Rangkaian Ekivalen Thevenin Dan Norton

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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/
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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

    http://hyperphysics.phy-astr.gsu.edu/hbase/electric/visource.htmlhttp://hyperphysics.phy-astr.gsu.edu/hbase/electric/visource.html
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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

    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/voldiv.htmlhttp://hyperphysics.phy-astr.gsu.edu/hbase/electric/voldiv.htmlhttp://hyperphysics.phy-astr.gsu.edu/hbase/electric/visource.htmlhttp://hyperphysics.phy-astr.gsu.edu/hbase/electric/thevenin.html
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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.html
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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.png
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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.

    http://hyperphysics.phy-astr.gsu.edu/hbase/electric/visource.htmlhttp://hyperphysics.phy-astr.gsu.edu/hbase/electric/visource.html
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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.

    http://hyperphysics.phy-astr.gsu.edu/hbase/electric/visource.htmlhttp://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/visource.html
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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.

    http://hyperphysics.phy-astr.gsu.edu/hbase/electric/norton.htmlhttp://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/norton.html
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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