Problem 11-3 Source:
Example 5.10 - page 188 - IRWIN, J. David -
Book: Análise de Circuitos em Engenharia - 4ª edição - Ed. Pearson - 2013.
In the circuit show in Figure 11-03.1, calculate:
a) The voltage of all the essencial nodes.
b) The voltage at the terminals a-b.
c) The currents ix , ik.
Figure 11-03.1
Solution of the problem using Node Voltage Methodclique aqui!
Solution of the problem using Superposition Methodclique aqui!
Solution of the Problem 11-3 -
Method of Transforming Sources
Let's do two transformations: one in the highlighted part in green and another
in the highlighted part in yellow. See the Figure 11-03.2.
Figure 11-03.2
Note that in the highlighted green part we have a current source in parallel
with a resistor. And in the part highlighted in yellow, one must add the two
3 ohms resistors that are in series, resulting in a single resistor with
a value equal to 6 ohms.
Figure 11-03.3
Thus, the above circuit is obtained, conform Figure 11-03.3 where two
resistors are in series and the sum of the two must be made.
We also have two voltage sources in series where we can substitute for a single
source of value equal to 36 volts, since both are opposite polarities and one
must subtract their numerical values .
Figure 11-03.4
In the Figure 11-03.4, again, we come to the problem 11-1. Therefore, the solution
follows the same steps, as shown below.
Figure 11-03.5
In the Figure 11-03.5 we have two sources of current pointing in the same direction, so we can add them together.
And calculating the parallel of the resistances of 8 and 3 ohms we arrive at the circuit below.
Figure 11-03.6
Now, by applying a current divider, the current ik is:
i k = 13.5 x 2.182 / (2.182 + 6) = 3.6 A
With the value of ik we calculate e2.
e2 = 6 i k = 6 x 3.6 = 21.6 volts
And as we know the relationship between e1 e e2 we reach the value of e1.
e1 = e2 + 12 = 21.6 + 12 = 33.6 volts
With the value of e2 and based on the circuit below we can calculate
ix using the circuit highlighted in blue. Therefore:
Figure 11-03.7
ix = ( e2 - 12 ) /8 = 1.2 A
Knowing ix and knowing the relationship between e2 e e3, we have:
e3 = e2 - 2 i x = 21.6 - (2 x 1.2) = 19.2 volts
Knowing ix we can calculate iy because we know that:
i y = i x + 2 = 1.2 + 2 = 3.2 A
Therefore, to calculate the voltage between the terminals a-b, simply apply the Ohm law on the resistance of 3 ohms,
which lies between the terminals a-b. Logo:
