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INC 111 Basic Circuit Analysis. Week 5 Thevenin’s Theorem. Special Techniques. Superposition Theorem Thevenin’s Theorem Norton’s Theorem Source Transformation. Linearity Characteristic. If R L change its value , how will it effect the current and voltage across it?. I.
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INC 111 Basic Circuit Analysis Week 5 Thevenin’s Theorem
Special Techniques • Superposition Theorem • Thevenin’s Theorem • Norton’s Theorem • Source Transformation
Linearity Characteristic If RL change its value , how will it effect the current and voltage across it?
I Voc = Voltage open-circuit Isc = Current short-circuit ISC V VOC For any circuit constructed from only linear components Not just RL, all resistors have this property.
Thevenin’s Theorem When we are interested in current and voltage across RL, we can simplify other parts in the circuit. Equivalent circuit
I ISC Slope = RL change value V VOC Voc = Voltage open-circuit Isc = Current short-circuit Rth = Rth equivalent
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.)
Why do we needequivalent 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)
Procedure • Remove RL from the circuit • Find voltage difference of the 2 opened connections. Let it equal VTH. • From step 2 findRTH by • 3.1 short-circuit voltage sources • 3.2 open-circuit current sources • 3.3 Look into the 2 opened connections. Find equivalent resistance.
Example Find Thevenin’s equivalent circuit and find the current that passes through RL when RL = 1Ω
6V 10V 6V 0V 0V 0V Find VTH
Find RTH Short voltage source RTH
Thevenin’s equivalent circuit If RL = 1Ω, the current is
Example Find Thevenin’s equivalent circuit
3V 5V 3V 0V 0V 0V Find VTH
Find RTH Open circuit current source RTH
Example: Bridge circuit Find Thevenin’s equivalent circuit
10V Find VTH 8V 2V 0V VTH = 8-2 = 6V
Find RTH RTH
Special Techniques • Superposition Theorem • Thevenin’s Theorem • Norton’s Theorem • Source Transformation
I ISC V VOC For any point in linear circuit
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.)
Example Find Norton’s equivalent circuit and find the current that passes through RL when RL = 1Ω
Find R total Find I total Current divider Find In
Find Rn Short voltage source RTH
Norton’s equivalent circuit IfRL = 1Ω, the current is
Relationship BetweenThevenin’s and Norton’s Circuit I ISC Slope = - 1/Rth V VOC
Norton’s equivalent circuit Thevenin’s equivalent circuit Same R value
Example Find Norton’s equivalent circuit
Current divider Find In
Find RTH Open circuit current source RTH
Thevenin’s equivalent circuit Norton’s equivalent circuit 0.2 x 15 = 3
Equivalent Circuits withDependent Sources We cannot find Rth in circuits with dependent sources using the total resistance method. But we can use
Example Find Thevenin and Norton’s equivalent circuit
I2 I1 KVL loop1 KVL loop2 Find Voc
I2 I1 Solve equations I1 = 3.697mA I2 = 3.678mA
I2 I1 I3 KVL loop1 KVL loop2 KVL loop3 Find Isc
I2 I1 I3 Find Isc I1 = 0.632mA I2 = 0.421mA I3 = -1.052 A Isc = I3 = -1.052 A
Norton’s equivalent circuit Thevenin’s equivalent circuit