superposition theorem

in this PPT you can understand how to prove superposition theorem using Proteus simulation

1. TOPIC Superposition theorem TO, RAVINDRA MEENA HOD Ist YEAR PRESENTED BY: Abdulqadir (k12430) Aj ay singh dhakad Aditi dadhich (k12582) IInd Sem , 1st Year, Section A A-1 Batch

2. CONTENTS Introduction to superposition theorem Applications of superposition theorem Working of superposition theorem Requirements Circuit Diagram

3. INTRODUCTION The  superposition theorem  for electrical circuits states that for a linear system the response (voltage or current) in any branch of a bilateral linear circuit having more than one independent source equals the algebraic sum of the responses caused by each independent source acting alone, where all the other independent sources are replaced by their internal impedances.

4. WORKING To ascertain the contribution of each individual source, all of the other sources first must be "turned off" (set to zero) by: Replacing all other independent voltage sources with a short circuit (thereby eliminating difference of potential i.e.  V =0; internal impedance of ideal voltage source is zero (short circuit). Replacing all other independent current sources with an open circuit (thereby eliminating current i.e.  I =0; internal impedance of ideal current source is infinite (open circuit)). This procedure is followed for each source in turn, then the resultant responses are added to determine the true operation of the circuit. The resultant circuit operation is the superposition of the various voltage and current sources.

5. APPLICATION The superposition theorem is very important in circuit analysis. It is used in converting any circuit into its Norton equivalent or  Thevenin equivalent. The theorem is applicable to linear networks (time varying or time invariant) consisting of independent sources, linear dependent sources, linear passive elements (resistors, inductors, capacitors) and linear transformers.

6. REQUIREMENTS CURRENT SOURCES VOLTAGE SOURCES RESISTORS

7. CIRCUIT DIAGRAM

8. result In the first circuit we calculated current is 0.84mA across 10k resistor. In the second circuit after replacing 5v voltage source with short circuit we obtain +1.20mA current across 10k resistor. In the third circuit after replacing 12v voltage source with short circuit we obtain -0.36mA current across 10k resistor. According to the superposition theorem current across 10k resistor in first circuit is equal to the algebraic sum of current across 10k resistor in 2 nd circuit and 3 rd circuit. Proof: current in first circuit = current in 2 nd + current in 3 rd +0.84mA = +1.20mA + (-0.36mA) + 0.84mA = +1.20mA -0.36mA + 0.84mA = +0.84mA

9. THANK YOU