(7+j4) – (-4-j4)(-4-j4) = 40 + j 28 = 48.83 < 34.99 º 1 = 20>0º -(15 >0º ) -4-j4 7 + j4 = (20>0º)(7+j4) - (-4-j4)(-(15 >0º )) = 80 + j20 =82.46< 14.04 º  EE602/ZAB/JUN2012
  • 10. 2 = 8 + j4 -4-j4 20>0º -(15 >0º ) = (8 + j4)(-(15 >0º )) - (-4-j4) (20>0º ) = -40 + j20 = 44.72 < 153.43º EE602/ZAB/JUN2012
  • 11. I1 = 1 /  =(80 + j20)  / (40 + j 28) = 1.58 - j0.60 Ampere I2= 2 /  =(-40 + j20)  / (40 + j 28) = -0.44+j 0.81 Ampere Ic= I1 – I2 = (1.58 - j0.60 ) –(-0.44+j 0.81 ) = 2.01-j1.42 Ampere EE602/ZAB/JUN2012
  • 12. 5Ω IA IB 4Ω IC 5+j3 Ω 100>0ºV + _ I I 1 + _ 2 Find the value of I 1 , I 2 and Ic EE602/ZAB/JUN2012 50>0ºV
  • 13.   The principles of Mesh Analysis is the algebraic sum of the voltage supply and the voltage drop across the load, around a closed path is equal zero. One of the method to solve the electrical circuits problems. EE602/ZAB/JUN2012