parallel power supply

Discussion in 'Homework Help' started by recca02, Apr 2, 2007.

  1. recca02

    Thread Starter Senior Member

    Apr 2, 2007
    1,211
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    this trivial question just popped up in my mind ,its not a hw question
    if we connect 2 ideal power sources ( 2 batteries for example of same voltage V)
    to a common load (a bulb of some power rating) how much current will
    flow through the bulb . if we consider that I amp flows if only one battery was
    there. if R is resistance of bulb assumed. (I=V/R).also assume conductors to have 0 resistance.

    i did not try it on paper but have given some thought to it
    -- parallel batteries should share the load and hence equal current should flow
    case1- since parallel batteries are connected voltage is equal to one battery
    acting alone hence current= V/R=I
    so shud I/2 current flow in each loop?
    case 2-
    since two batteries are connected each provides V/R=I
    and current thru bulb equals --2I
    case 3-
    apply superposition theorem
    since voltage sources are ideal while considering each source separately
    the removed voltage crkt is shorted hence current through bulb is zero in
    both consideration. so is the current thru bulb 0+0=0 amps

    please reply..
     
  2. beenthere

    Retired Moderator

    Apr 20, 2004
    15,815
    282
    If the batteries are in parallel, then the current through the bulb won't change. In series, the voltage will double. The current will not, as the filament has a non-linear response to current. It will most likely burn out, but, if it doesn't, the current will not be twice that for the original voltage.
    If the batteries are in series opposing, then not surrent flows through the bulb.
     
  3. thingmaker3

    Retired Moderator

    May 16, 2005
    5,072
    6
    With two identical batteries in parallel, half as much current is sourced by each battery. Ohm's Law still governs the current through the bulb.
     
  4. recca02

    Thread Starter Senior Member

    Apr 2, 2007
    1,211
    0
    thanx everybody.:)
    so for a series battery connection with a load having a linear response
    the current must double ,right?
     
  5. thingmaker3

    Retired Moderator

    May 16, 2005
    5,072
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    Yes. Current will always and forever be the product of voltage divided by impedence. That's why we call Ohm's Law a "Law" instead of a "suggestion.";)
     
  6. John Luciani

    Active Member

    Apr 3, 2007
    477
    0
    Your case three (superpostion) is correct.
    The "ideal" battery has a resistance of 0 and will shunt
    the current away from the bulb.

    (* jcl *)

    ---

    http://www.luciani.org
     
  7. JoeJester

    AAC Fanatic!

    Apr 26, 2005
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  8. recca02

    Thread Starter Senior Member

    Apr 2, 2007
    1,211
    0
    sir, if the circuit gets shorted and currrent is shunted will the bulb glow?:confused:
    (in view of my case 3 assumption)
    ne ways thanks again.
     
  9. John Luciani

    Active Member

    Apr 3, 2007
    477
    0
    The current takes the path of least resistance.

    An "ideal" short and an "ideal" voltage source each have a resistance of 0
    and will shunt current away from the bulb (which I am assuming has
    a resistance > 0).

    NB: If you try this with a "real" battery you will probably get a hot or exploding
    battery since your battery has a resistance > 0.

    (* jcl *)


    -------
    www.luciani.org
     
  10. recca02

    Thread Starter Senior Member

    Apr 2, 2007
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    0
    then what abt the case of two infinite buses (i mean main supplies )
    synchronized properly to work in parallel connected to same load
    i believe they can act as ideal sources (probably huge wire resistance
    cant be neglected so let us consider the load is having a very high resistance)
    will the load not get any supply, i might be wrong but for huge loads connecting
    two generators or two transformers (on distribution side) in parallel is a common practice.
     
  11. John Luciani

    Active Member

    Apr 3, 2007
    477
    0
    Paralleling transformer outputs is done to increase the output
    current.

    For superposition you replace an "ideal" voltage source with a short
    circuit. You do not replace a transformer output with a short circuit.

    (* jcl *)

    ----
    http://www.luciani.org
     
  12. thingmaker3

    Retired Moderator

    May 16, 2005
    5,072
    6
    So, no, reducing a circuit to an ideal shorted battery won't tell us anything. If we use an estimate of the resistance of the wire, superposition will work.
     
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