## Superposition Theorem with solved problems

Superposition theorem is used to solve the complex electric network, which consists of two or more sources and several resistances, by considering and analyzing all the sources individually.

Superposition theorem states that,

“In any linear, bilateral network having more than one source, the response across any element is the sum of the responses obtained from each source considered separately and all other sources are replaced by their internal resistance.”

It can also be stated as,

If a linear circuit consists of more than one independent source, then the current flowing through any part of the circuit is equal to the algebraic sum of the currents produced by each independent source, when it is considered separately.

## Illustration of Superposition theorem

Let us understand this theorem with the following circuit, which consists of three independent sources(I 1 , V 1 , V 2 ), several resistances(R 1 , R 2 , R 3 ) and a load resistor R L .

Now, to determine the current flowing through the load resistor R L , the circuit is analyzed by considering each source independently. While considering a single source, all other voltage sources in the circuit must be short-circuited and the current source, if any should be open-circuited.

By considering the current source I 1 alone acting in the circuit, as shown below, the current I L1 through the load is determined.

Now, this current source is open-circuited and the second source(Here, it is voltage source V 1 ) is connected in the circuit, as shown below. Considering this voltage source V 1 alone acting in the circuit, the current through the load I L2 is determined.

Similarly, the circuit is reconnected with the next voltage source V 2 , while the other voltage source is short-circuited and the current source is open-circuited. Considering this voltage source V 2 alone acting in the circuit, the current through the load I L3 is determined.

Now, according to the superposition theorem, the current flowing through any part of the circuit is equal to the algebraic sum of the currents produced by each independent source.

Thus the current flowing through the load R L is given by,

I L = I L1 + I L2 + I L3

## Steps to solve the circuits using Superposition Theorem

• Identify the load resistor (R L ) in the given problem.
• Considering a single source alone acting in the circuit, short-circuit the other voltage sources and open the current sources, if any.
• Calculate the current flowing through the load resistor R L  due to a single source.
• Repeat steps 2 and 3 for all other sources in the circuit.
• To find the total current through the load resistor, perform an algebraic sum of the currents produced by each independent source.

## Solved Problem 1

Find the current through 3 Ω resistor using superposition theorem.

Let I 1 and I 2 are the currents flowing through the 3 Ω resistor, due to the voltage sources 20 v and 40 v respectively.

(i) To find I 1 .

Consider 20 v voltage source alone . Hence, Short circuit the other voltage source and the circuit is redrawn as below,

Now, to find the current through 3 Ω resistor, it is necessary to determine the total current supplied by the source (I T ).

When you observe the circuit, 3 Ω and 6 Ω resistors are in parallel with each other. This parallel combination is connected in series with a 5 Ω resistor. Hence the equivalent or total resistance is obtained as below,

By applying Ohm’s law ,

Now, the current through 3 Ω resistor is determined by using current division rule . It is given by,

(ii) To find I 2 .

Consider 40 v voltage source alone . Hence, Short circuit the other voltage source and the circuit is redrawn as below,

When you observe the circuit, 3 Ω and 5 Ω resistors are in parallel with each other. This parallel combination is connected in series with a 6 Ω resistor. Hence the equivalent or total resistance is obtained as below,

The below figure shows the resultant circuit, which depicts the currents produced because of two voltage sources 20 v and 40 v acting individually.

By superposition theorem, the total current is determined by adding the individual currents produced by 20 v and 40 v.

Thus the current through 3 Ω resistor is = I 1 + I 2 = 1.904 + 3.174 = 5.078 A

## Solved Problem 2

Find the voltage across through 15 Ω resistor using superposition theorem.

Let V 1 , V 2 , V 3 , V 4 be the voltages across the 15 Ω resistor when each source (20v, 10v, 10A, 5A sources) are considered separately. Hence the resultant voltage is given by,

V T = V 1 + V 2 + V 3 + V 4

(i) To find V 1

Consider 20v source alone. Hence the other two current sources are open-circuited and a voltage source is short-circuited. The circuit is redrawn as below,

By applying KVL , the current through the 15 Ω resistor is given by,

Thus,the voltage across the 15 Ω resistor is given by,

(ii) To find V 2

Consider 10v source alone. Hence the other two current sources are open-circuited and a voltage source is short-circuited. The circuit is redrawn as below,

(iii) To find V 3

Consider 10A source alone. Hence the other two voltage sources are short-circuited and a current source is open-circuited. The circuit is redrawn as below,

Here, 15 Ω and 40 Ω resistors in parallel with each other, hence by using current division rule , current through the 15 Ω resistor is given by,

Thus, the voltage across the 15 Ω resistor is given by,

(iv) To find V 4

Consider 5A source alone. Hence the other two voltage sources are short-circuited and a current source is open-circuited. The circuit is redrawn as below,

Here, since the circuit has a low resistance path, a negligible amount of current flows through 15 Ω resistor. Hence, current through the 15 Ω resistor is given by,

Therefore, by the superposition theorem, the resultant voltage is given,

In the above expression, the negative sign in V 3 indicates that the direction of the current, in that case, is opposite to the assumed direction in the circuit.

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Hope you will solve more solutions

Solve problems on all theorems ?

In step iii, Since we need current through 15 ohm resistor, shouldn’t it be: I3= 10*15/(15+40) ?

No. that’s the wrong way. Since it is a parallel circuit, the current division rule is applied. Kindly go through the rule to get a clear idea.

very good notes.Easily understandable.Thank you

Glad to know that the contents are understandable. Thank you.

Sir actually i want to ask u a numerical question related to superposition theorem how i can sen u the question

You can post the question here or you can send it through the mail: [email protected]

Sir, in the second problem to actually verify superpositon theorem v total should also be calculated when all the sources are active which you didn’t mention there will you please explain the process involved there in detail?

The superposition theorem is used not to verify but to solve complex electric circuits. To verify this theorem, you can either simulate the circuit in any simulation software or you can do it practically. In the superposition theorem, the current or voltage due to each individual source is determined and then they are added to find the resultant(total) current or voltage across the given load. The same is followed for both the problems. kindly check the above steps to solve .

Am so glad because my problems in this area are now fully solved. Receive m gratitude

for problem 2, step (iv) can you please explain how/where the low resistance path works and why the 15 ohm resistor has a negligible resistance. thanks

in step (iv), for the 5A source, there are two paths for the flow of current. One path is through 15Ω resistance and the other path is the short-circuited path, which is across the current source. In an electrical circuit, a short circuit will have very low resistance. Since there is very low resistance in this path, all the current from the source rushes through the short circuit path, making a negligible amount of current to flow through 15Ω resistance. That is the reason, I4 =0. And in your question, you have asked, “why the 15 ohm resistor has a negligible resistance.” 15Ω resistor will never have negligible resistance. It has its own resistance of 15Ω.

In part iv, to find for V4, you said the other current source is open-circuited but on the re-drawing of the circuit it is short-circuited.

Yeah, you are right. It was corrected now and thanks for showing the mistake.

I don’t also understand why v4 in problem 2 is completely zero. I try using Ohms law for it, it was 0.060v. Can you help explain to me more better.

Could you please tell me how you applied Ohm’s law?

I don’t know how I can easily understand part iv, the reason for completely ignoring 5A, does it mean it’s not contributing any voltage to resistor 15, if it does then why not considering it. Secondly, I am asking, in part iii, when we connect to 10A, resistors 15 and 40 are considered to be in parallel, what about when on IV, when connected to 5A, are they in parallel or series?

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## Superposition Theorem – Circuit Analysis with Solved Example

Superposition theorem – step by step guide with solved example.

## What is Superposition Theorem?

The Superposition Theorem is used to solve complex networks with a number of energy sources. It is an important concept to determine voltage and current across the elements by calculating the effect of each source individually. And combine the effect of all sources to get the actual voltage and current of the circuit element.

Superposition theorem states that;

“In any linear bilateral network having a greater number of sources, the response (voltage and current) in any element is equal to the summation of all responses caused by individual source acting alone. While other sources are eliminated from the circuit.”

In other words, we will consider only one independent source acting at a time. So, we need to remove other sources. The voltage sources are short-circuited and the current sources are open-circuited for ideal sources. If the internal resistance of sources is given, you need to consider the circuit.

The superposition theorem is only applied to the circuit which follows Ohm’s law .

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## When to Use the Superposition Theorem?

The network must follow the below requirements to apply the superposition theorem.

• The components used in the circuit must be linear . It means, for resistors , the flow of current is proportional to the voltage ; for inductors, the flux linkage is proportional to current. Therefore, the resistor, inductor , and capacitor are linear elements. But the diode , transistor is not a linear element.
• The circuit components must be bilateral elements. It means, the magnitude of the current is independent of the polarity of energy sources .
• With the help of the superposition theorem, we can find the current passes through an element, voltage-drop of resistance, and node voltage. But we cannot find the power dissipated from the element.
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## Steps to Follow for Superposition Theorem

Step-1 Find out a number of independent sources available in the network.

Step-2 Choose any one source and eliminate all other sources. If the network consists of any dependent source, you cannot eliminate it. It remains as it is throughout the calculation.

If you have considered all energy sources are ideal sources, you need not consider internal resistance. And directly short-circuit voltage source and open-circuit current source. But in case, if internal resistance of sources is given, you have to replace internal resistance.

Step-3 Now, in a circuit, only one independent energy source is present. You need to find a response with only one energy source in the circuit.

Step-4 Repeat step-2 and 3 for all energy sources available in the network. If there are three independent sources, you need to repeat these steps three times. And every time you get some value of the response.

Step-5 Now, combine all responses by algebraic summation obtained by individual sources. And you will get a final value of response for a particular element of a network. If you need to find a response for other elements, you need to follows these steps again for that element.

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## Superposition Theorem Solved Example

Let’s understand the working of the superposition theorem by example. Find the current (I L ) passes through the 8Ω resister in the given network using the superposition theorem.

Step-1 As shown in the above network, one voltage source, and one current source is given. Therefore, we need to repeat the procedure two times.

Step-2 First we consider 28V voltage source is present in the network. So, you need to remove the current source by open-circuited terminals. As here, we consider the current source as an ideal current source. So, we need not connect the internal resistance.

The remaining circuit is as shown in the below figure.

Step-3 Find the current (I L1 ) passes through 8Ω resister. It gives the effect only of a voltage source.

Due to the open circuit of a current source, no current passes through the 10Ω resister. So, the network consists of only one loop .

Apply KVL to the loop;

28 = 6 I L 1 + 8 I L 1

28 = 14 I L 1

I L 1 = 28/14

Step-4 Now, we repeat the same procedure for the current source. In this condition, we remove the voltage source by short-circuiting. A remaining circuit is as shown in the below figure.

Here, we have to consider two loops. I 1 and I 2 are loop current. And find the current I L2 .

Apply KVL to the loop-1;

0 = 6 I 1 + 8 I 1 – 8 I 2

14 I 1 – 8 I 2 = 0

The current passes through the loop-2 is calculated from the current source. And it will be;

Put this value in the above equation;

14 I 1 – 8 (28) = 0

Now, the 8Ω resister branch is common in both loops. So, we need to find the resultant current (I L2 ) passes through the 8Ω resister.

I L 2 = I 1 – I 2

I L 2 = 16 – 28

I L 2 = -12A

Step-5 Now, we combine the effect of both sources by algebraic summation of current. So, the total current passes through the 8Ω resister are I L . Here, the direction of the current is most important. Current I L2 has a minus sign. It means during the 28A source, the current flows in opposite direction. And we cannot change the direction. That’s the reason while combining all sources, we are doing algebraic summation.

I L  = I L 1 – I L 2

I L = 2 + (-12)

Here, we have assumed that the current passes through the 8Ω resister are in the direction of the arrow shown in the figure. Minus sign shows the opposite direction. And the amount of current is 8A.

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## Superposition Theorem Experiment

The superposition theorem experiment can be done by following the below steps.

Determine the current flowing through the resistors and verify the superposition theorem.

Components Required

DC source, Resistors, DC multimeters, Connecting Wires, Breadboard, etc.

Superposition theorem is used in circuit analysis to find the current and voltage across the elements in a given circuit. This theorem is useful when the number of sources is more.

According to the superposition theorem, the response of elements is an algebraic summation of response from the individual energy sources.

First, we will calculate the current passes through resistors in a given network. When all sources are connected. After that, we remove sources and find currents for individual sources. At last, the summation of current measured from the individual sources is similar to the current measures with all sources connected.

We consider the linear bilateral circuit given in the below figure.

Connect all elements on the breadboard as given in the above circuit using connecting wires. Connect ammeter with all resisters to find the current passes through the resistors. Here, we have two voltage sources.

In the first case, measure the current that passes through all resisters when both voltage sources are present in the circuit. The current flowing through the resistor R 1 , R 2 , and R 3 is I 1 , I 2 , and I 3 respectively.

Now remove the 12V voltage source from the circuit. Short the terminals A’ and B’ as shown in the above figure. Here we have assumed that the voltage sources are ideal. So, no need to connect internal resistance. Now, measure the current I 1′ , I 2′  and I 3′ which is the current that passes through the resistor R1, R2, and R3 respectively.

Similarly, remove the 9V voltage source and the remaining circuit is as shown in the above figure. Again, measure the current flowing through all resisters and name it I 1 “ , I 2 “ , and I 3 “ .

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Put all values in below observation table.

Observation Table

Calculation :

I 1  = I 1′  + I 1″

I 2  = I 2′  + I 2″

I 3  = I 3′  + I 3″

To verify the superposition theorem, we compare the algebraic summation of current passes through resisters when an individual source is connected with the current measured when both sources are connected in a circuit.

If the above calculation satisfies, we can prove the superposition theorem.

In this experiment, we can prove that the current that passes through the resistance is an algebraic summation of current when an individual energy source is connected. And it proves the superposition theorem.

• Related Post: Kirchhoff’s Current & Voltage Law (KCL & KVL) | Solved Example

## Application of Superposition theorem

• Superposition theorem can be used for AC and DC networks.
• When the number of independent sources is more, it is easy to find a response of the network.
• It helps to calculate current passes and the voltage across the element by calculating the effect of each energy source individually. And after we can determine the combined effect on elements from all sources.

## Drawback of Superposition Theorem

• Superposition theorem cannot apply to the circuit having only a dependent source. It needed at least two independent sources.
• This theorem only applicable to the network which consists of linear elements. It cannot apply to the non-linear elements like a diode, transistor, etc.
• We cannot calculate power by this theorem. As the power is proportional to the square of voltage and current as it becomes non-linear.
• This theorem is not applicable in a condition where the resistance varies with voltage and current. For all energy sources, the value of resistance must remain constant.
• This theorem only applicable to the bilateral elements. If the response of the network depends on the direction of the current, this theorem is not applicable.

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## Automatic Street Light Control Circuit using LDR & Transistor BC 547

This was informative, just a little note that you made a mistake with the first example where you have IL2 = 16-28 = -10; it should be -12. I haven’t looked at the other question as yet but the process is correct, just that minor detail i recommend correcting.

The typo has been corrected. Thanks for correction.

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## What is a Superposition Theorem : Limitations & Its Applications

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))

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

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.

Apparatus/Required Components

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.

Circuit Diagram

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.

Observation Table

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.

## Limitations

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.

## Applications

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?

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## Solved Problems on Superposition Theorem

Home Blog Basic Electrical Engineering Solved Problems on Superpositi...

1) Find Current flowing through 1 \Omega resistor by using superposition theorem.

Consider 130 V Voltage source only.

-0.2I_1-0.2(I_1-I_2)+130 = 0

-0.4I_1 +0.2I_2 = -130 \ldots\ldots (1)

-I_1-0.2(I_2-I_1)-0.2I_2=0

0.2I_1-1.4I_2=0 \ldots\ldots (2)

I_1 = 350 A

I’_{1\Omega} = 50 A (\downarrow)

Consider 110 V only.

-0.2I_1 -0.2(I_1-I_2)-110 = 0

-0.4I_1+0.2I_2 = 110 \ldots\ldots(3)

-I_2+110-0.2(I_2-I_1)-0.2I_2=0

0.2I_1 – 1.4I_2 = -110 \ldots\ldots (4)

I_1 = -253.8467 A

I_2 = 42.3076 A

I’’_{1\Omega} = 42.3076 A (\downarrow)

\therefore I_{1\Omega} = I’_{1\Omega} + I’’_{1\Omega} = 50 + 42.3076 = 92.3076 A (\downarrow)

2)  Find current flowing through 5 \Omega .

2-0.1I_1-10I_1-5(I_1-I_2)=0

15.1I_1 + 5I_2=-2 \ldots\ldots (1)

-20I_1-0.2I_2-5(I_2-I_1)=0

5I_1-25.2I_2=0 \ldots\ldots (2)

I_1 = 0.1417 A

I_2 = 0.02812 A

I’_{5\Omega} = I_1-I_2 = 0.1158 A (\downarrow)

Consider 4 V only.

-0.1I_1-10I_1-5(I_1-I_2) = 0

-15.1I_1+5I_2 = 0 \ldots\ldots (3)

20I_2-0.2I_2-4-5(I_2-I_1)=0

5I_1-25.2I_2 = 4 \ldots\ldots (4)

I_1 = -0.05685 A I_2 = -0.1698 A

I’’_{5 \Omega} = I_2-I_1 = 0.1130 A (\downarrow)

I_{5 \Omega} =  I’_{5\Omega} +  I’’_{5\Omega} = 0.2288 A (\downarrow)

3) Determine current through 1 \Omega resistor.

Consider a 10 V voltage source only.

-2(I_1-I_2)-2I_1 = 0

-4I_1+2I_2=0 \ldots\ldots (1)

-10 -2(I_2-I_1)-I_2=0

2I_1-3I_2=10 \ldots\ldots (2)

I_1 = -2.5 A

I’_{1\Omega} = -5 A (\rightarrow)

Consider a 6 V voltage source only.

-2I_1-2(I_1-I_2)-6 = 0

-4I_1+2I_2=6 \ldots\ldots (3)

-I_2+6-2(I_2-I_1)=0

2I_1 – 3I_2 =-6 \ldots\ldots (4)

I_1 = -0.75 A

I’’_{1\Omega} = 1.5 A (\rightarrow)

Consider 8 V only.

8-2I_1-2(I_1-I_2)=0

-4I_1+2I_2=-8 \ldots\ldots (5)

-I_2-2(I_2-I_1)=0

2I_1-3I_2=0 \ldots\ldots (6)

I’”_{1\Omega} = 2 A (\rightarrow)

Total \; \; current \; \; I_{1 \Omega}=I’_{1 \Omega}+I’’_{1 \Omega}+I’’’_{1 \Omega}

I_{1 \Omega}= 2+1.5+(-5) = -1.5 A (\rightarrow)

I_{1 \Omega} = 1.5 A (\leftarrow)

4) Find current through 20 \Omega resistor.

For Mesh 1:

10-10I_1+10I_2=0

-10I_1+10I_2=-10 \ldots \ldots(1)

For Mesh 2:

-10I_2+10I_1-I_2-20I_2=0

10I_1-31I_2=0 \ldots\ldots(2)

I_1 = 1.4761 A

I_2 = 0.4761 A

I’_{20\Omega} = 0.4761 A (\downarrow)

Consider 8 V voltage source only.

I’’_{20\Omega} = -0.3809 A (\downarrow)

Consider a 12 V voltage source only.

-12-8I_1+8I_2=0

-8I_1+8I_2=12 \ldots\ldots(3)

-I_2-8I_2+8I_1-20I_2=0

8I_1-29I_2=0 \ldots\ldots(4)

I_1=-2.0714 A

I_2=-0.5714 A

I’’’_{20\Omega} = -0.5714 (\downarrow)

I_{20\Omega} = I’_{20\Omega}+I’’_{20\Omega}+I’’’_{20\Omega}

I_{20\Omega} = 0.4761+(-0.3809)+(-0.5714)

I_{20\Omega} = -0.4762 A  (\downarrow)

I_{20\Omega} = 0.4762 A  (\uparrow)

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## Superposition Theorem

In this article, we will learn all about the Superposition theorem along with its statement, steps to be followed, advantages and disadvantages, applications, numerical problems and frequently asked questions based on it. The Superposition Theorem allows you to analyze a circuit with multiple voltage or current sources by considering the effect of each source separately while treating the others as inactive.

## Statement of the Superposition Theorem

According to the Superposition Theorem, the response (voltage or current) at any point in a linear electrical network with multiple independent sources can be calculated by calculating the individual contributions of each source while assuming the other sources are “turned off” or replaced by their internal resistances.

## Steps of Superposition Theorem

• Turn Off All but One Source: Consider only one independent source (voltage or current source) active, while all other independent sources are turned off (replaced by their internal resistances, which are typically zero for ideal voltage sources and infinite for ideal current sources).
• Analysis of the Circuit: With only one source active, analyze the circuit using circuit analysis techniques such as Ohm’s law, Kirchhoff’s law, and other relevant methods.
• Calculation of the Response: Determine the voltage, current, or any other required parameters in the circuit due to the single active source.
• Repeat for Each Source: Repeat steps 1 to 3 for each independent source in the circuit.
• Combine Responses: After calculating the response for each individual source, you may determine the total response at the required  by adding or superimposing the individual response from each source.

The Superposition Theorem only applies to linear circuits in which the relationship between voltage and current is constant and does not include nonlinear components such as diodes or transistors.

The Superposition Theorem can significantly simplify the analysis of complex circuits, especially when there are many independent sources. However, this theorem can be time-consuming for circuits with a large number of sources, as you need to perform separate calculations for each source. In such cases, other circuit analysis techniques like nodal analysis or mesh analysis might be more efficient.

## Steps to Solve an Electronic Circuit in Superposition Theorem

Consider one source at a time from the various multiple sources. Replace all the other (except the selected source) impedances with their internal resistance. Calculate the current flowing through the source and the voltage drop across it. Repeat the above steps by taking the individual sources one by one. Take the algebraic sum once all the current sources and voltage drops have been calculated.

## 1. Find the current through 3 Ω resistor using superposition theorem.

Electric Circuit

Consider the 20 V voltage source alone. Short circuit the other voltage source.

To find the current through 3 Ω resistor, it is necessary to determine the total current supplied by the source (I T ).

If we observe the circuit, 3 Ω and 6 Ω resistors are in parallel with each other. This parallel combination is connected in series with a 5 Ω resistor. The equivalent or total resistance is obtained as below,

⇒R T = 5 + (3*6/9) = 7 Ω

By applying Ohm’s law,

⇒I T = V/R T = 20/7 = 2.857 A

Now, the current through 3 Ω resistor is determined by using current division rule. It is given by,

⇒I 1 = I T *(6/6+3) = 2.857*0.667 = 1.904 A

(ii) To find I 2 .

Consider the 40 V voltage source alone. Short circuit the other voltage source.

Now, to find the current through 3 Ω resistor, it is necessary to determine the total current supplied by the source (IT).

If we observe the circuit, 3 Ω and 5 Ω resistors are in parallel with each other. This parallel combination is connected in series with a 6 Ω resistor. Hence the equivalent or total resistance is obtained as below,

⇒R T = 6 + [(3*5)/(3+5)] = 7.875 Ω

⇒I T = V/R T = 40/7.875 = 5.079 A

⇒I 2 = I T *(5/5+3) = 5.079*0.625 = 3.174 A

The below figure shows the resultant circuit, which depicts the currents produced because of two voltage sources 20 V and 40 V acting individually.

By superposition theorem, the total current is determined by adding the individual currents produced by 20 V and 40 V.

Therefore, the current through 3 Ω resistor is = I 1 + I 2 = 1.904 + 3.174 = 5.078 A

## Applications of Superposition Theorem

Some applications of Superposition Theorem are:

• Superposition Theorem can be used to study the effects of each source separately on parameters such as voltage levels, current levels, etc.
• Superposition Theorem can be used to study how each element affects the overall circuit behavior in large electronic circuits. r.
• Superposition Theorem can be used how a single element like amplifiers, filters, and other electronic elements affect the large analog circuits when any defect arises.
• Superposition Theorem can be used in sensors evaluation where multiple sensors are involved an each sensor has to studied on the basis of their performance.
• Superposition Theorem can be used in Engineering Electromagnetics where the fault has to studied in transmission line analysis related to voltage, current, resistance, capacitance, inductance, etc.
• Superposition Theorem can be used for multiple-frequency application to analyze the effects of each frequency component for the overall filter response.

Some advantages of Superposition Theorem are:

• It makes the analysis of the circuit easy by breaking down the circuits into smaller parts for easy understanding.
• It makes the analysis of an individual source easy by making it easy to study one component of an electronic circuit without impacting the other components.
• Insight into Circuit Behavior: By studying the effect of each source independently, we can gain insight into how different sources interact with each other and the resulting circuit behavior.
• It makes the study of an electronic component about its behavior towards other components and the overall circuits.

Some disadvantages of Superposition Theorem are:

• Limited to Linear Circuits: The Superposition Theorem only applies to linear circuits when voltage and current have a constant relationship. It cannot be used in circuits including nonlinear components such as diodes and transistors.
• May Not Capture All Effects: In some cases, interactions between sources can lead to effects that the Superposition Theorem doesn’t fully capture. It assumes that the interaction between sources is negligible, which might not always be the case.
• Doesn’t Provide Overall Solution: The Superposition Theorem provides solutions for individual sources but doesn’t directly yield the overall response when all sources are active. Summing up individual solutions may not always result in an accurate representation of the circuit behavior when all sources are present.
• Not Suitable for Complex Networks: For circuits with interconnected components and complex feedback loops, applying the Superposition Theorem might not be practical or accurate in capturing all circuit dynamics.

The superposition theorem allows you to evaluate complex circuits with multiple voltage or current sources by evaluating each source individually and setting all other sources to zero (Voltage source is replaced by short circuit, Current source is replaced by open circuit). This allows you to break down a difficult circuit into simpler sub-circuits and more quickly solve for voltage, current, etc.

## FAQs on Superposition Theorem

Q.1: what is the principle of superposition theorem used in circuit analysis .

The Superposition Theorem is a principle used in circuit analysis to simplify the analysis of circuits with multiple independent sources. It states that the response at any point in a linear circuit can be determined by considering the effect of each source individually while treating the other sources as inactive.

## Q.2: When can the Superposition Theorem be applied?

The Superposition Theorem only applies to linear circuits where the voltage-current relationship is constant. It is incompatible with circuits incorporating non linear components such as diodes, transistors, or non linear resistors.

## Q.3: How does the Superposition Theorem work?

The theorem works by analyzing the circuit under the influence of each independent source one at a time while turning off the other sources or replacing them with their internal resistances. The total response is then obtained by superposing the individual responses due to each source.

## Q.4: What are the steps to apply the Superposition Theorem?

The steps involve turning off each independent source except the one being analyzed, calculating the response due to that source, and repeating the process for all sources. The final response is obtained by summing up the individual responses.

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## Superposition Theorem Example with Solution for AC Circuit

In the article Superposition Theorem Example with Solution for AC Circuit we will solve 10 different example of Superposition theorem for AC circuit. So let’s get started:

Find the current through j3Ω inductive reactance using the principle of superposition.

Using the principle of superposition, the given circuit is fed by each of the two sources at a time (figure 2 & 3).

and form figure 3,

$I_2 '' = \dfrac{10 \angle 60^\circ}{-j5 + j3} = \dfrac{10 \angle 60^\circ }{-j2} = 5 \angle 150^\circ A$

⸫ I 2 (following the principle of superposition),

$I_2 ' - I_2 ' = 2.5 \angle 120^\circ - 5 \angle 150^\circ$

Find, by the principle of superposition, the current through 5Ω resistor (figure 4).

Superposition theorem is applied to the given network as shown in figure 5 and 6 deactivating one source at a time.

In figure 5,

$I' = \dfrac{50 \angle 0^\circ}{(10||5) + 10} = \dfrac{50 \angle 0^\circ }{13.33} = 3.75 \angle 0^\circ A$

Again from figure 6,

$I'' = \dfrac{100 \angle 30^\circ }{10||5 + 10} = 7.5 \angle 30^\circ A.$

This gives,

$I_1'' = I''\dfrac{10}{10 + 5} = \dfrac{7.5 \angle 30^\circ \times 10}{10 + 5} = 5 \angle 30^\circ A$

Thus, the current through 5Ω resistor, using the principle of superposition is

$I_{5\Omega} = I_1' + I_1'' = 2.5 \angle 0^\circ + 5 \angle 30^\circ$

Find the current in the resistor (R L ) using the principle of superposition in figure 7.

Principle of superposition is applied in the given circuit taking each source at a time (figure 8 and 9).

In figure 8,

$I_1' = I\dfrac{j6}{j6 + 6 + (-j8)} = \dfrac{2 \angle 0^\circ \times 6 \angle 90^\circ }{6 - j2}$

On the other hand, in figure 9,

$I_1'' = \dfrac{V_1}{j6 + 6 - j8} = \dfrac{10 \angle 60^\circ }{6 - j2} = \dfrac{10 \angle 60^\circ }{6.32 \angle -18.43^\circ}$

⸫ I 1 the total current through R L is I = I 1 ’ + I 1 ’’ (since both are directed in the same direction)

$= 1.9 \angle 108.43^\circ + 1.58 \angle 78.43^\circ$

In the network of figure 10 the current flowing through the 5Ω resistor is equal when each of the voltages sources acts separately on the circuit. What is the value of the ratio of the two sources?

Principle of superposition is applied in the network taking each voltage source at a time (figure 11(a) and 11(b)).

Here in figure 11(a),

$Z' = [\dfrac{j5 \times 5}{j5 + 5} + (2 + j4)] \Omega$

where, Z’ is the impedance of the network across V 1

$= \dfrac{j25 + j30 -10}{(5 + j5)} = \dfrac{(-10 + j55)}{(5 + j5)}$

In figure 11(b), Z’’ (impedance across V 2 )

$[\dfrac{5(2 + j4)}{5 + 2 + j4} + j5]\Omega = \dfrac{10 + j20}{7 + j4} + j5$

As per the question,

$I_1' = I_1''$

Thus the ratio of voltages is 0.89 ∠-26.57 o only.

Obtain the steady state current through the 10V battery in time domain in the circuit of figure 12 using superposition theorem.

Superposition principle is applied in the circuit of figure 12 taking on source at a time as shown in figure 13.

Since j2Ω inductive reactance would act as short circuit while -j4Ω capacitive reactance would act as open circuit while encountering 10V d.c source, the currents I 1 ’ is equal to I’ and I 2 ’ = 0.

$\therefore I' (= I_1') = \dfrac{10}{5} = 2A$

In figure 13(b),

$Z'' = \dfrac{(j2\Omega)(5\Omega)}{j2 + 5} + (-j4)\Omega$

Where Z’’ is the impedance of the circuit shown in figure 13(b) across the a.c. source

$\therefore Z'' = \dfrac{j10}{5 + j2} - j4 = \dfrac{j10 - j20 + 8}{5 + j2}$

In time domain, the instantaneous value is

$i'' = \sqrt{2} \times 7.81 sin(\omega t + 141.34^\circ)$

⸫ Net current through the resistor, using the principle of superposition, is

$I = I' - i'' = 2 - 11 sin(\omega t + 141.34^\circ)A$

Thus, the battery current, in time domain, is

$(2 - 11 sin(\omega t + 141.34^\circ))A$

Find the current in the (-j6Ω) capacitive reactance using superposition theorem in the circuit of figure 14.

Let us apply superposition theorem (figure 15 and 16) deactivating one source at a time.

$I_2' = 2 \angle 30^\circ \dfrac{(2 + j4)}{(2 + j4) + (3 - j6)}$

In figure 16,

$I_2'' = 4 \angle 60^\circ \dfrac{(3 + 2 + j4)}{(3 + 2 + j4) - j6}$

Using the principle of superposition,

$I_2 = I_2' + I_2'' = 1.661 \angle 115.24^\circ + 4.754 \angle 120.46^\circ$

Thus, the current through the capacitor is 6.41 ∠119.117 o A.

Find the current through the capacitor of (-j5)Ω reactance in figure 17 using Superposition Theorem.

Superposition is applied in the given circuit as shown in figure 18 and 19 deactivating each source at time.

In figure 18 the impedance Z’ across 100 ∠0 o V is

$Z' = 5 + \dfrac{(5 - j5)(3 + j4)}{5 - j5 + 3 + j4}$

On the other hand, in figure 19,

$I_3'' = I \dfrac{5 \angle 0^\circ}{-j5 + \dfrac{(3+ j4)5}{3 + j4 + 5} + 5}$

Find the current through the capacitor by superposition theorem (figure 20).

Superposition principle is applied in the circuit as shown in figure 21 and 22 deactivating one source at a time.

In figure 21,

$I_1' = \dfrac{50 \angle 0^\circ }{-j5} = \dfrac{50 \angle 0^\circ }{5 \angle -90^\circ} = 10 \angle 90^\circ A.$

[I 1 ’ = current through the capacitor due to 50 ∠ 0 o V source].

In figure 22, I 1 ” (current through the capacitor due to 100 ∠ 0 o V) = 0, since the terminals are seen to be shorted.

⸫ Net current through capacitor

$= I_1' + I_1'' = 10 \angle 90^\circ A.$

In the circuit of figure 23, find the current through the inductor as well as through the capacitor using the principle of superposition.

Principle of superposition is applied in the given circuit by applying one source at a time and deactivating the other one (figure 24 and 25).

In figure 24,

$I_1' = \dfrac{10 \angle 0^\circ}{j1} = 10 \angle -90^\circ A$

In figure 25,

$I_1'' = 0$

Using the Principle of superposition,

$I_1 = I_1' + I_1'' = 10 \angle -90^\circ + 0$

Thus, using the principle of superposition, the current through j1Ω inductive reactance is 10 ∠-90 o A and that through the capacitor is zero.

Example: 10

In figure 26, E 1 = 10 sin(314t + 45 o )V and I 1 = 5 cos(314t + 30 o )A. Using the principle of superposition, find the power delivered to the 1Ω resistor in the central branch.

The principle of superposition is applied as shown in figure 27(a) and 27(b). Here E 1 = 10 ∠ 45 o V from the given data (figure 27(a)).

⸫ Power delivered to the load of 1Ω by the voltage source E 1 is

$\dfrac{1}{2}(1.414)^2 \times 1 = 1W [I_{rms}' = ( \dfrac{1.414}{\sqrt{2}} )^2]$

In figure 27(b),

$-I'' = \dfrac{4 + j6}{4 + j6 + 1 - j1} \times 5 \angle 30^\circ$

⸫ Power delivered to the load of 1Ω by the current source is

$\dfrac{1}{2}(5.1)^2 \times 1 = 13W$

By the principle of superposition, the power delivered to the load due to both the sources is (1W + 13W) = 14W.

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## Superposition Theorem Questions

The superposition theorem is one of the essential theorems applied to electrical networks. The superposition theorem helps in solving the network with two or more sources. It also aids in calculating the current through the resistor in two loops.

Superposition theorem states that “If more than one source acts simultaneously in an electric circuit, then the current through any one of the branches of the circuit is the summation of currents, which would flow through the branch for each source, keeping all the other sources dead.”

Note: When removing a voltage source, its value is set to zero.

Limitations of Superposition Theorem

• The superposition theorem does not apply to non-linear circuits.
• Applying the superposition theorem requires two or more sources in the circuit.
• While you sum the individual contributions of each source, cross-check the signs assigned to the quantities. It is suggested to assign a reference direction to each unknown quantity. If a contribution from a source has the same direction as the reference direction, the sum is positive. If in the opposite direction, then mention the negative sign.
• All the components must be linear, to use the superposition theorem.
• This theorem does not apply to power, as power is not a linear quantity.

Important Superposition Theorem Questions with Answers

1. State true or false: While removing a voltage source, the value of the voltage source is set to zero.

Explanation: The voltage source is replaced with a short circuit.

2. When removing a current source, its value is set to zero. This is done by replacing the current source with an _____

• Short circuit
• Open circuit

3. Superposition theorem is valid for

• Linear circuits
• Non-linear circuits
• Both linear and non-linear circuits
• None of the options

Explanation: The superposition theorem is valid for linear bilateral networks.

4. The superposition theorem converts circuit into

• Thevenin equivalent circuit
• Norton equivalent circuit
• Both a) and b)
• None of the option

Answer: c) Both a) and b)

5. For applying the superposition theorem, we need

• Only one source
• Two or more sources

Answer: c) Two or more sources.

Explanation: The application of the superposition theorem requires two or more sources in the circuit.

6. Choose YES or NO: The superposition theorem can be easily applied to unbalanced bridge circuits.

Explanation: The superposition theorem cannot be applied to unbalanced bridge circuits.

7. Choose the passive bilateral element

• All the above options

Answer: d) All the above options.

Explanation: Inductor, capacitor, and resistor are passive bilateral elements.

8. According to which law, is the current through a conductor between two points directly proportional to the voltage across the two points?

• Superposition theorem
• Newton’s law

9. According to Ohm’s law:

10. State true or false: The superposition theorem is not valid for AC circuits.

Explanation: The superposition theorem is valid for AC circuits.

Practice Questions

• State superposition theorem.
• List the limitations of the superposition theorem.
• Explain the superposition theorem using an example.
• State Ohm’s law.
• Define open circuit and close circuit.

See the video below to learn about the superposition theorem.

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1. Superposition Theorem with solved problems

Solved Problem 1 Find the current through 3 Ω resistor using superposition theorem. Let I 1 and I 2 are the currents flowing through the 3 Ω resistor, due to the voltage sources 20 v and 40 v respectively. (i) To find I1. Consider 20 v voltage source alone. Hence, Short circuit the other voltage source and the circuit is redrawn as below,

2. Superposition Theorem

Example 1: Find the current flowing through 20 Ω using the superposition theorem.

3. PDF Superposition Examples

This example illustrates the use of superposition with an op-amp circuit. The circuit is shown in Fig. 11. The object is to solve for By superposition of vA vO. With v2 = 0, it follows that vA = v1, vB = 0, and With v1 = 0, and vC = [1 + R4/ (R3kR5)] v1. vC, vO can be written R2 R2 R2 R4

4. Superposition Theorem

Step by Step Guide with Solved Example When to Use the Superposition Theorem? The network must follow the below requirements to apply the superposition theorem. The components used in the circuit must be linear. It means, for resistors, the flow of current is proportional to the voltage; for inductors, the flux linkage is proportional to current.

5. Superposition Theorem Example with Solution

In the article Superposition Theorem Example with Solution we had solved various kind of problem regarding Superposition Theorem. While solving these example we are assuming that you have knowledge of Superposition Theorem. Check the article on Superposition Theorem . Example 1: Find I in the circuit shown in figure 1.

6. Superposition Theorem : Examples, Problems, Applications & Limitations

We know that ohms law V= IR I1= V1/R Disable the primary voltage source i.e, V1, and calculating the flow of current I2 in the following circuit. When V1 Voltage Source is Disabled I2= -V2/R

7. Superposition Theorem for DC Circuits with examples

Step 1 At first, find the current through 2Ω resistor with 48V source acting alone. Hence replace the 24 V source by a short circuit. Here current I 2 flows through the load resistor. To find the load current, find the total current supplied by the source (I 1) with its total resistance.

8. Superposition Theorem

Step 1: Replace All of the Power Sources Except for One When replacing the power supplies in the circuits, we must follow these two rules: Replace all voltage sources with short circuits (wires) Replace all current sources with open circuits (breaks)

9. Superposition Theorem: Statement, Procedure, and Solved Examples

The Superposition Theorem is a principle used in electrical circuit analysis that states that the voltage or current in any one branch of a linear, passive network can be determined by considering the effects of each independent source separately. Get Unlimited Access to Test Series for 790+ Exams and much more. Know More ₹17/ month

10. Superposition (article)

f ( x 1 + x 2) = f ( x 1) + f ( x 2) To solve a circuit using superposition, the first step is to turn off or suppress all but one input. To suppress a voltage source, replace it with a short circuit. To suppress a current source, replace it with an open circuit. Then you analyze the resulting simpler circuits. Repeat for all inputs.

11. Solved Problems on Superposition Theorem

Solved Problems on Superposition Theorem Basic Electrical Engineering Home Blog Basic Electrical Engineering Solved Problems on Superpositi... 1) Find Current flowing through 1 \Omega 1Ω resistor by using superposition theorem. Consider 130 V Voltage source only. -0.2I_1-0.2 (I_1-I_2)+130 = 0 −0.2I 1 −0.2(I 1 −I 2)+130 = 0

12. Superposition Theorem

Example: 1. Find the current through 3 Ω resistor using superposition theorem. Electric Circuit Answer: (i) To find I1. Consider the 20 V voltage source alone. Short circuit the other voltage source. To find the current through 3 Ω resistor, it is necessary to determine the total current supplied by the source (IT).

13. Superposition Theorem Worksheet

Question 2 Suppose we have a single resistor powered by two series-connected voltage sources. Each of the voltage sources is "ideal," possessing no internal resistance: Calculate the resistor's voltage drop and current in this circuit.

14. Maximum Power Theorem Example with Solution

Solution: Let us first convert the "I" source to "V" source and remove R from x-y terminal, the voltage at these terminals being Vo.c. With reference to figure 5, Thus in the left loop, or, Again, with reference to figure 6, RTh (internal resistance of the circuit looking through x-y) is obtained as. As per maximum power transfer theorem,

15. Superposition Theorem

Solved Example of Superposition Theorem Limitations of Superposition Theorem Definition of Superposition Theorem The superposition theorem is an analytical method used in electrical engineering to solve complex networks where multiple sources are interconnected.

16. 5.3: Superposition Theorem

We will solve this using superposition and compare the results to those of Example 5.2.3 which used source conversion. Example 5.3.1 5.3. 1. For the circuit of Figure 5.3.2 5.3. 2, determine vb v b using superposition. I = 2E − 3∠90∘ I = 2 E − 3 ∠ 90 ∘ amps peak and E = 10∠0∘ E = 10 ∠ 0 ∘ volts peak.

17. Superposition Theorem Example with Solution for AC Circuit

The principle of superposition is applied as shown in figure 27 (a) and 27 (b). Here E1 = 10 ∠ 45oV from the given data (figure 27 (a)). By the principle of superposition, the power delivered to the load due to both the sources is (1W + 13W) = 14W. In the article Superposition Theorem Example with Solution for AC Circuit we will solve 10 ...

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Download PDF. The Superposition Theorem The Superposition theorem is used to find the solution to networks with two or more sources that are not in series or parallel. It treats each source independently, and the algebraic sum is found to determine a particular unknown quantity of the network. The Superposition Theorem states the following: The ...

21. PDF Superposition Theorem Examples With Solutions

When superposition theorem example problems can get rth. This website uses cookies to ensure you get the best experience on our website. For superposition theorem with each wave has expired or voltage. Cancellations take effect at the next billing period. Try again the theorem with superposition. From the nomenclature the alkyne group should be ...

22. 4.3: Superposition Theorem

As useful as the source conversion technique proved to be in Example 6.2.3, it will not work for all circuits. Thus, more general approaches are needed. One of these methods is superposition. Superposition allows the analysis of multi-source series-parallel circuits. Superposition can only be applied to networks that are linear and bilateral.

23. Superposition Theorem Questions

Answer: a) TRUE Explanation: The voltage source is replaced with a short circuit. 2. When removing a current source, its value is set to zero. This is done by replacing the current source with an _____ Light bulb Short circuit Open circuit Resistor Answer: c) Open circuit. 3. Superposition theorem is valid for Linear circuits Non-linear circuits