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INC 112 Basic Circuit Analysis Week 5 Thevenin’s Theorem

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Special Techniques Superposition Theorem Thevenin’s Theorem Norton’s Theorem Source Transformation

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Linearity Characteristic If R L change its value, how will it effect the current and voltage across it?

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I V V OC I SC For any circuit constructed from only linear components Not just RL, all resistors have this property. Voc = Voltage open-circuit Isc = Current short-circuit

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Thevenin’s Theorem When we are interested in current and voltage across RL, we can simplify other parts in the circuit. Equivalent circuit

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I V V OC I SC Voc = Voltage open-circuit Isc = Current short-circuit R = R equivalent Slope = 1/R

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Thevenin’s Equivalent Circuit Thevenin’s equivalent circuit VTH = Voc (by removing RL and find the voltage difference between 2 pins) RTH (by looking into the opened connections that we remove RL, see how much resistance from the connections. If we see a voltage source, we short circuit. If we see a current source, we open circuit.)

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Why do we need equivalent circuit? To analyze a circuit with several values of RL For circuit simplification (source transformation) To find RL that gives maximum power (maximum power transfer theorem)

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Procedure 1. Remove RL from the circuit 2. Find voltage difference of the 2 opened connections. Let it equal VTH. 3.From step 2 find RTH by 3.1 short-circuit voltage sources 3.2 open-circuit current sources 3.3 Look into the 2 opened connections. Find equivalent resistance.

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Example Find Thevenin’s equivalent circuit and find the current that passes through RL when RL = 1Ω

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0V 10V 0V 6V Find VTH

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Short voltage source RTH Find RTH

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Thevenin’s equivalent circuit If RL = 1Ω, the current is

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Example Find Thevenin’s equivalent circuit

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Find VTH 0V 5V 0V 3V

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Open circuit current source RTH Find RTH

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Thevenin’s equivalent circuit

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Example: Bridge circuit Find Thevenin’s equivalent circuit

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Find VTH 0V 10V 8V 2V VTH = 8-2 = 6V

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Find RTH RTH

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Thevenin’s equivalent circuit

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Special Techniques Superposition Theorem Thevenin’s Theorem Norton’s Theorem Source Transformation

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I V V OC I SC For any point in linear circuit

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Thevenin’s Equivalent Circuit

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Norton’s Equivalent Circuit In= Isc from replacing RL with an electric wire (resistance = 0) and find the current Rn = RTH (by looking into the opened connections that we remove RL, see how much resistance from the connections. If we see a voltage source, we short circuit. If we see a current source, we open circuit.)

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Example Find Norton’s equivalent circuit and find the current that passes through RL when RL = 1Ω

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Find In Find R total Find I total Current divider

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Short voltage source RTH Find Rn

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Norton’s equivalent circuit If RL = 1Ω, the current is

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Relationship Between Thevenin’s and Norton’s Circuit I V V OC I SC Slope = - 1/Rth

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Norton’s equivalent circuit Thevenin’s equivalent circuit Same R value

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Example Find Norton’s equivalent circuit

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Find In Current divider

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Open circuit current source RTH Find RTH

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Norton’s equivalent circuit

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Thevenin’s equivalent circuit Norton’s equivalent circuit 0.2 x 15 = 3

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Equivalent Circuits with Dependent Sources We cannot find Rth in circuits with dependent sources using the total resistance method. But we can use

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Example Find Thevenin and Norton’s equivalent circuit

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Find Voc I1 I2 KVL loop1 KVL loop2

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I1 I2 Solve equations I1 = 3.697mAI2 = 3.678mA

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Find Isc I1 I2 I3 KVL loop1 KVL loop2 KVL loop3

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Find Isc I1 I2 I3 I1 = 0.632mA I2 = 0.421mA I3 = -1.052 A Isc = I3 = -1.052 A

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Thevenin’s equivalent circuit Norton’s equivalent circuit

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