Problem 12-2 Source:
Problem 4.12 - page 96 - NILSSON & RIEDEL -
Book: Circuitos Elétricos - 8ª edição - Ed. Pearson Education do Brasil - 2010.
In the circuit show in Figure 12-02.1, calculate the voltages in the nodes and the currents i1 , i2 , i3 and i4.
Figure 12-02.1
Solution of the problem using Method of Transforming SourcesClick here!
Solution of the problem using Superposition MethodClick here!
Solution of the problem using Thévenin/Norton MethodClick here!
Solution of the Problem 12-2 -
Nodal Voltage Method
Item a
Starting from the node e1 we can write the equation:
(e1 - 144) /4 + (e1) /10 + (e1 - e2) /80 = 0
Starting from the node e2 we can write the equation:
(e2 - e1) /80 + e2 /5 - 3= 0
Developing the node equation e1, we arrive at the equation:
e1 = 17 e2 - 240
And developing the node equation e2, we have the equation:
29 e1 - e2 = 2 880
Now, if we replace in this last equation the value of e1, already calculated, we were able to calculate the e1 and e2, that is:
e1 = 100 volts
e2 = 20 volts
Item b
As we already know e1 e e2, we are able to
calculate the currents that circulate in the circuit. To calculate
i1 we simply apply the law of Ohm in the resistor of 4 ohms. Then:
i1 = 144 - e1 /4 = 144 - 100 /4 = 11 A
In the same way we do for the calculation of i2, or:
i2 = e1 /10 = 100 /10 = 10 A
For the calculation of i3, we can use the law of Nodes for the node
e1. So:
i3 = i1 - i2 = 11 - 10 = 1 A
And finally to calculate i4, we also use the law of Nodes for the node
e2. Then:
i4 = i3 + 3 = 1 + 3 = 4 A
We can also calculate i4 dividing the voltage of the node
e2 by 5 ohms, which is the value of the resistor that is
connected between the node
e2 and the common point of the circuit, reaching the same value
found above
( 20 /5 = 4 A ).
Power Balance
To check if the results are correct, let's take stock of the circuit's power. First let us
compute the dissipated powers on all resistors, or:
P4 = 4 ( i1 )2 = 4 x 112 = 484 W
P10 = 10 ( i2 )2 = 10 x 102 = 1 000 W
P10 = 80 ( i3 )2 = 80 x 12 = 80 W
P5 = 5 ( i4 )2 = 5 x 42 = 80 W
Now, summing all the dissipated powers in the resistors we find the value of:
P+ = 484 + 1 000 + 80 + 80 = 1 644 W
For the sources, note that the current goes out by the positive pole of the source
of 144 volts, that is, it is supplying power to the circuit, so its
value is negative, or:
P144 = - 144 i1 = - 144 x 11 = - 1 584 W
For the current source of 3 A the voltage on it is the node voltage
e2 of 20 volts and the positive polarity is at the tip
of the arrow. So it is also supplying power
(negative) to the circuit, or:
P3 = - 3 e2 = - 3 x 20 = - 60 W
>Now we can add algebraically the powers provided by the sources and we find:
P - = - 1 584 - 60 = - 1 644 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.
