Problem 15-4 Source:
Problem 3.58 - page 140 - IRWIN, J. David -
Book: Análise de Circuitos em Engenharia - 4ª edição - Ed. Pearson - 2013.
In the circuit shown in Figure 15-04.1, calculate:
The voltages of the essencial nodes, i1 , i2
and the voltage Vo.
Solution of the problem using Method of Transforming SourcesClick here!
Solution of the problem using Nodal VoltageClick here!
Solution of the Problem 15-4 -
Thévenin/Norton Method
To use the Thevenin theorem, let's remove the resistor of 2 ohms that
is under the potential V0 at the output of the circuit. Removing
this resistor from the circuit we see that i3 = i2, because no current flows through the
voltage source 64 volts.
On the other hand, the two sources of current arriving at the node e4,
implies that i3 = i2 = 14 A. See the Figure 15-04.2.
And knowing the values of these variables, we conclude that:
e2 - e1 = 4 x 14 = 56 volts
And to calculate Vth, we need to know
e1 ( but e1 = 6 i1 )
in relation to the reference point or ground (marked on the circuit). Note that in the first circuit we have:
i3 = i1 + i = 14 A
Thus, if we make the mesh involving the voltage source of 12 volts , the resistors
of 6 ohms and 3 ohms , we can calculate i1.
Not forgetting that the current flowing through the resistor of 3 ohms
is equal to i - 6 , we have:
12 + 3 ( i - 6 ) - 6 i1 = 0
Remembering that i = 14 - i1 and solving the equation,
we found i1, or:
i1 = 4.0 A
Then
e1 = 6 i1 = 6 x 4.0 = 24.0 volts
Therefore, by making the outer loop of the circuit marked by the line highlighted in green , we find:
- Vth - 64 + ( 4 x 14 ) + e1 = 0
Solving the equation, we find the value of Vth, or:
Vth = 16 volts
Now, we only need to find the Thévenin resistance . To do this, simply remove
all the independent sources of the circuit. Voltage sources are
short circuited and current sources are open circuits .
See the circuit modified in the Figure 15-04.3.
Looking at the circuit in the figure above, we realize that the resistors of 3 and
6 ohms are in parallel and these in series with 4 ohms . Doing the calculations:
Rth = [( 6 x 3 ) / (6 + 3 )] + 4 = 6 ohms
So, as we already know Vth and Rth, we are
able to draw the Thévenin equivalent of the circuit. The 2 ohms
resistor, which we had removed from the circuit to calculateVth
and Rth, was replaced. The Figure 15-04.4 shown
the new configuration of the circuit.
Therefore, we can calculate i, or:
i = Vth / ( Rth + 2 ) = 16 / 8 = 2 A
Then, the value of V0 is:
V0 = 2 i = 2 x 2 = 4 volts
Note that knowing the value of V0, we conclude that the current
through the voltage source of 64 volts is 2 A . Then the current
i3 is:
i3 = 14 - i = 14 - 2 = 12 A
We can calculate the values of e1 and e2, because applying
the Ohm's law for e2, we have:
e2 = 64 + V0 = 64 + 4 = 68 volts
e1 = e2 - 4 i3 = 68 - 4 x 12 = 20 volts
And to calculate i1, we can write by the law of Ohm that:
i1 = e1 /6 = 20 /6 = 3.33 A
Now let's calculate the values of e3 and e4, or:
e3 = e2 - 12 = 20 - 12 = 8 volts
e4 = e2 + 12 i2 = 68 + 168 = 236 volts
Therefore, we can summarize the values found.
Vo = 4 volts
e1 = 20 volts
e2 = 68 volts
e3 = 8 volts
e4 = 236 volts
i1 = 3.33 A
i2 = 14 A
i3 = 12 A
Finally, in the Figure 15-04.5 we see the complete circuit with all the
indicated currents.
Notice that the whole circuit satisfies the laws two nodes and meshes .
We can verify if it obeys law of energy conservation , making a
power balance.