# Superposition Theorem

The superposition theorem is a very important concept in the circuit theory. If a circuit has two or more independent sources, one way to determine the value of a specific variable (voltage or current) is to use nodal or mesh analysis. Another way is to determine the contribution of each independent source to the variable and then add them up. The latter approach is known as the superposition.
In general, the theorem can be used to do the following:
• Analyze networks that have two or more sources that are not in series or parallel.
• Reveal the effect of each source on a particular quantity of interest.
• For sources of different types (such as dc and ac, which affect the parameters of the network in a different manner) and apply a separate analysis for each type, with the total result simply the algebraic sum of the results.
The superposition theorem states the following:
The current through, or voltage across, any element of a network is equal to the algebraic sum of the currents or voltages produced independently by each source.
In other words, this theorem allows us to find a solution for a current or voltage using only one source at a time. Once we have the solution for each source, we can combine the results to obtain the total solution.
If we are to consider the effects of each source, the other sources obviously must be removed. Setting a voltage source to zero volts is like placing a short circuit across its terminals. Therefore,
when removing a voltage source from a network schematic, replace it with a direct connection (short circuit) of zero ohms. Any internal resistance associated with the source must remain in the network.
Setting a current source to zero amperes is like replacing it with an open circuit. Therefore,
when removing a current source from a network schematic, replace it by an open circuit of infinite ohms. Any internal resistance associated with the source must remain in the network.
The above statements are illustrated in Fig. 1.
Fig. 1: Removing a voltage source and a current source to permit the application of the superposition theorem.
Since the effect of each source will be determined independently, the number of networks to be analyzed will equal the number of sources. If a particular current of a network is to be determined, the contribution to that current must be determined for each source. When the effect of each source has been determined, those currents in the same direction are added, and those having the opposite direction are subtracted; the algebraic sum is being determined. The total result is the direction of the larger sum and the magnitude of the difference.
Similarly, if a particular voltage of a network is to be determined, the contribution to that voltage must be determined for each source. When the effect of each source has been determined, those voltages with the same polarity are added, and those with the opposite polarity are subtracted; the algebraic sum is being determined. The total result has the polarity of the larger sum and the magnitude of the difference.
Analyzing a circuit using superposition has one major disadvantage: it may very likely involve more work. If the circuit has three independent sources, we may have to analyze three simpler circuits each providing the contribution due to the respective individual source. However, superposition does help reduce a complex circuit to simpler circuits through replacement of voltage sources by short circuits and of current sources by open circuits.
Keep in mind that superposition is based on linearity. For this reason, it is not applicable to the effect on power due to each source, because the power absorbed by a resistor depends on the square of the voltage or current. If the power value is needed, the current through (or voltage across) the element must be calculated first using superposition.
Example 1:
a. Using the superposition theorem, determine the current through resistor $R_2$ for the network in Fig. 2.
b. Demonstrate that the superposition theorem is not applicable to power levels.
Fig. 2: For example 1.
Solution:
a. In order to determine the effect of the 36 V voltage source, the current source must be replaced by an open-circuit equivalent as shown in Fig. 3. The result is a simple series circuit with a current equal to $$\begin{split} I'_2 &={E \over R_T} \\ &= {E \over R1 + R2}\\ &= {36 V \over 12 Ω + 6 Ω} = {36 V \over 18 Ω} = 2 A \end{split}$$
Fig. 3: Replacing the 9 A current source in Fig. 2 by an open circuit.
Examining the effect of the 9 A current source requires replacing the 36 V voltage source by a short-circuit equivalent as shown in Fig. 4. The result is a parallel combination of resistors $R_1$ and $R_2$. Applying the current divider rule results in $$\begin{split} I''_2 &= {R_1(I) \over R_1 + R_2} \\ &= {(12 Ω)(9 A) \over 12 Ω + 6 Ω} = 6 A \end{split}$$
Fig. 4: Replacing the 36 V voltage source by a short-circuit.