Nodal Analysis

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

2. 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

4. Z1
IA V
IB
Z2
IC
V1
+
_
Z3
EE602/ZAB/JUN2012
I

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