EE302Lecture19

# EE302Lecture19 - S rposition of Powe no upe r S rposition m...

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1 Superposition of Power? - no Superposition may be used to find any current or voltage in a circuit You cannot use a superposition approach to calculate power directly since power is proportional to voltages and currents squared.

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2 When do I need to use superposition? Superposition is required only when the independent sources in a circuit are fundamentally different Example: a combination of DC and AC sources
3 Why learn to apply superposition? Superposition is an important and powerful principle that will be used on more complex linear systems that are driven simultaneously be different types of energy sources. EE302 is preparing you with this fundamental understanding of how to apply superposition

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4 Reminder: Source Transformations + - R I L R I L Open circuit voltage: V s Short circuit current: V s /R Voltage across R L : V s R L /(R+R L ) Open circuit voltage: (V s /R)*R = V s Short circuit current: I s = V s /R Voltage across R L : V s R L /(R+R L ) These circuits are equivalent!!
5 Types of Equivalent Circuits Thévenin Equivalent Circuit Norton Equivalent Circuit We will focus on the Thévenin Equivalent Circuit I N = V TH /R TH R N = R TH

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6 V OC and I SC Method This method applies for all circuits with at least one independent source – same method you have already learned: Remove load resistor, if present Calculate the open-circuit voltage ( V oc ) using any method. Replace the load resistor with a short circuit (i.e., a wire) Calculate the short circuit current ( I sc ) using any method. sc I and oc Th OC Th V R V V = =
7 Dependent Source Example I C 5I C 2A 1 Ω 5 Ω 10 Ω 10V - + A B Treat the 5-ohm resistor as the load resistor and determine the Thévenin Equivalent. We solved this same circuit earlier. V B = 20 volts

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8 Dependent Source Example: V OC I C 5I C 2A 1 Ω 10 Ω 10V - + A B KCL at Node B: -I c + 5I c = 0 --- The only possible value of I C is zero. Therefore, V OC = 10V. V OC - +
9 Dependent Source Example: I SC Ω 2.5 - = 4 - 10 = I V = R 4A - = 4I - = I 0 = I + 5I + I - : B Node at KCL A 5 = 5I A 1 = 10Ω 10V = I SC OC TH C SC SC C C C C I C 5I C 2A 1 Ω 10 Ω 10V - + A B I SC Can R Th really be negative? What does this mean?

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10 Dependent Source Example I C 5I C 2A 1 Ω 5 Ω 10 Ω 10V - + 5 Ω -2.5 Ω 10V - + V L - + V V L 20 5 . 2 5 5 10 = - = Negative R Th is possible with dependent sources. (More on this later.)
Test Source Method for R Th This method can be used to find the Thévenin Equivalent for ANY linear circuit . Test source method: Remove the load from the circuit. Turn OFF all independent sources. Attach a test source to the output terminals: Either a voltage source or a current source. A one amp current source is often convenient.

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EE302Lecture19 - S rposition of Powe no upe r S rposition m...

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