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Assumed polarities in KVL

  1. Sep 6, 2011 #1
    i generally get confused in polarities of inductors and capacitors while doing KVL.
    eg about a series RLC circuit ,the equation turns out to be ( Ldi/dt + iR + Vc = 0 )
    this is the equation about assumed 'i' if we assume both capacitor and inductor in charging mode....
    ....
    what if I assume inductor is discharging....then how can i reach the above equation......?

    similar problem happens to me with circuits having only 1 active component..........
    .
    .
    .
    can anyone enlighten me on this ..?

    thanks
     
  2. jcsd
  3. Sep 6, 2011 #2

    gneill

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    Staff: Mentor

    Writing KVL around a loop is all about recording the voltage "responses" of the components to the loop current. You take an assumed direction for the loop current and follow that direction around the loop. If you are given specifics about a given component being in the process of charging or discharging, then that only serves to determine the direction that you should assume for the current.

    Resistors respond to the instantaneous value of the current with a voltage drop I*R. Inductors respond to the rate of change of the current and produce a drop of L*dI/dt. Note that the assumed direction of the current also specifies the direction (sign) of dI/dt --- positive is an increase in current in the direction of the assumed current.
     
  4. Sep 6, 2011 #3
    how does assumed direction of current specifies di/dt...?
     
  5. Sep 6, 2011 #4

    gneill

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    Staff: Mentor

    An increase in current, di, is an increase in the same direction as the assumed current flow. A decrease is a change in the opposite direction. It's these increases or decreases which provoke the component's response in terms of voltage change, and thus the polarity of those changes.
     
  6. Sep 6, 2011 #5
    i m not able to get....
    in LCR circuit after having assuming direction of current say clockwise;
    what will be the assumed polarities of inductor and capacitor knowing say initial capacitor voltage Vo
     
  7. Sep 6, 2011 #6
    with initial capacitor voltage and direction of assumed current i can get sign of dq/dt
    (depending upon current entering or exiting from positive plate ) but what about di/dt for inductor...?
     
  8. Sep 6, 2011 #7

    gneill

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    The definition of the characteristics of the inductor give the answer:

    attachment.php?attachmentid=38608&stc=1&d=1315320909.gif

    An increase in the current i causes a voltage to appear across the inductor as shown. If di/dt were negative instead, then the resulting voltage polarity would be reversed, too.
     

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  9. Sep 6, 2011 #8
    in the attached figure.., let i be in clockwise.
    so how can i get di/dt ..
    again when battery is removed how can i get di/dt..
    cause it is what di/dt that will determine polarity of inductor
     

    Attached Files:

  10. Sep 6, 2011 #9
    in this figure with assumed polarities and current how can i get the corresponding differential equation...?
     

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  11. Sep 6, 2011 #10

    gneill

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    When you're working from given initial conditions then you need to know something about the properties of the components.

    Suppose that you being with the battery disconnected, no charge on the capacitor, and (obviously) no current at all flowing in the circuit. When the battery is first applied, the inductor resists change in current -- no current flows at the very first instant. But the current will be changing, increasing in the 'forward' direction, being driven by the polarity of the battery. This di/dt will cause a voltage to appear across the inductor according to v = L*di/dt. Initially this voltage will be equal in magnitude to the supply voltage.

    The capacitor, being initially uncharged, will have zero volts across it at the first instant. As the current begins to flow, it will gradually build up a voltage. Eventually the capacitor will be fully charged with a voltage equal to the supply voltage. No further current will flow. With no current, the voltage across the resistor will be zero. With no change in current (di/dt) the voltage across the inductor will be zero.

    When the battery is removed there will be no closed circuit -- no closed path for current to flow. So the capacitor will just sit there, fully charged.
     
  12. Sep 6, 2011 #11

    gneill

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    You need to state the initial conditions. Is there a current already flowing? Is the capacitor charged? If so, what is the polarity of the voltage on the capacitor?

    The initial conditions will determine the 'directions' in which things happen.
     
  13. Sep 6, 2011 #12
    ok i will try on a few questions , if doubt persists i will ask..
    thanks
     
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