Superposition Theorem
• The superposition theorem extends the use of Ohm’s law to
circuits with multiple sources.
• Superposition theorem is very helpful in determining the voltage
across an element or current through a branch when the circuit
contains multiple number of voltage or current sources
• Superposition is a general principle that allows us to determine
the effect of several energy sources (voltage and current sources)
acting simultaneously in a circuit by considering the effect of
each source acting alone, and then combining (superposing)
these effects.
Statement:
• In a linear, bilateral d.c. network containing more than one energy
source, the resultant potential difference across or current through
any element is equal to the algebraic sum of potential differences or
currents for that element produced by each source acting alone with
all other independent ideal voltage sources replaced by short circuits
and all other independent ideal current sources replaced by open
circuits (non-ideal sources are replaced by their internal resistances).
• In order to apply the superposition theorem to a network, certain
conditions must be met:
1. All the components must be linear. (Ex: the current is proportional to the
applied to the applied voltage (for resistor), flux linkage is proportional to
current (in inductor)). Etc
2. All the components must be bilateral (the current is the same amount
for opposite polarities of the source voltage)
3. Passive components may be used. (Ex: Resistors, inductors and
capacitors that do not amplify or rectify).
4. Active components may not be used. (Active components includes
transistors, semiconductor diodes and electron tubes. Such components
are never bilateral and seldom linear)
Procedure for applying Superposition Theorem
• Step 1 − Find the response in a particular branch by considering one independent
source and eliminating the remaining independent sources present in the
network.
• If Circuits containing only Ideal Independent Sources.
• Consider only one source to be active at a time and Remove all other Ideal Voltage Sources
in the circuit by Short Circuit & all other Current Sources by Open Circuit.
• If the circuit contains Practical Sources, Practical source are replaced with their
internal resistance.
Step 2 − Repeat Step 1 for all independent sources present in the network.
Step 3 − Add all the responses in order to get the overall response in a particular
branch when all independent sources are present in the network.
Example:
Find the current flowing through 20 Ω resistor of the following circuit using superposition
theorem.
• Step 1 − Let us find the current flowing through 20 Ω resistor by considering
only 20 V voltage source. In this case, we can eliminate the 4 A current
source by making open circuit of it. The modified circuit diagram is shown in
the following figure.
Therefore, the current flowing through 20 Ω resistor is 0.4 A, when only 20 V voltage
source is considered.
• Step 2 − Let us find the current flowing through 20 Ω resistor by
considering only 4 A current source. In this case, we can eliminate the 20
V voltage source by making short-circuit of it. The modified circuit
diagram is shown in the following figure.
Therefore, the current flowing through 20 Ω resistor is 1.6 A, when only 4
A current source is considered
• Step 3 − We will get the current flowing through 20 Ω resistor of the
given circuit by doing the addition of two currents that we got in step
1 and step 2. Mathematically, it can be written as
• I=I1+I2
• Substitute, the values of I1 and I2 in the above equation.
• I=0.4+1.6=2A
• Therefore, the current flowing through 20 Ω resistor of given circuit is 2 A
