Objectives
Introduction
If the circuit has more than one independent (voltage and/or current) sources, one way to determine the value of variable (voltage across the resistance or current through a resistance) is to use nodal or mesh current methods as discussed in detailed in lessons 4 and 5. Alternative method for any linear network, to determine the effect of each independent source (whether voltage or current) to the value of variable (voltage across the resistance or current through a resistance) and then the total effects simple added. This approach is known as the superposition. In lesson-3, it has been discussed the properties of a linear circuit that satisfy (i) homogeneity property [response of output due to input = αu(t) equals to α times the response of output due to input = u(t) , S (αu(t)) = αS (u(t)) for all α ; and = input to the system] (ii) additive property [that is the response of u1 (t) + u2 (t) equals the sum of the response of u1(t) and the response of u2(t) ,S(u1(t) + u2(t)) = S (u1(t)) + S (u2 (t))]. Both additive and multiplicative properties of a linear circuit help us to analysis a complicated network. The principle of superposition can be stated based on these two properties of linear circuits.
Statement of superposition theorem
In any linear bilateral network containing two or more independent sources (voltage or current sources or combination of voltage and current sources ), the resultant current / voltage in any branch is the algebraic sum of currents / voltages caused by each independent sources acting along, with all other independent sources being replaced meanwhile by their respective internal resistances.
Superposition theorem can be explained through a simple resistive network as shown in fig.7.1 and it has two independent practical voltage sources and one practical current source.
One may consider the resistances R1 and R3 are the internal resistances of the voltage sources whereas the resistance R4 is considered as internal resistance of the current source. The problem is to determine the response I in the in the resistor R2 . The current I can be obtained from
according to the application of the superposition theorem. It may be noted that each independent source is considered at a time while all other sources are turned off or killed. To kill a voltage source means the voltage source is replaced by its internal resistance (i.e. or R1 R3 ; in other words E1 or E2 should be replaced temporarily by a short circuit) whereas to kill a current source means to replace the current source by its internal resistance (i.e. ; in other words Is should be replaced temporarily by an open circuit).
Remarks: Superposition theorem is most often used when it is necessary to determine the individual contribution of each source to a particular response.
Procedure for using the superposition theorem
Step-1: Retain one source at a time in the circuit and replace all other sources with their internal resistances.
Step-2: Determine the output (current or voltage) due to the single source acting alone using the techniques discussed in lessons 3 and 4.
Step-3: Repeat steps 1 and 2 for each of the other independent sources.
Step-4: Find the total contribution by adding algebraically all the contributions due to the independent sources.
Application of superposition theorem
Example- Consider the network shown in fig. 7.2(a). Calculate Iab and Vcg using superposition theorem.
Solution: Voltage Source Only (retain one source at a time):
First consider the voltage source Va that acts only in the circuit and the current source is replaced by its internal resistance (in this case internal resistance is infinite (∞)). The corresponding circuit diagram is shown in fig.7.2(b) and calculate the current flowing through the ‘a-b’ branch.
Now current through a to b, is given by
Voltage across c-g terminal :
Vcg = Vbg+ Vcb = 2 × 1.043 + 4 × 0.13 = 2.61volts (Note: we are moving opposite to the direction of current flow and this indicates there is rise in potential). Note ‘c’ is higher potential than ‘g’
Current source only (retain one source at a time):
Now consider the current source Is = 2 A only and the voltage source Va is replaced by its internal resistance which is zero in the present case. The corresponding the simplified circuit diagram is shown below (see fig.7.2(c)& fig.7.2(d)).
Current in the following branches:
4 Ω resistor = 2− 1.217 = 0.783 A
Voltage across 3Ω resistor (c & g terminals) Vcg = 1.217×3 = 3.651 volts
The total current flowing through 1Ω resistor (due to the both sources) from a to b = 0.913 (due to voltage source only; current flowing from ‘a’ to ‘b’) – 0.522 (due to current source only; current flowing from ‘b ’to ‘a’) = 0.391A.
Total voltage across the current source Vcg = 2.61volt (due to voltage source ; ‘c’ is higher potential than ‘g’) + 3.651 volt (due to current source only; ‘c’) is higher potential than ‘g’) = 6.26 volt.
Example For the circuit shown in fig.7.3(a), the value of Vs and Is are fixed. When , the current Vs2 = 0 I = 4 A . Find the value of I when Vs2 = 32V.
Solution: Let us assume that the current flowing 6Ω resistors due to the voltage and current sources are given by (assume circuit linearity)
where the parametersα , β, and η represent the positive constant numbers. The parameters α and β are the total conductance of the circuit when each voltage source acting alone in the circuit and the remaining sources are replaced by their internal resistances. On the other hand, the parameter η represents the total resistance of the circuit when the current source acting alone in the circuit and the remaining voltage sources are replaced by their internal resistances. The expression (7.1) for current I is basically written from the concept of superposition theorem. From the first condition of the problem statement one can write an expression as (when the voltage source Vs1 and the current source Is acting jointly in the circuit and the other voltage source Vs2 is not present in the circuit.)
(Note both the sources are fixed)
Let us assume the current following through the 6Ω resistor when all the sources acting in the circuit with Vs = 32V is given by the expression (7.1). Now, one can determine the current following through 6Ω resistor when the voltage source Vs2 = 32V acting alone in the circuit and the other sources are replaced by their internal resistances. For the circuit shown in fig.7.3 (b), the current delivered by the voltage source to the resistor is given by
The current following through the 6 Ω due to the voltage source Vs2 = 32V only is 2A (flowing from left to right; i,e. in the direction as indicated in the figure 7.3(b)). Using equation (7.1), the total current I flowing the 6 Ω resistor can be obtained as
Example : Calculate the current Iab flowing through the resistor3 as shown in fig.7.4(a), using the superposition theorem.
Solution: Assume that the current source 3A (left to the 1 source) is acting alone in the circuit and the internal resistances replace the other sources. The current flowing through 3Ω resistor can be obtained from fig.7.4(b)
and it is given by
I1(due to 3 A current source)
Current flowing through 3Ω resistor due to 2V source (only) can be obtained from fig.7.4(c)
and it is seen from no current is flowing.
I2(due to 2V current source) = 0 A(a to b)
Current through 3Ω resistor due to 1V voltage source only (see fig.7.3(d)) is given by
I3(due to 1V current source)
Current through 3Ω resistor due to 3 A current source only (see fig.7.3(e)) is obtained by
I4(due to 3A current source)
Current through 3Ω resistor due to 2V voltage source only (see fig.7.3(f)) is given by
I5(due to 2V current source)
Resultant current Iab flowing through 3Ω resistor due to the combination of all sources is obtained by the following expression (the algebraic sum of all currents obtained in eqs. (7.4)-(7.8) with proper direction of currents)
1ab = I1(due to 3 A current source) + I2(due to 2 V voltage source) + I3(due to 1V voltage source) + I4(due to 3 A current source)+I5(due to 2 V voltage source)
Limitations of superposition Theorem
Example: Consider the circuit diagram as shown in fig.7.5.
Using superposition theorem one can find the resultant current flowing through 12Ω resistor is zero and consequently power consumed by the resistor is also zero. For power consumed in an any resistive element of a network can not be computed using superposition theorem. Note that the power consumed by each individual source is given by
PW1(due to 12Vsource(left)) =12 WattS , Pw2(due to l2y source(right)) =12 WattS
The total power consumed by 12Ω = 24 watts (applying superposition theorem). This result is wrong conceptually. In fact, we may use the superposition theorem to find a current in any branch or a voltage across any branch, from which power is then can be calculated.
57 docs|62 tests
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1. What is the superposition theorem in the context of DC voltage and current sources acting in a resistive network? |
2. How does the superposition theorem apply to a resistive network with DC voltage and current sources? |
3. Can the superposition theorem be used for non-linear circuits? |
4. Are there any limitations to applying the superposition theorem? |
5. How can the superposition theorem be practically applied in circuit analysis? |
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