nodal analysis

13
By Pn Zabidah Binti Haron Jabatan Kejuruteraan Elektrik, PSA EE602/ZAB/JUN2012

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Nodal Analysis

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Page 1: Nodal Analysis

By Pn Zabidah Binti Haron Jabatan Kejuruteraan

Elektrik, PSA

EE602/ZAB/JUN2012

Page 2: Nodal Analysis

EE602/ZAB/JUN2012

Page 3: Nodal Analysis

Kirchoff’s Current Law states that the the branch currents incident at a node must sum to zero.

i2+i3-i1-i4 = 0Note : nodes" (points where elements or branches connect) in an electrical circuit in terms of the branch currents.

EE602/ZAB/JUN2012

Page 4: Nodal Analysis

Z1Z2

I Z3

V1 + _

IA IB

IC

EE602/ZAB/JUN2012

V

Page 5: Nodal Analysis

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

Page 6: Nodal Analysis

Find the value of I 1 , I2 and Ic

I1

4+j4 Ω

I2

+ _

+ _

IA IB

IC

20>0ºV 15>0ºV

EE602/ZAB/JUN2012

Page 7: Nodal Analysis

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 = 20>0º

(8+ j4 ) I 1 + (- 4 - j4 ) I 2 = 20>0º ----- ( 1 )

 

EE602/ZAB/JUN2012

Page 8: Nodal Analysis

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 I 1 = 20>0º-4-j4 7 + j4 I 2 -(15 >0º )

EE602/ZAB/JUN2012

Page 9: Nodal Analysis

= 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 -4-j4 -(15 >0º ) 7 + j4

= (20>0º)(7+j4) - (-4-j4)(-(15 >0º )) = 80 + j20=82.46< 14.04 º 

EE602/ZAB/JUN2012

Page 10: Nodal Analysis

2 = 8 + j4 20>0º -4-j4 -(15 >0º )

= (8 + j4)(-(15 >0º )) - (-4-j4) (20>0º )= -40 + j20= 44.72 < 153.43º

EE602/ZAB/JUN2012

Page 11: Nodal Analysis

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

Page 12: Nodal Analysis

Find the value of I 1 , I2 and Ic

I1

5+j3 Ω

I2

+ _

+ _

IA IB

IC

100>0ºV 50>0ºV

EE602/ZAB/JUN2012

Page 13: Nodal Analysis

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