Problem 12-4 Source:
Problem 3.58 - page 140 - IRWIN, J. David -
Book: Análise de Circuitos em Engenharia - 4ª edição - Ed. Pearson - 2013.
In the circuit show in Figure 12-04.1, calculate:
a) The voltages of the essential nodes , i1 and i2.
b) The voltage Vo.
Figure 12-04.1
Solution of the problem using Method of Transforming SourcesClick here!
Solution of the problem using Thévenin/Norton MethodClick here!
Solution of the Problem 12-4 -
Nodal Voltage Method
In this circuit we have four essential nodes identified as e1 , e2 ,
e3 and e4. In addition, the points e1 and e3
form a super-node, as well as the points e2 and Vo. So we can write the equation
for the super-node e1 and e3.
e1 /6 + (e1 - e2) /4 + e3 /3 + 6 = 0
From the circuit, making the mesh through e1 and e3 we concluded that:
e1 = e3 + 12
Substituting this last equation in the first, there is a relation between e2
and e3. After some algebraic adjustments:
3 e3 - e2 = - 44
eq. 12-4.1
Let's develop the equation for the node e2, remembering the super-node
e2 and Vo.
( e2 - e1 ) /4 + ( e2 - e4 ) /12 + V0 /2 = 0
eq. 12-4.2
Note that making a mesh through e2 and for V0,
we get the relation:
e2 = V0 + 64 ⇒ V0 = e2 - 64
On the other hand, it should be noted that the two sources of current in the
circuit, of 6 A and that of 8 A together, produce a current of
14 A, which must pass through the resistor of 12 ohms. This
current is i2. As a consequence, we have a voltage drop on
this resistor of 12 x 14 = 168 volts. Therefore, the relationship between
e2 and e4 is:
e2 = e4 - 168 ⇒ e2 - e4 = - 168
Substituting these last two equations into equation
eq. 12-4.1 and, after
some algebraic arrangements, we obtain the following relation:
e3 = 3 e2 - 196
This is another equation that relates e2 and e3. Therefore,
we now have a system of two equations with two unknowns. So we can solve the system. If we replace
this last equation in equation eq. 12-4.2, we find the values of the voltages of the nodes.
3 ( 3 e2 - 196 ) - e2 = - 44
Solving this equation, we find the values of the voltages of the nodes.
e2 = 68 volts
e3 = 8 volts
e4 = 236 volts
e1 = e3 + 12 = 20 volts
Knowing e1, easily calculated i1, or:
i1 = e1 /6 = 20 /6 A
And i2 we have already calculated, since it is the sum of the two current sources, or:
i2 = 14 A
On the other hand, as we know e2, we have the value of V0.
V0 = e2 - 64 = 68 - 64 = 4 volts
Power Balance
To verify the accuracy of the calculations, let's make the Power Balance.
See the Figure 12-04.2 for the circuit with the indication of the currents in the branches.
Figure 12-04.2
Initially we will calculate the powers dissipated in the resistors, remembering that,
by convention, these POSITIVE powers are considered.
P6 = 6 x 3.332 = + 67 W
P3 = 3 x 2.662 = + 21 W
P4 = 4 x 122 = + 576 W
P12 = 12 x 142 = + 2,352 W
P2 = 2 x 22 = + 8 W
For the calculation of the powers in the sources we will remember
that voltage sources with the current coming out of the positive pole are considered NEGATIVE powers, because they are
supplying power to the circuit. Otherwise,
will be POSITIVE powers. The same goes for current sources.
