What is a Superposition Theorem : Limitations & Its Applications
- Jun 02, 2022
For every electrical circuit, there are two or additional independent supplies like the current, voltage, or both sources. For examining these electrical circuits, the superposition theorem is widely utilized and mostly for time-domain circuits at various frequencies. For instance, a linear DC circuit consists of one or more independent supply; we can get the supplies like voltage and current by using methods like mesh analysis and nodal analysis techniques. Otherwise, we can employ the “superposition theorem” that includes every individual supply result on the worth of the variable to be decided. This means the theorem assumes that every supply in a circuit independently discovers the rate of the variable, and lastly produces the secondary variable by inserting the variables which are reasoned by the effect of every source. Even though the process of it is very difficult but still can be applied for every linear circuit.
What is a Superposition Theorem?
The superposition theorem is a method for the Independent supplies present in an electrical circuit like voltage & current and that is considered as one supply at a time. This theorem tells that in a linear n/w comprising one or more sources, the flow of current through a number of supplies in a circuit is the algebraic calculation of the currents when acting the sources like independently.
The application of this theorem involves simply linear n/ws, and also in both the AC & DC circuits where it assists to build the circuits like “Norton” as well as “Thevenin” equivalent circuits.
For instance, the circuit which has two or more supplies then the circuit will be separated into a number of circuits based on the statement of the superposition theorem. Here, the separated circuits can make the entire circuit seem very simple in easier methods. And, by merging the separated circuits another time after individual circuit modification, one can simply discover factors like node voltages, voltage-drop at every resistance, currents, etc.
Step-by-Step Methods of Superposition Theorem Statement
The following step-by-step methods are used to discover the response of a circuit in a specific division by superposition theorem.
- Calculate the response in a specific branch of a circuit by allowing for one independent supply as well as removing the residual independent supplies the current in the network.
- Do again the above step for all voltage and current sources there in the circuit.
- Include all the reactions in order to obtain the total response in a specific circuit when all the supplies are there in the network.
What are the Conditions for Applying Superposition Theorem?
The following conditions must be met to apply this theorem to a network
- The circuit components must be linear. For instance, the flow of current is proportional to the voltage for resistors which is applied to the circuit; the flux linkage can be proportional to current for inductors.
- The circuit components must be bilateral which means that the flow of current in the opposite polarities of the voltage source must be the same.
- The components used in this network are passive because they do not amplify otherwise rectify. These components are resistors, inductors & capacitors.
- The active components should not be used because they never seldom linear as well as never bilateral. These components mainly include transistors, electron tubes, and semiconductor diodes.
Superposition Theorem Examples
The basic circuit diagram of the superposition theorem is shown below, and it is the best example of this theorem. By using this circuit, calculate the flow of current through the resistor R for the following circuit.
Disable the secondary voltage source i.e, V2, and calculating the flow of current I1 in the following circuit.
We know that ohms law V= IR
Disable the primary voltage source i.e, V1, and calculating the flow of current I2 in the following circuit.
According to the superposition theorem, the network current I = I1 + I2
I = V1/R-V2/R
How to Use Superposition Theorem?
The following steps will tell you how to apply a superposition theorem to solve a problem.
- Take one source in the circuit
- Remaining independent sources must be set to zero by replacing voltage sources through short circuit whereas current sources with open circuit
- Leave the independent sources
- Calculate the flow of current direction as well as magnitude throughout the required branch as an outcome of the single source preferred in the first step.
- For every source, repeat the steps from the first step to the fourth until the required branch current has been measured because of the source acting alone.
- For the required branch, add all the component current using directions. For the AC circuit, the phasor sum needs to be done.
- The same steps need to follow to measure the voltage across any element in the circuit.
Superposition Theorem Problems
The following circuit shows the basic DC circuit for solving the superposition theorem problem such that we can get the voltage across the load terminals. In the following circuit, there are two independent supplies namely current and voltage.
Initially, in the above circuit, we keep only voltage supply is acting, and the remaining supply like the current is changed with inside resistance. So the above circuit will become an open circuit as shown in the below figure.
Consider the voltage across the load terminals VL1 with voltage supply performing alone, then
VL1 =Vs (R3/(R3 + R1))
Here, Vs= 15, R3= 10 and R2-= 15
Please substitute the above values in the above equation
VL1 = Vs × R3 / (R3 + R2)
= 15 (10 / (10 + 15))
= 6 Volts
Hold the current supply only and change the voltage supply with its inside resistance. So the circuit will become a short circuit as shown in the following figure.
Consider the voltage across the load terminals is ‘VL2’ while only current supply performing. Then
VL2= I x R
IL = 1 x R1/(R1+R2)
R1 = 15 RL= 25
= 1 × 15 / (15 +25) = 0.375 Amps
VL2 = 0.375 × 10 = 3.75 Volts
As a result, we know that the superposition theorem states that the voltage across the load is the amount of VL1 & VL2
VL = VL1 + VL2
6 + 3.75 = 9.75 Volts
Prerequisites of the Superposition Theorem
The superposition theorem simply applicable to the circuits which are reducible toward the combinations of series or parallel for every power source at a time. So this is not applicable for examining an unbalanced bridge circuit. It simply works wherever the fundamental equations are linear.
The linearity requirement is nothing but, it is only appropriate to determine voltage & current. This theorem is not used for the circuits where the resistance of any component varies through the current otherwise voltage.
Therefore, the circuits including components such as gas-discharge or incandescent lamps otherwise varistors could not be evaluated. Another requirement of this theorem is that the components which are used in the circuit should be bilateral.
This theorem uses in the study of AC (alternating current) circuits as well as semiconductor circuits, where alternating current is frequently mixed through DC. As the AC voltage, as well as current equations, is linear similar to direct current. So this theorem is used to examine the circuit with a DC power source, after that with an AC power source. Both the results will be combined to tell what will happen with both the sources in effect.
Superposition Theorem Experiment
The experiment of the superposition theorem can be done like the following. The step by step of this experiment is discussed below.
Verify the superposition theorem experimentally using the following circuit. This is an analytical method used to determine currents within a circuit using more than one source of supply.
The apparatus of this circuit are a breadboard, connecting wires, milli-ammeter, resistors, etc.
Theory of the Experiment
The superposition theorem is simply used when the circuit includes two or more sources. This theorem is mainly used to shorten the calculations of the circuit. This theorem states that, in a bilateral circuit, if a number of energy sources are used like two or above, then the flow of current will be there at any point and it is the sum of all currents.
The flow will be at the point where every source was separately considered & other sources will be changed at the time through impedance which is equivalent to their internal impedances.
The step by step procedure of this experiment is discussed below.
- Connect DC power supply across terminals of 1 & I1 & the voltage applied is V1= 8V and likewise, apply across terminals where the voltage supply V2 is 10 volts
- Measure the flow of current throughout all branches and they are I1, I2 & I3.
- First, connect the voltage source V1 = 8V across the terminals of 1 to I1 & short circuit terminals across 2 to I2 is V2 = 0V.
- Calculate flow of currents in all branches for V1 = 8V and V2=10V through a milli-ammeter. These currents are denoted with I1’, I2’& I3’.
- Likewise connect the only V2 =10 volts across 2 to I2 terminals as well as short circuit terminals 1 & I1, V1=0. Calculate flow of current throughout all branches for the two voltages with the help of a milliammeter and these are denoted with I1”, I2” & I3”.
To verify the superposition theorem,
I1= I1’+ I1”
I2= I2’+ I2’
Measure the theoretical currents values and these must be equivalent to the values which are measured for currents.
The values of I1, I2, I3 when V1= 8V & V2 =10V, the values of I1’, I2’ & I3’ when V1= 8V and V2=0 and for the values, I1’’, I2’’ & I3’’ when V1=0 & V2=10V.
In the above experiment, the branch current is nothing but the algebraic sum of currents because of the separate voltage source once the remaining voltage sources are short-circuited; thus this theorem has been proved.
The limitations of the superposition theorem include the following.
- This theorem is not applicable for measuring power but it measures voltage and current
- It is used in linear circuits but not used in nonlinear
- This theorem is applied when the circuit must have above one source
- For unbalanced bridge circuits, it is not applicable
- This theorem is not used for power calculations because the working of this theorem can be done based on the linearity. Because the power equation is the product of current & voltage otherwise square of the voltage or current but not linear. Therefore the power utilized through the element within a circuit using this theorem is not achievable.
- If the load option is changeable otherwise the load resistance varies regularly, then it is required to achieve every source contribution for voltage or current & their sum for each transform within load resistance. So this is a very difficult process for analyzing difficult circuits.
- The superposition theorem cannot be useful for power calculations but this theorem works on the principle of linearity. As the power equation is not linear. As a result, the power used by the factor in a circuit with this theorem is not achievable.
- If the load selection is changeable, then it is necessary to achieve each supply donation and their calculation for each transform in load resistance. So this is a very difficult method to analyze compound circuits.
The application of the superposition theorem is, we can employ only linear circuits as well as the circuit which has more supplies.
From the above superposition theorem examples, this theorem cannot be used for non-linear circuits, but applicable for linear circuits. The circuit can be examined with a single power source at a time, the
Equivalent section currents and voltages algebraically included discovering what they will perform with every power supply in effect. To cancel out all except one power supply for study, substitute any power source with a cable; restore any current supply with the break.
Thus, this is all about an overview of the superposition theorem which states that by using this theorem, at a time we can analyze the circuit using one power source only, the related component currents, as well as voltages, can be added algebraically to observe what they will achieve using all power sources effectively. To cancel out all, but one source of power for analysis, then change any voltage source with a wire and change any current source through an open (break). Here is a question for you, what is KVL?