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Voltage Polarity Example

Discussion in 'Electronic Basics' started by [email protected], Nov 13, 2005.

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  1. Guest


    I'm pretty new to this stuff. I'm trying to teach myself with "Lessons
    In Electric Circuits" over at

    Anyway, I've made it to Chapter 6 in Book 1, and I'm a little confused
    about a voltage polarity example. An image of the circuit I'm confused
    about can be found at:

    The author says that the voltage between points 3 and 4 is +/- 32,
    depending on whether you're measuring point 3 with respect to point 4,
    or vice versa. This doesn't seem very intuitive to me...

    It seems like the voltage between the two points should be
    (35V-20V)-(25V-13V) = 15V-12V = 3V. I got that number by calculating
    the voltage of point 3 with respect to point 8 (20V drop across
    resistor), and the voltage of point 4 with respect to point 5 (13V drop
    across resistor), then subtracting them because "voltage is relative."

    Then again, I don't understand why you have to connect points 8 and 9
    (while not connecting points 3 and 4) in order to use Kirchhoff's Law,
    or how this circuit would actually work in real life, so that's
    probably why I just don't get it. Can somebody help me out?

    Thank you very much.
  2. The bottom line (points 7,8,9,10) is the common point - all points on
    that line are at the same voltage - so you should call that line
    "common", or "ground", and calculate all voltages relative to that
    line. You calculated the voltage at 4 relative to 5, rather than
    relative to 9, and mixed up signs somewhere.

    If you calculate the voltages at 3 and 4 relative to 7,8,9,10, it
    should work out correctly.

    Point 3 is 20 volts positive from the common point (7,8,9,10), and
    point 4 is 12 volts negative from the common point, so there is 32
    volts between them, with 3 being more positive. So, if you declare 3
    as "zero volts), 4 is at -32. But if you declare 4 as "zero volts", 3
    is at +32.
    Well, you could connect any point in one loop to any point in the
    other - but you would get a different problem for each pair of points
    you connect.

    Before you can measure the voltage between two points, there must be
    some path for current to flow between those points. In this case, the
    author chose to make the path between points 8 and 9.

    If you remove the connection between 8 and 9, you won't measure any
    voltage between 3 and 4, or between 8 and 5, or in general between any
    point on the left loop, and any point on the right loop. You would,
    however, still be able to measure voltage differences within either
    Peter Bennett, VE7CEI
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  3. Chris

    Chris Guest

    Just try visualizing the circuit a little differently, and it might
    become clear (view in fixed font or M$ Notepad):

    | +|
    | .-.
    | | | 15V
    | | |
    | '-'
    +| -|
    35V --- 3o---------------.
    - +| |
    | .-. |
    | | | 20V |
    | | | |
    | '-' |
    | -| / \
    o----------o ( V )
    |7/8/9/10 +| \_/
    | .-. |
    | | | 12V |
    | | | |
    +| '-' |
    25V --- -| |
    - 4o---------------'
    | +|
    | .-.
    | | | 13V
    | | |
    | '-'
    | -|
    created by Andy´s ASCII-Circuit v1.24.140803 Beta

    This is exactly the same circuit, but by adjusting your perspective a
    little, it's apparent that there's 32V between the two points of the
    voltmeter. "Adjusting your perspective" is a lot easier with a just
    little practice.

    You applied the math incorrectly. In fact, if you visualize node
    7/8/9/10 as 0V, then node 1/2 is at +35V, and node 5/6 is -25V. That
    would make node 3 at +20V, and node 4 at -12V relative to the
    artificial GND at node 7/8/9/10. The difference is 32V.

    This also works no matter what node you call 0V/GND. I'll leave that
    as an instructional exercise for you. And since you've evoked the
    spirit of J.S. Mill, you might also want to ponder the philosophical
    implications of this business of arbitrarily putting 0V/GND at any node
    in the circuit, and everything still works the same. You will greatly
    profit from this, which would make the neoliberal spirit of Mr. Mill

    Good question, and good luck in your studies.
  4. Pooh Bear

    Pooh Bear Guest

    You mean (35V-15V) - (25V-13V) except that the polarity of the second term
    is reversed too.


    So it's (35V-15V) - ( - ( 25V-13V))

    or 20V + 12V = 32V.

    And the way the meter's drawn it'll actually read -32V since the postive
    terminal is connected to the negative voltage.
    If 8 and 9 aren't connected there is no common current path and there would
    be effectively no indication. It helps a bit if you think of the meter
    needing to draw some current to get this idea straight.

  5. Graham Knott

    Graham Knott Guest

  6. Jasen Betts

    Jasen Betts Guest

    The voltage source on the right is "upside down."
    points 7,8,9,and 10 are connected by conductors of negligable resistance.
    (that's what the straight black lines represent)

    for that reason they have the same potential (voltage)

    point 3 is 20V more positive than point 8 , point 9 has the same potential
    as 9 and point 9 is 12V more positive than point 4

    thus point 3 is 32V more positive than point 4

    if either voltage source was reversed the voltages on the resistors in that
    half of the circuit would reverse and the 2V you predicted would be present
    between points 3 and 4.

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