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:
**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: **
**Fig. 3: **Replacing the 9 A current source in Fig. 2 by an
open circuit.
**Fig. 4: **Replacing the 36 V voltage source by a short-circuit.

- 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.

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.

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
$$\begin{split}
I'_2 &={E \over R_T} \\
&= {E \over R1 + R2}\\
&= {36 V \over 12 Ω + 6 Ω} = {36 V \over 18 Ω} = 2 A
\end{split}$$

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
$$\begin{split}
I''_2 &= {R_1(I) \over R_1 + R_2} \\
&= {(12 Ω)(9 A) \over 12 Ω + 6 Ω} = 6 A
\end{split}$$