At the end of this lesson, the students should be able to:
• State superposition principle
• Explain superposition principle
• Solve simple problems on superposition principle
2.11 SUPERPOSITION PRINCIPLE
The superposition principle is a method that allows us to determine the current through or the voltage across any resistor or branch in a network.
The advantage of using this approach instead of Kirchhoff’s laws is that it is not necessary to use determinant or matrix algebra to analyze a given circuit.
The theorem state the following:
In a circuit containing several sources of e.m.f., the resultant current in any branch is the algebraic sum of the currents in that branch that would be produced by each e.m.f, acting alone, all other sources of e.m.f being replaced meanwhile by their respective interval resistances.
In order to apply the superposition principle it is necessary to remove all sources other than the one being examined. In order to ‘’Zero’’ a voltage source, we replace it with a short circuits, since the voltage across a short circuit is zero volts.
A current source is zeroed by replacing it with an open circuit, since the current through an open circuit is zero amps.
Example 8.1
Solve the same problem in example 3.5 to determine all the branch currents using the superposition principles.
Solution
To solve this problems we shall follow the steps.
Step I, Re-draw the given circuit diagram with E1 = 15V and E2 = OV, as shown in Fig 9.1 below
Fig. 9.1
Step II: Next, determine all the branch currents I1’, I2’ and I3’ indicated in Fig 9.1 (b),
With E1 = 15V and E2 = OV.
Step III: we find the equivalent resistance of 3Ω resistor and 2Ω resistor connected in parallel.
i.e. R3Ω//R2Ω = 3 X 2 = 6 = 1.2Ω
3 + 2 5
Step IV: The circuit diagram of Fig 9.1(b) reduces to the form shown in Fig 9.1
Step V: we can apply the Kirchhoff’s Voltage law to the circuit diagram of Fig 9.2 to obtain,
Step VI: Turn back to fig 9.1(b) and bear in mind that I1 is the total current that flows into the 3Ω resistor branch and 2Ω resistor branch which are in parallel. Applying the principle of current division rule, we get
Step VII: Draw the corresponding circuit diagram if the original circuit diagram of Fig 9.1(a) has E1 = OV and E2 = 10V, to produce the circuit diagram shown in Fig. 9.3(b)
Fig. 9.3
(i) R4Ω//R3Ω = 4 X 3 = 12 Ω = 1.714Ω
4 + 3 7
(ii) The resulting equivalent circuit diagram after combing R4Ω and R3Ω in parallel gives Fig 9.4.
(iii) Applying Ohm’s law to the circuit diagram of Fig 9.4, we get
I”3 = E2 = 10 ≅ 2.693A
2 + R4sΩ // R3Ω (2 + 1.714)
(iv) with reference to Fig 9.3, we see that if I”3 is to serve as the total current then according to KCL then I”1 and I”2 must be seen to flow into node A. For
(v) this to happen the direction I”2 must be opposite to what is indicated in the diagram of Fig 9.3. This means that I”2 must bear a negative value. Only I”1 flows into node A.
Therefore, I”1 = 3 X 2.693 = 1.154A
(4 + 3)
Similarly, I”2 = _ 4 X 2.693 = - 1.539A
(4 + 3)
Step VIII: The total branch currents can be obtained by adding up algebraic as follows:
I1 = I”1 + I’1 = 2.885 + 1.154 = 4.039A
I2 = I’2 + I”2 = 1.154A – 1.538A = - 0.384A
I3 = I’3 + I”3 = 1.731A + 2.693A = 4.424A
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