SUPERPOSITION THEOREM

IMPORTANCE OF SUPERPOSITION THEOREMPROPERTIES OF SUPERPOSITION THEOREMLIMITATIONS OF SUPERPOSITION THEOREMRELATIONSHIP WITH MESH AND NODAL ANALYSISPROPORTIONALITY IN ELEMENTSADDITIVITY PROPERTYAPPLICATION OF SUPERPOSITION THEOREMAN EXAMPLE FOR USING SUPERPOSITION THEOREMANOTHER WORKED EXAMPLESUMMARY

IMPORTANCE OF SUPERPOSITION THEOREM

Network theorems provide insight into the behaviour and properties of electricalcircuits. Superposition theorem is of theoretical importance, because it isfundamental to linear circuit analysis. A circuit is linear only when it behaves inaccordance with superposition theorem. This theorem states that the linearresponses in a circuit can be obtained as the algebraic sum of responses, due toeach of the independent sources acting alone. This theorem defines the behaviourof a linear circuit. Within the context of linear circuit analysis, this theoremprovides the basis for all other theorems. Given a linear circuit, it is easy to seehow mesh analysis and nodal analysis make use of the principle of superposition.

PROPERTIES OF SUPERPOSITION THEOREM

There are two guiding properties of superposition theorem. The first is theproperty of homogeneity or proportionality, and the second is the property ofadditivity.

LIMITATIONS OF SUPERPOSITION THEOREM

As stated earlier, the linear responses in a circuit can be obtained using thistheorem as the algebraic sum of responses, due to each of the independent sourcesacting alone. Current and voltage associated with an element are linear responses.On the other hand, power in an element is not a linear response. It is a non-linearfunction, varying proportionately either with the square of voltage across theelement or with the square of current through the element. Hence it is not possibleto apply superposition theorem directly to determine power associated with anelement. In addition, application of superposition theorem does not normally lead

to simplification of analysis. It is not the best technique to determine all currentsand voltages in a circuit, driven by multiple of sources.

RELATIONSHIP WITH MESH AND NODAL ANALYSIS

Superposition theorem is valid for linear circuits and analysis of linear circuits isrelatively easy. On the other hand, the principle of superposition is not valid for non-linear circuits. And the analysis of non-linear circuits is quite complexcomplex and difficult. It is possible to apply mesh and nodal analysis to non-linear circuits. However, within the context of linear circuits, mesh or nodalanalysis of a circuit illustrates how the principle of superposition is ever sopervasive in defining the behaviour of linear circuits.

PROPORTIONALITY IN ELEMENTS

A linear circuit consists of linear elements. The passive elements, the dependentsources and the independent sources used in a linear circuit are linear. Let us howlinearity is defined for each type of element.

In the case of a resistor, the voltage across a resistor varies proportionally with itscurrent. The ratio of voltage to current is resistance. Power in a resistor varieswith the square of its voltage or its current.

In the case of an inductor, the linearity is between the flux linkage and the current.The ratio of flux linkage to current is inductance. The product of inductance andthe rate of change of current is the inductor voltage. In an ideal inductor, the core

of inductor does not get saturated. In an inductor with a magnetic core, saturationof flux occurs at some level, depending on the material used for the core.

In the case of a capacitor, the linearity is between the charge stored and thecapacitor voltage. The ratio of charge to voltage is capacitance. The product ofcapacitance and the rate of change of voltage is the capacitor current. In an idealcapacitor, the capacitor voltage can be very high. In a real capacitor, the dielectricwithin the capacitor breaks down a magnetic core at some level, depending on thedielectric.

In the case of dependent source, the output variable varies proportionately with thecontrolling variable. In the case of ideal independent source, linearity implies thata voltage can supply any current at constant voltage, and that a current source cansustain any load voltage while supplying constant current.

ADDITIVITY PROPERTY

The principle of additivity can be explained with the help of a sketch. When thereare two sources in a circuit, the contribution due to each source can be found outseparately and the response due to both sources is the algebraic sum ofcontributions due to each source acting alone. This aspect is illustrated by thecircuit shown above. The current through the resistor is the sum of currents due

1 2 1to source V and source V , acting alone. When the contribution due to source V

2is to be calculated, the contribution due to source V should be zero. Hence source

2 2V can be replaced by a short circuit. When source V is replaced by a shortcircuit, its contribution has to be nil. Similarly, when the contribution due to

2 1source V is to be calculated, the contribution due to source V should be zero.

1Hence source V can be replaced by a short circuit. The relevant circuits areshown above.

APPLICATION OF SUPERPOSITION THEOREM

If the direct application of superposition theorem is not easy, the question ariseswhen it is suitable to use superposition theorem. It is best to use superpositiontheorem to find a particular current or voltage in a circuit, when the circuit hasmultiple independent sources. This theorem states that the linear responses in acircuit can be obtained as the algebraic sum of responses, due to each of theindependent sources acting alone. A voltage source that makes no contribution isreplaced by a short-circuit, whereas a current source that makes no contributionis replaced by an open-circuit. The internal resistance of the source is left in thecircuit, as it is where it is. The worked examples are used to show how to applysuperposition theorem to circuits.

AN EXAMPLE FOR USING SUPERPOSITION THEOREM

A simple circuit is used to illustrate how the principle of superposition can be usedto obtain the current through the resistor in the circuit shown in Fig. 1.

When superposition theorem is used, the response due to only one independentsource is obtained at a time. The other sources are replaced, by either open-circuits or short-circuits, as the case may be. In this circuit, there are two sources,

1 2 1voltage V and voltage V , . When response due to source V is calculated,

2 1source V is replaced by a short-circuit. Let the current through the resistor be I ,

2 1as shown in Figure 2. When response due to source V is calculated, source V

2is replaced by a short-circuit. Let the current through the resistor be I .

Without using superposition theorem, it can be stated that the voltage across theresistor is as shown by equation (1). Then we can express the current through theresistor by equation (2), and it can be split into two fractions, as shown byequation (3). We see next how we can apply superposition theorem.

1 2The expressions for currents I and I are marked in Figure 2. Using superposition

1theorem, the current through the resistor is obtained as the sum of currents I and

2I . Equation (4) is the same as the previous equation obtained for current throughthe resistor.

It is seen that the total current in the resistor is expressed as the sum of tworesponses, due to each source acting alone. That is, when the response due to

1 12voltage source V is to be calculated, the contribution of voltage source V

2should be zero and then source V is replaced by a short circuit. A voltage sourcewith zero volts allows current to pass through it, and does not contribute anythingto a circuit. That is, when contribution of an ideal voltage source is to be zero, it

2can be replaced by a short circuit. Similarly, when the response due to source V

1is to be calculated, the contribution of source V should be zero and it is replacedby a short circuit.

1 2Equation (4) is a linear equation. In this equation, V and V are the independent

1 variables. For example, if value of V changes from 6 Volts to 12 Volts, itscontribution to current through resistor doubles. Hence the response due to anindependent variable varies proportionately.

That is, the ratio of independent variable to the response is a constant. In this case,the Ohms Law applies, as shown by equation (5).

Equation (5) expresses the property of proportionality of a linear relationship. Theother property of a linear relationship is the additive property, expressed byequation (4). Here the total response is expressed as the algebraic sum ofresponses, due to each independent source acting alone. Equation (4) is formedbased on the principle of superposition.

The current through resistor R in the circuit shown in Fig. 1 can be determinedusing the superposition theorem, and the power absorbed by this resistor isdetermined in the end, after finding the resultant current in the resistor due to boththe sources.

The values of components present in the circuit in Fig. 1 are presented by equation(6).

Current through the resistor is obtained, as shown by equation (7).

Once the current through the resistor is known, the power absorbed by it can befound out, as shown by equation (8). Next, it is shown why superposition theoremcannot be applied directly to determine the power absorbed by the resistor Rpresent in the circuit in Fig. 1.

Based on the circuits presented in Fig.2, the power absorbed by the resistor, dueeach of the sources acting alone, can be obtained. Equation (9) states the power

1 2absorbed by the resistor when V = 12 Volts, and V = 0 V.

1Equation (10) states the power absorbed by the resistor when V = 0 Volts, and

2V = 6 V.

The power absorbed by the resistor due to both sources is presented by equation

1 2(8), which is 18 Watts. On the other hand, the sum of P and P is

190 Watts. In this circuit, the power associated with source V is - 36 Watts, sinceit is delivering power. The resistor absorbs 18 Watts, and the remaining power is

2.absorbed source V ANOTHER WORKED EXAMPLE

2A circuit has been presented in Fig. 3. Determine current I . The values ofcomponents have been specified.

Solution:

It is possible to apply the superposition theorem directly. But in this case, themesh equations are formed, and then the components in the solution can be splitinto two parts, to show how the principle of superposition is in operation, even ifit is not applied directly.

Given the circuit in Fig. 3, the mesh equation that can be formed is of the typeshown by equation (12). The values of the elements in the loop resistance matrixcan be obtained by inspection of the circuit.

11 22 33The elements a , a , and a along the main diagonal reflect the self-resistanceof the loops, whereas the other elemenst are the mutual resistances, common totwo loops. Note that all the loops are marked in the clockwise direction, and thecurrent through any element is either the loop current or the difference of two loopcurrents. In this context, the mutual resistances have negative values. It can alsobe seen that the loop-resistance matrix is symmetric.

Let the determinant if the loop resistance matrix be calculated, as shown by

2equation (14). Then we can obtain the value of current I using Cramer’s rule, as

2shown by equation (15). The value of current I can be expressed in terms of

1 2cofactors of V and V , as shown by equation (16).

2 1 2Contribution to current I due to V and V can be calculated separately, as shown

2above, and the value of I is obtained to be 2 A. It can be seen that the use of meshanalysis reflects the application of superposition theorem.

SUMMARY

The analysis of linear circuits is founded on the superposition theorem. Even

though the direct use of superposition theorem is not always easy, it remains theguiding principle for the behaviour of linear circuits. Next we take up Thevenin’stheorem.