Be the Figure 13-01 where currents arrive and leave the A node. We will follow the pattern of currents established by technical literature:
will have POSITIVE values the currents that leave the node, and values NEGATIVES
the currents that enter the node.
Figure 13-01
We see in the next figure that the currents I1 and I4arrive at the node,
therefore negative values. And the I2 and I3 currents leave the node,
therefore, positive values. In this way, we can write the equation that defines the behavior
of the currents to the A node. See the equation below.
eq. 13-01
So let's enunciate the law of nodes as follows:
Figure 13-02
In the Figure 13-02 we see a circuit with three current sources. Note that the currents of the 14 and 32 A
sources arrive at the node e1, while the 4 A leaves the node. Therefore, by adding algebraically the currents
we find the value of 42 A.
This means that we could substitute the three current sources for a single
of value equal to 42 A, pointing upwards.
Given this data, we easily calculate the value of e1. Just make the parallel of the 3 and 6 ohms
resistors. Performing the calculation we find the value of 2 ohms. Therefore, by applying the Ohm's law to the circuit,
we have e1 = 2 42 = 84 volts .
The Law of Meshes deals with the decrease of tensions between the various components that
form a closed loop . Like this:
Figure 13-03
In the Figure 13-03 we see a circuit with four voltage sources, which form a series circuit and a closed loop with
the resistor of 3 ohms. Note that three sources are with positive polarity pointing to point a. Therefore,
we can add the value of these three sources resulting in 40 volts. And the source of 10 volts points in the
opposite direction. In this case, we must subtract it from the previous result, finding a value of 30 volts.
As a final result, we find a single source with a value equal to 30 volts. With this value, by applying the Ohm's law
to the circuit, we find the electric current that flows through it, that is, I = 30/3 = 10 A. Thus, on the resistor we
have a voltage drop Vab = 30 volts. Note that this voltage has polarity opposite that of the voltage source.
Therefore, by adding algebraically the two values we find zero as a result. That is, fully in accordance with the law of meshes.
A super-node is characterized when there is a voltage source interconnecting two nodes. In this case,
since we do not know how to determine the current through the voltage source, then we apply the law of nodes to the two nodes as if they were one.
Figure 13-04
In the Figure 13-04 we see a circuit with a voltage source interconnecting nodes a e b.
Let's analyze these nodes by making the equations of the currents that arrive and leave each node.
Initially, let's write the equation for node a.
iab + iy - ix = 0 ⇒
iab = ix - iy
And for node b we can write:
ik - iab - 10 = 0 ⇒
iab = ik - 10
We see clearly that we have two equations relating iab with the
other currents. Thus, relating the equalities, we have:
ik - 10 = ix - iy ⇒
ik + iy - ix - 10 = 0
Now, note that if we do the equation of the node, considering nodes a and b as if it were a single node, we would get the same equation as above. This means that when we have a source of voltage interconnecting two nodes, we can consider them as one. This is what we call SUPER NODE.
But do not forget that the voltage between them is different. In our case, there is a potential difference of 15 volts between them.
A super-mesh is characterized when there is a current source interconnecting two meshes. Thus, we reduce from
two meshes to a single one, bypassing the current source. If the current source is on the periphery of the circuit, we must
ignore the mesh to which the current source belongs.
Figure 13-05
In the Figure 13-05 we see a circuit where we have a current source in the
periphery of the circuit. Notice that the current I1 is the actual value of the current source. So we can focus only on the other two meshes, finding two equations with two unknowns, this system of easy solution. In this way, we can solve the problem by solving the problem: I1 = 2 A , I2 = 0.68 A and I3 = 3 A.
To do so, we use the equations below:
- 10 I1 - 15 I2 + 30 I3 = 60
25 I2 - 15 I3 = - 28
Do not forget that in the first equation we know the value of I1 = 2 A. Therefore, we can simplify the system for two equations to two unknowns, as was said above.
To construct the equations of a circuit using Kirchhoff's laws, let's take the circuit shown in the Figure 13-06 as an example.
Figure 13-06
Notice that this circuit has three meshes. This leads us to conclude that to solve this circuit we must construct a system of three equations with three unknowns. To do so, let's start with the mesh containing the current I1. Let us remember that we must follow the established convention regarding the polarity of voltages.
In this case we will start from the ground point.
Passing through the voltage source of 20 volts (from the bottom up) we have the minus sign indicating the polarity of the source. Therefore, for the source we will use the value of - 20. Following the arrow clockwise, we find the resistor of 5 ohms. Like the current
I1 enters the left side of the resistor, this causes a positive signal relative to the voltage drop on it. The same is true for the 10 ohms resistor.
Then, due only to I1, we have
5 I1 + 10 I1 = 15 I1. On the other hand, notice that current I2 runs on the resistor of 10 ohms in the opposite direction to I1. This will cause a voltage drop with opposite polarity. So we have - 10 I2. And finally, using the same
line of reasoning in relation to I3, we have - 5 I3.
With all this information, we can write the equation referring only to the mesh I1. Note the positive signal on the portion of I1 and the signal
negative in the plots referring to I2 and I3.
- 20 + 15 I1 - 10 I2 - 5 I3 = 0
We must proceed in the same way for the currents I2 and I3 to find the three equations we need to solve the problem.
In the mesh referring to I2 we have no source of voltage. Therefore, the portion relating to it will be zero . Starting from ground, the current rises by the resistor of 10 ohms giving
a positive sign of the voltage drop on this resistor. The same happens with the other two resistors that make up the I2 mesh. In this way, we have the
15 I2. On the other hand, notice that current I1 cycles over the resistor 10 ohms in the opposite direction to I2. This will cause a voltage drop with opposite polarity. Therefore, this parcel will be - 10 I1. And finally,
for I3 we have the same situation as I1, that is, the current flows in the opposite direction, giving the parcel - 4 I3. Thus, equation
complete is:
- 10 I1 + 15 I2 - 4 I3 = 0
Again, look at the positive signal in the portion of I2 and the negative
sign in the parcels referring to I1 e I3.
Now let's write the equation for the last mesh, that is, referring to I3. Starting from the point a we have the resistor of 6 ohms. The voltage drop on this resistor will be +6 I3. The same will happen for the other two resistors that are in the I3 mesh. Therefore, the plot will be + 15 I3. And as in the two previous cases, we will have the plots -5 I1 and -4 I2. Let's not forget to add the voltage source of + 10 volts. Then the equation will be:
+ 10 - 5 I1 - 4 I2 + 15 I3 = 0
Rearranging the equations we obtain the following system of three equations to three unknowns.
15 I1 - 10 I2 - 5 I3 = 20
- 10 I1 + 15 I2 - 4 I3 = 0
- 5 I1 - 4 I2 + 15 I3 = - 10
Solving
this system in a software such as "Octave" we find the following values for the currents
of the circuit.
I1 = 3.5233 A
I2 = 2.6744 A
I3 = 1.2209 A
Here we try to show how to assemble the system of equations from a given circuit. For circuits with a maximum of
three meshes it is possible to solve the system with traditional methods, such as
Cramer rule, by replacing variables, etc ... However, for systems with more than three meshes, we must
use a computer program that makes it easier to find the solution, because trying to solve a system of
these "by hand" is extremely laborious and time-consuming.
This theorem establishes the power relationship in an electrical circuit. It is valid for DC circuits as well as AC circuits.
It is worth noting that this theorem is valid for every circuit that obeys
Kirchhoff's voltage and current laws. The statement of the theorem
is:
"The algebraic sum of the powers involved in all branches of a circuit, at any instant, is always equal to zero."
eq. 13-02
This theorem can be written mathematically as shown in eq. 13-02 above. In the problems tab there are several problems where
are asked to perform a power balance. This is done by applying Tellegen's Theorem.
