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    Series/Parallel Circuit Simplification:

    Kirchoff, Thevenin & Norton

    Session 1d of Basic Electricity

    A Fairfield University E-CoursePowered by LearnLinc

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    3/30/2002 Basic Electricity 3

    Basic Electricity

    (Continued) Text: Electricity One-Seven, Harry Mileaf,

    Prentice-Hall, 1996, ISBN 0-13-889585-6(Covers several Modules and more)

    References: Digital Mini Test: Principles of Electricity Lessons

    One and Two, SNET Home Study Coordinator, (203)771-5400

    Electronics Tutorial (Thanks to Alex Pounds [email protected] )

    Electronics Tutorial (Thanks to Mark Sokos at

    [email protected] )

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    3/30/2002 Basic Electricity 4

    Section 1:

    Electron Flow and Resistance

    0BJECTIVES: This section introduces fivebasic electrical concepts as well as the underlyingatomic structure of electrical materials.

    Conductance(G),

    Resistance (R),

    Current (I), Power (P), and

    Electromotive force (E) or voltage (V).

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    3/30/2002 Basic Electricity 5

    Section 1 Schedule:Text 1.1 1.39

    Text 1.40 1.68

    Text 2.1 2.52

    Text 2.53 2.98

    2.99 2.115

    2.116 2.133

    1.42, 1.63, 2.5, 2.129

    Sokos

    Atoms, Charge and Current

    Conductivity (G), Electric Fields

    and Electromotive Force (EMF)

    Resistance (R), Conductance (G),

    Ohms Law () & Power (Watts)

    Resistors in Series and Parallel andWorking with Equations

    Series / Parallel Simplification

    Kirchoff, Thevenin & Norton

    Review: The Water Model

    Session a 03/04

    03/06 & 03/08

    were Math

    Tutorials

    Session b 03/11

    Session c 03/13(lab - 03/16, sat.)

    Session d 03/18

    Session e 03/20

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    3/30/2002 Basic Electricity 7

    Series-Parallel Circuits

    A mixture of series and parallel circuit elements A sequence of small steps will find an

    equivalent circuit.

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    3/30/2002 Basic Electricity 8

    S-P Step 1

    Redraw the circuit to show series and parallelelements clearly.

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    3/30/2002 Basic Electricity 9

    S-P Step 2

    Combine some elements to simplify the circuit Here R6 and R7 (parallel) are replaced with theirequivalent resistance ( 2 Ohms)

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    3/30/2002 Basic Electricity 10

    S-P Step 3

    Now we can add R5 and R6-7 (series) to furthersimplify the circuit

    Now we have three resistors in parallel

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    3/30/2002 Basic Electricity 11

    S-P Step 4

    Replacing R3, R

    3and R

    5-6-7(parallel) with their

    equivalent resistance (1 Ohm) yields a simple

    series circuit which simplifies by adding and

    were done.

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    3/30/2002 Basic Electricity 12

    S-P: Example 1 1/30 + 1/20 = 5/60 = 1/12

    8 + 10 + 12 = 30 Ohms

    Itotal = 30 volts / 30 ohms = 1 amp

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    3/30/2002 Basic Electricity 13

    S-P: Example 2

    6

    3

    5

    Reff= 6.667

    and I = 6 amps

    Find the equivalent resistance and use Ohms law to get

    the total current in the battery.

    2/3

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    3/30/2002 Basic Electricity 14

    S-P Example 3

    Small steps to find ER3

    Req = 50 Ohms2, Itotal = 2 Amps

    IR3 = 1 Amp (1/2 Itotal)

    ER3 = 20 Volts 60 30

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    3/30/2002 Basic Electricity 15

    Voltage Divider

    Since the total currentflows through both

    resistors, the bigger

    resistor has the largershare of the total

    voltage. E2 = Ein* R2 / Rtotal

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    3/30/2002 Basic Electricity 16

    Current Divider

    The smaller resistor gets the larger share of the current.

    I1 = Itotal * 1/R1 / (1/R1 + 1/R2) or

    I1 = Itotal * R2 / Rtotal

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    3/30/2002 Basic Electricity 17

    Kirchoffs Voltage Law

    The sum of all thevoltages around aloop is zero

    Be careful to take signsinto account

    Starting at the top left

    corner and goingclockwise:

    20*I 75 + 10*I + 60 = 0

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    3/30/2002 Basic Electricity 18

    Kirchoffs Current Law

    The sum of all currents into a node equals zero.

    Again watch out for signs

    (direction of current flow)

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    3/30/2002 Basic Electricity 19

    Using Kirchoff

    Use voltage divider or,

    Kirchoffs Voltage Law and a current divider, or

    Set up and solve Loop Equations

    2Itot

    = 6/4 Amps

    V2 = 3 volts

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    3/30/2002 Basic Electricity 20

    Superposition

    Linear systems (R, L and C circuits are linear) You can deal separately with each power source andthen add the resulting currents to get the total result

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    3/30/2002 Basic Electricity 21

    Thevenin Equivalent Circuits

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    Thevenin(Continued)

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    Norton Equivalent Circuits

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    Next Class

    Module Review: Electron Flow and Resistance

    Conductance(G),

    Resistance (R),

    Current (I), Power (P),

    Electromotive force (E) or voltage (V), and

    DC Circuit Analysis.

    Quiz (via email to see how were doing)


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