ELE 205
Lecture # 6 – Superposition Theorem
(Floyd – page 288)
The superposition theorem is a way to determine currents and
voltages in a linear circuit that has multiple sources by taking
one source at a time and algebraically summing the results.
Linear circuit
- A circuit in which relationships are proportional.
- In a linear circuit, current is proportional to voltage.
Procedure (Steps):
1.
Select one source and replace all other sources with their
internal resistances.
- Voltage source = 0 Ω (short circuit)
- Current source = ∞ Ω (open circuit)
2.
Determine the level and direction of the current flowing
through the desired branch as a result of the single
source acting alone.
3.
Repeat steps 1 and 2 using each source in turn until all
the sources have been used.
4.
Algebraically sum the component currents to obtain the
actual branch current.
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Disadvantages of using the Superposition Theorem.
- If we have N sources, we have to solve N individual circuits
(i.e. with each of the sources acting alone).
- There is a chance for making an algebraic mistake if we do
not keep the same current reference direction each time.
Example 1 (Floyd – page 290 -295)
• Using the superposition theory, determine the current
through the resistor R2 for the network shown below:
• To do this we must determine the contributions to this
current first from the voltage source alone and then from the
current source alone and finally sum the two.
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1.
To determine the contribution of the voltage source
alone we replace the current source with its open-circuit
equivalent:
The result is a simple series circuit with a current given by:
2.
To determine the contribution of the current source alone we
replace the voltage source with its short-circuit equivalent:
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In this case R1 and R2 are in parallel and the current through
R2 is given by (using proportionality – current divider rule):
Now we have to take careful account of the directions of the
two contributing currents, in this case they are in opposite
directions.
3.
By convention we take IR2V to be in the positive direction and
IR2I to be in the negative direction, the total solution for the
current IR2 is then simply the sum of these currents:
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Example 2
Find I2 using superposition
Current due to V S alone
VS
3V
I' 2 =
=
= 0 . 25 A
R1 + R 2 8Ω + 4 Ω
Current due to I S alone
R2
4Ω
I' ' 2 =
IS =
2 A = 0 . 66 A
R1 + R 2
8Ω + 4Ω
I 2 = I' 2 + I' ' 2 = 0 . 25 A + 0 . 66 A = 0 . 91 A
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