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Theveninand Norton
Equivalent Circuitshttp://hyperphysics.phy-
astr.gsu.edu/hbase/electric/thevenin.html
http://www.calvin.edu/~svleest/circuitExamples/Theven
inNorton/http://www.allaboutcircuits.com/vol_1/chpt_10/8.html
http://www.calvin.edu/~svleest/circuitExamples/TheveninNorton/http://www.calvin.edu/~svleest/circuitExamples/TheveninNorton/http://www.calvin.edu/~svleest/circuitExamples/TheveninNorton/http://www.calvin.edu/~svleest/circuitExamples/TheveninNorton/http://www.calvin.edu/~svleest/circuitExamples/TheveninNorton/http://www.calvin.edu/~svleest/circuitExamples/TheveninNorton/http://hyperphysics.phy-astr.gsu.edu/hbase/electric/thevenin.htmlhttp://hyperphysics.phy-astr.gsu.edu/hbase/electric/thevenin.htmlhttp://www.calvin.edu/~svleest/circuitExamples/TheveninNorton/http://www.calvin.edu/~svleest/circuitExamples/TheveninNorton/http://www.allaboutcircuits.com/vol_1/chpt_10/8.htmlhttp://www.allaboutcircuits.com/vol_1/chpt_10/8.htmlhttp://www.calvin.edu/~svleest/circuitExamples/TheveninNorton/http://www.calvin.edu/~svleest/circuitExamples/TheveninNorton/http://hyperphysics.phy-astr.gsu.edu/hbase/electric/thevenin.htmlhttp://hyperphysics.phy-astr.gsu.edu/hbase/electric/thevenin.htmlhttp://hyperphysics.phy-astr.gsu.edu/hbase/electric/thevenin.htmlhttp://www.calvin.edu/~svleest/circuitExamples/TheveninNorton/http://www.calvin.edu/~svleest/circuitExamples/TheveninNorton/http://www.calvin.edu/~svleest/circuitExamples/TheveninNorton/http://www.calvin.edu/~svleest/circuitExamples/TheveninNorton/http://www.calvin.edu/~svleest/circuitExamples/TheveninNorton/http://www.calvin.edu/~svleest/circuitExamples/TheveninNorton/http://www.calvin.edu/~svleest/circuitExamples/TheveninNorton/http://www.calvin.edu/~svleest/circuitExamples/TheveninNorton/http://www.calvin.edu/~svleest/circuitExamples/TheveninNorton/http://www.calvin.edu/~svleest/circuitExamples/TheveninNorton/8/9/2019 Rangkaian Ekivalen Thevenin Dan Norton
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Thevenin's Theorem
Any combination of batteries and resistances with two terminals can be replaced
by a single voltage sourceeand a single series resistor r. The value of eis the
open circuit voltage at the terminals, and the value of ris edivided by the current
with the terminals short circuited
Thevenin Equivalent Circuit
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Thevenin Voltage
The Thevenin voltage VTh.used in Thevenin's Theoremis anideal voltage sourceequal to the open circuit voltage at the terminals. In
the example below, the resistance R2 does not affect this voltage and
the resistances R1 and R3 form a voltage divider
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Thevenin/Norton Resistance
The Thevenin resistance RTh.used in Thevenin's Theoremis the
resistance measured at terminals AB with all voltagesourcesreplaced by short circuits and all current sourcesreplaced
by open circuits. It can also be calculated by dividing the open circuit
voltage by the short circuit current at AB, but the previous method is
usually preferable and gives
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/thevenin.htmlhttp://hyperphysics.phy-astr.gsu.edu/hbase/electric/visource.htmlhttp://hyperphysics.phy-astr.gsu.edu/hbase/electric/visource.htmlhttp://hyperphysics.phy-astr.gsu.edu/hbase/electric/visource.htmlhttp://hyperphysics.phy-astr.gsu.edu/hbase/electric/visource.htmlhttp://hyperphysics.phy-astr.gsu.edu/hbase/electric/visource.htmlhttp://hyperphysics.phy-astr.gsu.edu/hbase/electric/visource.htmlhttp://hyperphysics.phy-astr.gsu.edu/hbase/electric/thevenin.html8/9/2019 Rangkaian Ekivalen Thevenin Dan Norton
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Examples
Vt= Vo- RoIL
http://upload.wikimedia.org/wikipedia/commons/b/ba/Thevenin_and_norton_step_1.pnghttp://upload.wikimedia.org/wikipedia/commons/b/ba/Thevenin_and_norton_step_1.pnghttp://upload.wikimedia.org/wikipedia/commons/b/ba/Thevenin_and_norton_step_1.pnghttp://upload.wikimedia.org/wikipedia/commons/b/ba/Thevenin_and_norton_step_1.pnghttp://upload.wikimedia.org/wikipedia/commons/b/ba/Thevenin_and_norton_step_1.pnghttp://en.wikipedia.org/wiki/File:Thevenin_step_2.pnghttp://en.wikipedia.org/wiki/File:Thevenin_step_2.pnghttp://en.wikipedia.org/wiki/File:Thevenin_step_2.pnghttp://en.wikipedia.org/wiki/File:Thevenin_step_2.pnghttp://en.wikipedia.org/wiki/File:Thevenin_step_2.pnghttp://en.wikipedia.org/wiki/File:Thevenin_step_4.pnghttp://en.wikipedia.org/wiki/File:Thevenin_and_norton_step_3.pnghttp://en.wikipedia.org/wiki/File:Thevenin_step_2.pnghttp://upload.wikimedia.org/wikipedia/commons/b/ba/Thevenin_and_norton_step_1.png8/9/2019 Rangkaian Ekivalen Thevenin Dan Norton
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Problem 1
Find the Thevenin equivalent with respect to the 7k ohm
resistor.
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Solution 1
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Problem 2
Let's suppose that we decide to
designate R2 as the load resistor
in this circuit
Problem: Determine the
Thevenin equivalent circuit !
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Solution 2We can determine the voltage across
the open load terminals by applying
the rules of series circuits, Ohm's
Law, and Kirchhoff's Voltage Law:
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AC Thevenin's Theorem
Any combination of sinusoidal AC sources and impedances with two
terminals can be replaced by a single voltage sourcee and a single series
impedance z.
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Norton's Theorem
Any collection of batteries and resistances with two terminals is electrically
equivalent to an ideal current sourceiin parallel with a single resistor r. The
value of ris the same as that in the Thevenin equivalentand the current ican
be found by dividing the open circuit voltage by r.
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Norton Current
The value ifor the current used in Norton's Theoremisfound by determining the open circuit voltage at theterminals AB and dividing it by the Norton resistancer.
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Problem 3
2. Find the Norton Equivalent
with respect to the 20uF
capacitor.
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In order to find the Norton Short-circuitcurrent, short the terminalswhere the
capacitor used to be. Since we are finding the
Norton Superposition analysis to find the
current through the short means we solve
several simpler circuits (one for each source.
Since there are three sources, we have threesimpler circuits to solve.
Isc = Isc15 + Isc12 + Isc20
Solution 3
I-Norton
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2). Isc12 = -12V/6k = -2mA
3). Isc20 = 20mA
Isc = Isc15 + Isc12 + Isc20= 2.5mA - 2mA + 20mA= 20.5 mA
1). Isc15 = 15V/6k = 2.5mA
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R Norton = R TheveninIn order to find the Norton Short-circuit current, short the terminals
where the capacitor used to be, since we are finding the Norton
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Mesh-current
analysis to find the
current through the
short.
I3 is the current we are particularly interested in.
Here are the mesh current equations:
KVL for i1:
12k*i1 -1k*i4 -12 = 0KVL for i2:
-15 + 4k*i2 - 4k*i3 = 0
KVL for i3:
-4k*i2 + 10k*i3 -6k*i4 + 12 = 0
KVL for i4: We have trouble writing the voltage
over the current source, so we either must addanother variable, or simply write:
i4 = 20mA
Now solve the system of equations.
Solving the second equation for i2, we get:
i2 = (15 + 4k*i3) / 4k
Now rewrite equation 3, plugging in our formulasfor i2 and i4:
-4k*(15 + 4k*i3)/4k + 10k*i3 - 6k*20mA + 12 = 0
Solve for i3:
-15 - 4k*i3 + 10k*i3 -120 + 12 = 0
i3 = 123/6k = 20.5mA