Superposition Theorem By Dr Vaibhav Jain Associate Professor

Superposition Theorem By Dr. Vaibhav Jain Associate Professor, Dept. of Physics, D. A. V (PG) College, Bulandshahr, U. P. , India.

Objective of Lecture • Introduce the superposition principle. – Fundamentals of Electric Circuits • Provide step-by-step instructions to apply superposition when calculating voltages and currents in a circuit that contains two or more power sources. – Any combination of voltage and current sources.

Superposition • The voltage across a component is the algebraic sum of the voltage across the component due to each independent source acting upon it. • The current flowing through a component is the algebraic sum of the current flowing through component due to each independent source acting upon it.

Usage • Separating the contributions of the DC and AC independent sources. Example: To determine the performance of an amplifier, we calculate the DC voltages and currents to establish the bias point. The AC signal is usually what will be amplified. A generic amplifier has a constant DC operating point, but the AC signal’s amplitude and frequency will vary depending on the application.

Steps 1. 2. 3. 4. Turn off all independent sources except one. Redraw circuit. Solve for the voltages and currents in the new circuit. Turn off the active independent source and turn on one of the other independent sources. 5. Repeat Steps 2 and 3. 6. Continue until you have turned on each of the independent sources in the original circuit. 7. To find the total voltage across each component and the total current flowing, add the contributions from each of the voltages and currents found in Step 3.

Turning Off Sources � Voltage sources should be replaced with short circuits. � A short circuit will allow current to flow across it, but the voltage across a short circuit is equal to 0 V. � Current sources should be replaced with open circuits. � An open circuit can have a non-zero voltage across it, but the current is equal to 0 A.

A Requirement for Superposition • Once you select a direction for current to flow through a component and the direction of the polarity (+ /_ signs) for the voltage across a component, you must use the same directions when calculating these values in all of the subsequent circuits.

Example #1

Example #1 (con’t) Replace Is 1 and Is 2 with open circuits

Example #1 (con’t) Since R 2 is not connected to the rest of the circuit on both ends of the resistor, it can be deleted from the new circuit. Redraw circuit without R 2 in it.

Example #1 (con’t) Vs

Example #1 (con’t) Replace VS with a Short Circuit and Is 2 with an Open Circuit Redraw circuit. IS 1

Example #1 (con’t) IS 1 Note: The polarity of the voltage and the direction of the current through R 1 has to follow what was used in the first solution.

Example #1 (con’t) IS 1

Example #1 (con’t) IS 1

Example #1 (con’t) Replace VS with a Short Circuit and Is 1 with an Open Circuit IS 2

Example #1 (con’t) IS 2 R 2 and I 2 are not in parallel with R 1 and R 3. Since V 1 = -V 3, but I 1 must equal I 3, the only valid solution is when I 1 = I 3 = 0 A.

Example #1 (con’t) IS 2

Example #1 Currents and voltages in original circuit with all sources turned on. I 1 I 2 I 3 Vs on +42. 9 m. A 0 +42. 9 m. A Is 1 on +0. 286 A -1 A -0. 714 A Is 2 on 0 A 2 A 0 A Total +0. 329 A +1 A -0. 671 A V 1 V 2 V 3 +2. 14 V 0 V 0. 857 V +14. 3 V -30 V -14. 3 V 0 V + 60 V 0 V 16. 4 V +30. 0 V -13. 4 V

Summary �Superposition can be used to reduce the complexity of a circuit so that the voltages and currents in the circuit can be determined easily. �To turn off a voltage source, replace it with a short circuit. �To turn off a current source, replace it with an open circuit. �Polarity of voltage across components and direction of currents through the components must be the same during each iteration through the circuit. �The total of the currents and voltages from each iteration is the solution when all power sources are active in the circuit.

Thank You