Basic DC Circuits Analysis and Theory

Discussion in 'Electronics Homework Help' started by jkele, Nov 21, 2010.

  1. jkele

    jkele

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    Hello All,

    Trying to get my head around some theory to prepare for my catch up exam on electronics stuff. Anyway, i have attached an image since in circuit analysis we assume that the voltage is the same everywhere in a wire (before we reach a circuit element), this should imply that both the Va are at the same voltage right? Lets say its 4V, hence the potential difference between these two points is 0 right? So we have no electric force therefore shouldn't there be no current flow? - However doesn't this contradict the meaning of short circuit which says that if the voltage between two points is 0 then a lot of current flows.

    To summarize the question - If two points in a wire are at the same voltage then the difference between them is 0 and this is a short circuit which causes a lot of current to flow however if we have 0V between two points then we have no electric force hence no current flows?

    All help is appreciated.
     

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    jkele, Nov 21, 2010
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  2. jkele

    barathbushan VIP Member

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    Untitled.png

    THINK ABOUT YOUR CONCEPT OF SHORT CIRCUIT AGAIN !!

    HINT: current chooses the least resistive path (R=0) , just apply ohms law
     
    Last edited: Nov 21, 2010
    barathbushan, Nov 21, 2010
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  3. jkele

    barathbushan VIP Member

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    PUZZLE : OK lets think of potential energy in terms of physics , in a horizontal pipe , containing water , the water doesn't flow if left freely , since the pipe is horizontal [a.k.a same potential between the two ends] , now if you connect a water pump or motor , you can circulate the water , but even now the ends of pipe are at same potential , "BUT THE WATER FLOWS" , HOW ???
    THINK !!
     
    barathbushan, Nov 21, 2010
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  4. jkele

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    You make a number of incorrect assumptions.

    Firstly you assume that the voltage will be the same at the points you refer to as Va and Va, and Vb and Vb. You make a series of errors here, since you reference those voltages to nothing. However, let me assure you that in this circuit, there WILL be a voltage drop across Va and Va, and Vb and Vb because wires have resistance. Practically speaking the resistance, and therefore the voltage drop is almost always too small to be concerned with, and for most purposes you can treat them as zero.

    Even if the wires had no resistance (or the 2 points Va were coincident), your assertion fails because the function for ohms law is undefined for R = 0 where you're attempting to find I.

    The easiest solution is to assume it's a series circuit with a known voltage across Vload, and determine the current from that. Because it's a series circuit, the currents in the other parts is easily calculated.
     
    (*steve*), Nov 21, 2010
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  5. jkele

    Laplace VIP Member

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    @jkele: "If two points in a wire are at the same voltage then the difference between them is 0 and this is a short circuit which causes a lot of current to flow however if we have 0V between two points then we have no electric force hence no current flows?"

    You ask the question as though the wire between two points is the whole circuit. But a circuit is a circle and the wire between two points is only part of the circle. Instead of thinking of the wire, you need to think of the other part of the circle, i.e., what is the Thevenin or Norton equivalent circuit for the other part of the circle.
     
    Laplace, Nov 21, 2010
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  6. jkele

    lechesa

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    Mr

    your analysis is correct jkele. but u jst misd a litle theory remember, any circuit element has some resistance. part of the wire btwn Va's has resistance whose its value is given by [ R= p(L/A) ]. do u remember ths relationship now?
     
    lechesa, Apr 27, 2011
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  7. jkele

    lechesa

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    your analysis is correct jkele. but u jst misd a litle theory remember, any circuit element has some resistance. part of the wire btwn Va's has resistance whose its value is given by [ R= p(L/A) ]. do u remember ths relationship now?
     
    lechesa, Apr 27, 2011
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