Problem 12-10 Source:
Adapted from the 1st test - Circuit Analysis I - Prof. Dr. Dênis Fernandes - PUCRS - 2018.
In the circuit show in Figure 12-10.1, calculate the voltage
Vab , v1 and v2.
Figure 12-10.1
Problem solving using Transformations Sources Methodclique aqui!
Solution of the Problem 12-10 -
Nodal Voltage Method
To reduce the amount of variables involved in the problem, let's take the
point b as
reference, that is, Vb = 0 and rename the important points. So,
we have the circuit show in Figure 12-10.2 below already with the modifications.
Figure 12-10.2
Let's apply the nodal method to determine what the problem is asking for. We begin by applying the law of nodes to the node a . We consider positive the currents that exit from node and negative otherwise. Thus, we have:
Va / 2 + (Va - v1) / 4 + (Va - v2) / 1 = 0
By performing the operations and rearranging the terms, we arrive at the first equation, that is:
- v1 - 4 v2 + 7 Va = 0
eq. 12-10.1
For the node 1 , we find the following relation:
v1 / 2 - (Va - v1) / 4 + 3 = 0
Working the relationship above, we find the second equation necessary to solve the problem.
3 v1 - Va = - 12
eq. 12-10.2
And to find the third equation (since we have three variables), we apply the same method to
node 2 , or:
- (Va - v2) / 1 + v2 / 2 - 3 = 0
We rearrange the terms, we arrive at the following relation:
3 v2 - 2 Va = 6
eq. 12-10.3
With the equations eq. 12-10.1,
eq. 12-10.2 and eq. 12-10.3, we construct a system of three equations with three unknowns.
Your solution can be obtained by any method. We find the following values:
Vab = Va = 1 volt
v1 = -3.67 volts
v2 = 2.67 volts
Power Balance
First, let us compute the powers absorbed by all resistors, or:
P4 = 4 (1.363)2 = 5.45 W
P1 = 1 (1.67)2 = 2.78 W
P2 = 2 (1.83)2 = 6.73 W
P'2 = 2 (1.33)2 = 3.54 W
For the case of the dependent source whose current value is Vab / 2 = 0.5 A , we observe that
the greatest potential is in the opposite direction to that of the arrow source. Then it is absorbing energy
from the circuit and the value of its power will be positive . Like this:
Pab = 1 x 0.5 = +0.5 W
Now summing all the absorbed powers by the resistors and dependent source, and naming it as
P+, we find the value of:
P+ = 5.45 + 2.78 + 6.73 + 3.54 + 0.5 = +19 W
For the current source of 3 A we must first calculate the potential difference on it. Let's call it V21.
By doing the mesh we have
-v1 - V 21 + v 2 = 0 . Hence,
V21 = 6.34 volts results. Note that the highest potential is in the same direction as the current
source arrow. Therefore this source is supplying power to the circuit. So its value will be
negative and let's call it P -.
P - = - 3 x 6.34 = - 19 W
Finally, we know that the algebraic sum of the powers supplied and dissipated in a circuit, shall be equal
to ZERO , since it must satisfy the Law of Conservation of Energy.
