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  1. Join Date
    Oct 2002
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    2,716
    #41
    After learning Thevenin's Theorem ... they also teach Norton's Theorem which states

    Any linear electrical network with voltage and current sources and only resistances can be replaced at terminals A-B by an equivalent current source Ino in parallel connection with an equivalent resistance Rno.

  2. Join Date
    May 2013
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    5
    #42
    OT na hehe

    Sent from my HTC One X+ using Tapatalk 2

  3. Join Date
    Oct 2011
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    26,787
    #43
    Prove first that the battery is a resistive load. kung hindi, parallel connection yan.

  4. Join Date
    Oct 2002
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    #44
    when you only have two components in a circuit, e.g, a battery and a motor, the connection is both series and parallel

    series because they have the same current, parallel because they have the same voltage

  5. Join Date
    Jun 2012
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    1,490
    #45
    Quote Originally Posted by Retz View Post
    Kung ang dead battery ay considered na load edi magiging simple series circuit yan. But could we consider the dead battery as a load? Baterya or source supple un eh mahina lang nga voltage.
    We could not consider the dead/low battery as a source kasi current is flowing into it. Saka sa set-up na jump-start, physically naroon lang yung dead/low battery but electrically, it is not functioning as a battery but another resistor or load.

    BTT: I think yung mga commercial jumper ratings are good enough. Ang importante, maganda maganda ang quality ng cable at maganda pagka crimp ng clamp.

  6. Join Date
    Oct 2011
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    #46
    ^

    Granting na considered as resistive load ang dead battery, edi tama nga mga mekaniko daig pa nila ang ee or ece graduate.

  7. Join Date
    Oct 2002
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    2,716
    #47
    maybe ... old folks say "series" because it is easier to remember

    btt: on cheaper jumper cables, the usual fault is the crimping ... soldering the copper conductor to the clamp will fix it

  8. Join Date
    Jun 2012
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    1,490
    #48
    Quote Originally Posted by Retz View Post
    Prove first that the battery is a resistive load. kung hindi, parallel connection yan.
    Opps, balik tayo sa Physics 101.

    Parallel batteries:
    I(total)= I(batt1) + I(batt2)


    In the case of a jump start:
    I(total) = I(batt1) - I(batt2)

    Where batt2 is the dead/low battery.

  9. Join Date
    Nov 2009
    Posts
    3,474
    #49
    Haha.. pinag debatehan pa talaga ang series o parallel

    In Series, the current throughout each components is the same

    While in Parallel, the Voltage is the same across all components be a load or a second battery.

    As for the battery in parallel, it will equalize kung medyo mahina yung isa.


    Kung AC circuits yan dyan kayo mag debate. The battery will now consist of capacitor resistor and inductor (battery impedance), while the starter will be purely inductive.

  10. Join Date
    Oct 2011
    Posts
    26,787
    #50
    ^

    Sir, question lang. Pag ba sinabing dead battery as in zero(0) ang voltage reading nyan?

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