Problem + Hard 13-1 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 shown in Figure 13-01.1.
Figure 13-01.1
Solution of the Problem + Hard 13.1
As we can see in the circuit show in Figure 13-01.1, we have a current source of 2 A
in series with a resistor of 2 ohms between points a - e. This one
resistor can be removed from the circuit as it will not affect the calculations. On the other hand,
we have two more current sources of 2 A that interconnect 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 13-01.2
You can see how the circuit turned out in the Figure 13-01.2. 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 chain of 2 - I0. So obviously Vx = 2 - I0.
The current that passes through the 1 ohm resistor that connects the point a was named I
to the ground. Let's replace the dependent current source 2. Vx by
4 - 2 I0. The circuit with the identification of the other currents is shown in the Figure 13-01.3.
Figure 13-01.3
Note that there are 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 13-01.4, we draw two loops: one is highlighted by the arrow
green and the other is highlighted by the purple arrow.
Figure 13-01.4
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:
