Although Kirchhoff's law is a very useful tool, in practice it proves to be a bit cumbersome, since when we have an
electrical circuit with many meshes it can result in a rather complex mathematical system to find its solution . In this case,
only with computer programs or calculators of specific use.
Let us present another tool for solving electric circuits of very practical use for certain situations.
This tool is the Superposition Theorem. This theorem states that we can calculate, for example, the current or voltage
at a given point in the electric circuit, using a voltage or current source, at a time. To do so, we must eliminate the other
sources of voltage or current, remaining only one source. Let's remember that to eliminate a voltage source we must
short-circuit it . And to eliminate a current source we must remove it from the circuit, resulting in an open circuit.
Thus, in the Superposition theorem we obtain a number of equations according to the number of sources in the circuit. Also we must define in which variable we are interested.
To understand how this method works, let's look at a simple circuit shown in the Figure 14-01.
Figure 14-01
The circuit has two voltage sources, V1 and V2.
We want to calculate the current I that runs through the resistor R2. Then, by the Superposition theorem the current I will be given by:
eq. 14-01
Of course, this equation, with two parcels, is for the particular case presented here as an example, since we have only two voltage sources. If there were more sources, we would have a number of parcels equal to the number of sources in the circuit.
Therefore, to compute I we must first calculate K1 and K2.
To compute K1, we will short-circuit V2, so we will be calculating
the fraction of I that corresponds to the voltage source V1. We will call this fraction
of I1. Subsequently, we shorted V1 and we calculated the fraction
of I that corresponds to the voltage source V2. This fraction, in turn, will be called
I2. So, to get the final value of the current I, we simply add I1 with I2, or:
I = I1 + I2
It should be noted that in this case, the values of I1 and I2
are given by:
I1 = K1 V1
I2 = K2 V2
Figure 14-02
In the Figure 14-02 we present the circuit with the short-circuited voltage source V2.
For this circuit we must develop an equation that relates the current I1 to the voltage source
V1. In other words,
we must compute K1. Note that with the short circuit, resistor R2
was paralleled by R3.
For the calculation of I1, we must calculate the equivalent resistance of the circuit.
Note that the parallel of R2 and R3 is in series with R1, then:
Req = (R1 R2 + R1 R3 + R2 R3) /
(R2 + R3)
Keep in mind that by dividing V1 by Req, we are calculating the current that flows through the source and by the R1 resistor, arriving at the b node.
Therefore, to calculate I1, we will apply a current divider to the b node.
Using the basic equation, already presented in the chapter 10 - law of Ohm, and after some algebraic
arrangements, we arrive at:
eq. 14-02
Note that K1, in the equation above, is this equation, but without the presence of
V1. To calculate K2, let's proceed in the same way.
Figure 14-03
In the Figure 14-03 we present the circuit with the short-circuited voltage source V1. For this circuit we must develop an equation that relates the current I2 to the voltage source V2. In other words,
we must compute K2. Note that with the short circuit, resistor R2 is now in parallel with R1.
In this case, the equivalent resistance that the circuit offers to V2
is given by:
Req = (R1 R2 + R1 R3 + R2 R3) /
(R2 + R1)
In the same way we solve for V1, let's solve for V2. Then, if we divide the value of V2 by Req, we find the value of the current supplied by the V2 and it will arrive at the b node. This current arriving at the b node, forks between R2 and R1. By applying a current divider to this node, we easily calculate the current I2 passing through R2.
See below, how the equation was after we made the algebraic arrangements.
eq. 14-03
Analyzing the two equations that determine I1 and I2, we notice that the
denominators are equal. Therefore, to find I, which is the algebraic sum of
I1 and I2, we simply add the numerators since the denominators are the same. Then, we see below, the equation that determines the value of I.
eq. 14-04
Note the fact that this equation is showing that the two parcels of the numerator represent the parcels of I1 and I2, respectively. We must not forget
that this equation was obtained with the circuit containing two voltage sources and both are with positive poles facing upwards.
What happens if one or both of the sources do not have positive poles facing up?
Well, in this case just change the polarity of the parcel, in the numerator of the equation, corresponding to the source
which is with the polarity reversed. So we have four cases to consider. Note that in all cases
the denominator will always be the same. To simplify the equation, let's call the denominator
of Rsp, or resistor sum-product, according to equation below:
Rsp = R1 R2 + R1 R3 + R2 R3
Case-by-Case Analysis
1º Case: - The polarity positiva de V1 and V2
are facing upwards. Then, the normal equation is used, as has already been deduced previously.
See the equation below.
eq. 14-05
2º Case: - The polarity of V1 is inverted (the positive pole
is facing down), but V2 does not.
In this case, we must put the NEGATIVE sign on the first portion of the numerator
(-V1 R3). See below, how the equation was.
eq. 14-06
3º Case: - The polarity of V2 is inverted (the positive pole points down),
but V1 does not. In this case, we must put the NEGATIVE signal in the second portion
of the numerator (- V2 R1). See below, how the equation was.
eq. 14-07
4º Case: - Both sources have inverted polarities (the positive poles
pointing down). In this case, we must put the NEGATIVE sign in front of the entire equation,
because surely the value of I will be NEGATIVE.
See below, how the equation was.
The most pertinent question right now is: why study this particular circuit in detail? The answer is simple: anyone who knows how to solve this circuit, is able to solve 80% of the problems that appear in textbooks about electrical circuits. Just do transformation of sources until we get to the circuit studied above. Do we need to "decorate" all these equations? In fact, not. Let's present a very practical way of solving the circuit without worrying about "memorizing" equations.
Practice Rule
Look at the Figure 14-04 where our circuit is reproduced. The first step is to multiply the resistors that make up the circuit, two by two, and sum the results by finding Rsp, or:
Rsp = R1 R2 + R1 R3 + R2 R3
Look, is enough looking at the circuit to calculate the value of Rsp .
Figure 14-04
Now that we have the Rsp value, which is the denominator of the equation, let's calculate the numerator. In the Figure 14-05 it is sufficient to cross-multiply V1 with R3 and V2
with R1, as indicated by the blue and red arrows, respectively. These two calculated values will be
positive,
because the two voltage sources are facing upwards. If not, refer to the Case-by-Case Analysis item above. Then, summing these values we obtain the numerator of the equation.
Now just divide the numerator by the denominator and we will have the value of I that circulate through the resistance
R2. With the value of I we can compute Vb = R2. I. And, of course, with the value of Vb, we easily calculate the values of the currents that circulate through the R1
and R3 resistors, applying the law of Ohm. Thus, we find the complete circuit solution.
In this example problem, we first have the "traditional" solution and in the end we present the solution by the
Basic Circuit Method, where the simplicity of the method is evident.
When we use the Superposition method in a circuit that contains one or more dependent sources, we must take into account that we can not eliminate the dependent source(s) of the circuit.
To exemplify how we use this method when the circuit contains dependent sources, let's refer to the example that appears on page 127 of Fundamentals of Electrical Circuits - Charles K. Alexander,
Mathew N. O. Sadiku - 5th Edition - Ed. McGraw Hill - 2013. Transcribed in the Figure 14-06 the circuit. Let's calculate the value of ix in the circuit.
Figura 14-06
In this circuit, we have two independent sources and one dependent source. Notice that the dependent source is tied to the value of ix, which is the electric current that flows through the 2 ohms resistor. To apply the Superposition theorem we must eliminate one of the independent sources.
Figure 14-07
In the Figure 14-07 we see the circuit in which we eliminate the current source of 3 A.
With this, we have an open circuit in place of it. In this way, we can easily calculate
what is the portion of ix for the voltage source of 10 volts.
This portion will call ixa.
Since we have a single mesh, we simply apply the law of Ohm and find the value of ixa.
- 10 + 3 ixa + 2 ixa = 0
Performing the calculation we find the value of ixa.
ixa = 2 A
Figura 14-08
Now let's compute the portion of ix for the current source of 3 A. To do so, we will eliminate the
voltage source, replacing it with a short-circuit, as shown in the figure on the right.
This plot will be called ixb. Note that by the resistor of
1 ohm will pass a current of 3 + ixb.
Therefore, by looping the short-circuit, by the resistors
1 and 2 ohms and by the dependent source, we have:
2 ixb + 1 (3+ ixb) + 2 ixb = 0
Performing the calculation we find the value of ixb.
ixb = - 0.6 A
Now we must add algebraically the values of ixa and ixb and find the value of
ix, or:
ix = ixa + ixb = 2 - 0.6 = 1.4 A
In order to solve the problem, it is important to give different names the
variables involved in each modification that is performed in the circuit,
because we are making it clear that each calculation is due to a situation
different from the previous one.
