Problem + Hard 12-2
Source: Problem 4.28 - page 170 - IRWIN, J. Davis -
Book: Analysis of Circuits in Engineering - 4th Edition - Ed. Pearson - 2013.
Calculate the value of Vx and the voltage Vo in the circuit of the Figure 12-02.1.
Figure 12-02.1
Solution of the Problem + Hard 12.2
As you can see in the circuit shown in Figure 12-02.1 , there is a current source of 2 A
in series with a 2 ohms resistor between points a - e. This one
resistor can be removed from the circuit as it will not affect the calculations. On the other hand,
there are two more current sources of 2 A that connect the points
e - f and the points f - g. In this way, it is evident that by the resistor
of 1 ohm that connects the points b - e and by the resistor of
2 ohms that interconnect the c - f points, no current will pass
electric. Then they can be taken out of the circuit.
Figure 12-02.2
In the circuit shown in Figure 12-02.2, you can see how the circuit turned out. The current was named I0 that passes through the 1 ohm resistor and on which the voltage V0 is to be determined.
In this way, through the 1 ohm resistor that connects the points g-d will circulate
a current of 2 - I0. So obviously Vx = 2 - I0.
It was called I, the current that passes through the 1 ohm resistor that connects point a to ground. The dependent current source 2 Vx is replaced by
4 - 2 I0. In the Figure 12-02.3 we have the circuit with the identification of the other currents.
Figure 12-02.3
Note that we have two unknown chains: I and I0. Therefore,
To solve the system, two different equations must be obtained that relate these
incognitos. To do so, in the Figure 12-02.4, we draw two loops: one is highlighted by the arrow
green and the other is highlighted by the purple arrow.
Figure 12-02.4
Applying Kirchhoff's law for voltages to these loops and remembering that the
sum of the voltages in the components that form the circuit must be equal to zero, we have
for the path highlighted by the green arrow the equation:
I + 2 I - 4 + 2 + I - 2 I0 + 6 + I - 3 I0 = 0
For the path highlighted by the purple arrow, we have the equation:
-6 - I + 3 I0 - 12 - 2 + I0 + 3 I0 = 0
Rearranging these two equations, we find a system of two
equations with two unknowns of easy solution, that is:
5 I0 - 5 I = 4
7 I0 - I = 20
Solving the system we find the values of the currents, or:
I0 = 3.2 A e I = 2.4 A
With these values in hand, we can easily calculate the values
of Vx and V0, because:
