Connect with us

Basic Circuit Theory Question, independent loop definition wrong?

Discussion in 'Electronic Basics' started by ShamShoon, Jun 5, 2005.

Scroll to continue with content
  1. ShamShoon

    ShamShoon Guest

    Hello there,
    In the book "Fundamentals of Electric Circuits" by Alexander/Sadiku,
    the text says that "A loop is said to be independent if it contains at
    least one branch which is not part of any other independent loop", I
    find this definition misleading, since in the example on the previous
    page there's a circuit that has a voltage source and a resistance in
    series and those in parallel with a resistor, in parallel with another
    resistor, and then in parallel with a current source.
    If we apply the definition to the current source and voltage source, we
    get two independent loops one on the right and one on the left, however
    the middle loop shares its two elements (the two parallel resistors),
    with both loops, yet it's still independent, which shows that the
    definition is not accurate. I found it hard to find a definition of an
    independent loop.

    Any ideas? Am i missing something?
     
  2. John Bokma

    John Bokma Guest

    1
    +-----+-----+-----+
    | | | |
    V R R I
    | | | |
    R | | |
    | | | |
    +-----+-----+-----+
    2
    ?

    Has been ages and ages, but I would say that 1 and 2 are both branches
    that don't belong to the left and the right loops.
     
  3. That is a commonly accepted definition.
    That description mentions three resistances. Yet below
    you mention "the two parallel resistors". I get all three
    in parallel from the above description. For discussion, I
    propose the following diagram. (View with fixed-width font.)

    B1 B12 B2
    .--------o--------o-------o-------.
    | | | | |
    /+\ .-. .-. .-. / \
    ( ) | | | | | | ( | )
    Es \-/ R1| | R3| | R2| | Is \^/
    | '-' '-' '-' |
    | | | | |
    '--------o--------o-------o-------'
    Using the above diagram so we can refer to resistors easily,
    and pretending R3 does not exist, one set of independed loops is:
    (Es B1 R1), (R1 B12 R2), (R2 B2 Is).
    Another set would be:
    (Es B1 B12 R2), (Is B2 B12 R1), (R1 B12 R2).
    Another set would be:
    (Es B1 B12 B2 Is), (Es B1 R1), (Is B2 R2).
    There are a few more. (factorial(3) altogether)

    Let's consider just the first set, which is the one you
    appear to find confusing.
    The (Es B1 R1) loop is at least one independent loop.
    The (R1 B12 R2) loop is independent because it contains at
    least one branch (B12 or R2) which is not part of any other
    independent loop.
    The (R2 B2 Is) loop is independent because it contains at
    least one branch (B2 or Is) which is not part of any other
    independent loop.
    There are no more because all the possible branches
    have been included in the loops already listed.

    Let's consider the members in the order that leaves
    you wondering about the definition.
    The (Es B1 R1) loop is at least one independent loop.
    The (R2 B2 Is) loop is independent because it contains at
    least one branch (Is, B2 or R2) which is not part of any
    other independent loop.
    The (R1 B12 R2) loop is independent because it contains
    at least one branch (B12) which is not part of any other
    independent loop.
    You found the definition you quoted.
    Apparently, you have mssed the presence of branch
    B12. Branches B1 and B2 are redundant with Es
    and Is, so you can eliminate them from consideration.
    But branch B12 is not uniquely associated with any
    circuit element, so you can eliminate it only if you are
    willing to forget what a branch is.

    Another clue to your difficulty is this: The order I
    first elaborated satisfies the definition even if the B?
    branches are left out. Yet the order you mentioned
    does not when the B? branches are left out. So it
    must be the case that branches and circuit elements
    are not the same thing.
     
  4. John Bokma

    John Bokma Guest

    Yup, the 1st one is "voltage source and a resistance in series"

    e.g.:

    +-----+-----+-----+
    | | | |
    V R R I
    | | | |
    R | | |
    | | | |
    +-----+-----+-----+
     
  5. That's certainly a more sensible way to interpret those words.
     
  6. John Bokma

    John Bokma Guest

    I had to read it a few times, and draw it on paper :)
     
  7. Don Kelly

    Don Kelly Guest

    The problem is that you have a current source on one side. You only have two
    independent equations to solve for the two unknown currents. The third
    current is known. One approach is to replace the current source with a
    voltage source equivalent then write the loop equations - you will then have
    2 independent loops with two known voltage sources and two unknown currents.
    This is the easy way
    -----R1-----o--------R2-------o
    | -->I1 | -->I3 |
    V1 R3 V2 R2 in parallel
    with current source Is becomes
    |-------------|-------------------| V2 =R2*Is in series with
    R2

    V1=(R1+R3)I1 -R3*I3
    -V2= -R3*I1 +(R2+R3)I3 or V2=R3*I1 -(R2+R3)I3

    It can be solved using the current source but the "Ideal " current source is
    known -eliminating the 3rd loop equation.
    Put it this way with the original circuit as I see it.
    -----R1-----o----------o---------
    | -->I1 | --> | |
    V1 R3 I3 R2 Is ^
    |-------------|------------|---------|

    Left loop V1=(R1+R3)I1 -R3*I3 where I3 is the loop current circulating
    between the parallel resistors.
    Center loop 0 =-R3I1 +(R3+R2)I3 +R2*Is
    Right loop: V2 =R2*Is +R2*I3
    V2 is unknown and Is is known. so the right loop equation simply gives the
    unknown voltage V2 in terms of I3 and Is This leaves you with the two
    equations, left and center in terms of the two unknown currents. . Knowing
    I2 then allows you to solve the left and center loop equations
    simultaneously- that is - there are only two independent loop equations in
    fact as only two of the 3 current variables are unknown. The right side
    equation can be used to solve for the voltage V2 but it isn't necessary once
    you solve for I1, I3
    Note that these become
    center: R2*Is =V2 =R3*I1 -(R3+R2)I3 which is the same as above
    left equation is unchanged from above.

    Anytime you have a current source and are using loop equations, the known
    current source eliminates a variable and hence eliminates a simultaneous
    equation (making life easier)
     
  8. John Bokma

    John Bokma Guest

    Use mono spaced fonts for this. Easier, and if someone uses a proportional
    font, he/she can fix it by copy pasting it to an editor. The other way
    around requires access to the same font :-(
     
  9. Don Kelly

    Don Kelly Guest

    Thanks. the dumn thing is that I knew this.
     
Ask a Question
Want to reply to this thread or ask your own question?
You'll need to choose a username for the site, which only take a couple of moments (here). After that, you can post your question and our members will help you out.
Electronics Point Logo
Continue to site
Quote of the day

-