Example: Calculate the current Io.
((V1 -6)/3k) + (V1/6k) =2mA
2V1 -12 + V1 =12
3V1 =24
V1=8 volts
B
V2
Io
I1
Io = (8 -6)/3k
=0.667 mA
I1 =6/4k
=1.5 mA
V1 I
s
4k
Example
Io= V1/10k
We want to calculate the node voltages.
So, for node A
(V1/10k) + ((V1- V2)/10k) = 4mA
Io
2Io
(V1/10k) + (V1/10k) – (V2/10k) = 4mA
Io + Io – (V2/10k) = 4mA
2I0 - (V2/10k) = 4mA
20Io –V2 =40mA
For node 2
(V2/10k) + ((V2 – V1)/10k) = -2Io
(V2/10k) + (V2/10k) – (V1/10k) = -2I0
(V2/10k) +(V2/10k) – I =-2I
Io
2Io
2V2 + 10I0 = 0
Io
2Io
Equating equations of node 1 and 2
40Io - 2V2 = 80mA
10I0 + 2V2 = 0
50Io =80mA
hence Io =8/5 mA
Io
2Io
So, V1
Io = (V1/10k)
8/5 mA = (V1/10k)
V1 = 16 volts
From node 2
2V2 + V1 =0
2V2 =-V1
V2 = -16/2 = -8volts
SUPER NODE
A node which emerges as a result of
combination of two ordinary nodes
around a voltage source.
Constraint or coupling
Equation.
This is an equation which describes
a super node mathematically,instead
of writing equations individual
ordinary nodes of the super node.
Example
Calculate the
current source.
power
supplied
by
Power = 1A(V1)
Applying KCL to node 1
(V1/10) - Ix = 1
V1
V2
3
Ix
Apply KCL to node 2
0 = Ix+(V2/5)+((V2-2)/ 7)
Now, 3 unknown quantities can not be
calculated from 2 equations, so try
super node technique,
Equation for super node
1 = (V1/10) + (V2/5) + ((V2-2)/7)
Simplifying it, we can write
35 V1+120 V2 = 450
Also the constraint or coupling
equation is
V1 - V2 = 6
Solving the two equation
simultaneously,
350 = 35V1 + 70V2 + 50 – 100
Or
35V1+120V2=450
V1 – V2 = 6
Therefore,
V1 = 7.55 V
& V2 = 1.55 V
Hence, power
P = 1(A) (V1)
= 7.55 W
Example: Calculate I1 from the given
circuit.
We can redraw the circuit as
Constraint equation for super node
V1 – V2 = 3
V1
V2
I1
6k
KCL equation for super node
(V1/3k) + (V2/6k) = 2 x 10-3
2V1 + V2 =12
Now from constraint equation
V2 = V1 - 3
V1
V2
I1
6k
Putting this value in the KCL equation of
super node
2V1 + V1 – 3 = 12
3V1 = 15
V1 = 5 volts
V1
V2
I1
6k
Now
I1 = V1/3k
=5/3k
=1.6 mA
Example: Find I1 from the circuit.
3V
I1
We will redraw this circuit as
V1
V2
3V
I1
Constraint equation for the super node
V2 – V1 = 6V
V1
V2
3V
I1
KCL equation for the super node
((V1 – 3)/6k) + (V1/12k) + ((V2 +3)/12k)+(V2/12k) =0
2V1 -6 + V1 +V2 + 3 + 2V2 =0
V1
V2
I1I1
3V1 + 3V2 = 3
V2 + V1 =1
3V
V1
V2
3V
I1
Adding constraint equation and KCL
equation
V2 – V1 =6
V2 + V1=1
2V2 = 7 or V2 = 3.5V
V1
V2
3V
I1
I1 = 3.5/6k
=35/60k
=7/12 mA
Example
V1
I1
V2
6k
We want to calculate the value of current I1
For node 1
(V1/12k) +(V1/4k) +(V1/6k) – (6/6k) = 2 mA
((1/12k) +(1/4k) +(1/6k))V1 –(6/6k) = 2 mA
V1
V2
I1
6k
(V1/2k) – (1/1k) = 2 x 10-3
V1 – 2 = 4
V1 =6V
V1
V2
I1
6k
Now
I1 =V1/4k
= 6/4k
= 1.5 mA