Superposition Theorem.

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Presentation transcript:

Superposition Theorem

Objective of Lecture Introduce the superposition principle. Provide step-by-step instructions to apply superposition when calculating voltages and currents in a circuit that contains two or more power 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.

Steps Turn on all independent sources , one at a time. Redraw circuit according to source type turned off. Solve for the voltages and currents in the new circuit. Repeat Steps 1 to 3. To find the total voltage across each component and the total current flowing, add the contributions from each of the voltages and currents found.

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 0V. 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 0A.

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

Example #1

Example #1 (con’t) Replace Is1 and Is2 with open circuits

Example #1 (con’t) Since R2 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 R2 in it.

Example #1 (con’t) Vs

Example #1 (con’t) IS1 Redraw circuit. Replace VS with a Short Circuit and Is2 with an Open Circuit Redraw circuit. IS1

Example #1 (con’t) Note: The polarity of the voltage and the direction of the current through R1 has to follow what was used in the first solution. IS1

Example #1 (con’t) IS1

Example #1 (con’t) IS1

Example #1 (con’t) Replace VS with a Short Circuit and Is1 with an Open Circuit IS2

Example #1 (con’t) IS2 R2 and I2 are not in parallel with R1 and R3. Since loop of I1 and I3 are connected at the same point, V across the terminal is zero. therefore I1 = I3 = 0A.

Example #1 (con’t) IS2

Example #1 Vs on Is1 on Is2 on Total I1 +42.9mA +0.286A 0A +0.329A I2 Currents and voltages in original circuit with all sources turned on. Vs on Is1 on Is2 on Total I1 +42.9mA +0.286A 0A +0.329A I2 -1A 2A +1A I3 -0.714A -0.671A V1 +2.14V +14.3V 0V 16.4V V2 -30V + 60V +30.0V V3 0.857V -14.3V -13.4V

Pspice Simulation

Summary Superposition can be used to find the voltages and currents in the circuit. 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.

QUESTIONS???