# Superposition Theorem: Example Problems, Limitations and Applications

For every electrical circuit, there are two or additional independent supplies like current, voltage or both the 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 in 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 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.

**Superposition Theorem Examples **

The basic circuit diagram of 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

I1= V1/R

Disable the primary voltage source i.e, V1, and calculating the flow of current I2 in the following circuit.

I2= -V2/R

According to superposition theorem, the network current I = I1 + I2

I = V1/R-V2/R

**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))

15(10/25)

= 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

**Applications & Limitations of Superposition Theorem**

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 superposition theorem** is, we can employ only for 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 the single power source at a time, the

Equivalent section currents and voltages algebraically included discovering what they will perform with every power supplies 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.

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