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anonymous

  • one year ago

Logic in the following circuit I will post it

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  1. anonymous
    • one year ago
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  2. anonymous
    • one year ago
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    |dw:1444493174253:dw|

  3. anonymous
    • one year ago
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    @ganeshie8 @Lyrae

  4. amistre64
    • one year ago
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    the only thing that comes to mind for me is superposition. remove all but one of the power supply, and compute the parts. do this for each power supply and then add the results for each part. not too sure if its applicable tho, its the only thing that comes to mind tho.

  5. Lyrae
    • one year ago
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    \(I_{ab} = I_1+I_2 \) is (probably) because of Kirchhoff's current law, i.e. the current entering any junction is equal to the current leaving that junction. |dw:1444511469312:dw| If we analyse this circuit with KCL we have two incoming currents in junction a, \(I_1\) and \(I_2 \). The only way for these currents to flow given they "choose" the path of least resistance is towards b. In a more complex circuit you might have to calculate the currents (and their direction) by using both KCL and Kirchhoff's voltage law to create and solve a set of linear equations. A small guess is also that understating Norton and Thévenin's equivalents might be helpful for that picture in your first post.

  6. Lyrae
    • one year ago
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    |dw:1444512913493:dw|

  7. radar
    • one year ago
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    |dw:1444531879314:dw|Not for sure what logic is supposed to be demonstrated here, just the fact that the various voltmeters, and ammeters are working as analysis would show.

  8. anonymous
    • one year ago
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    I mean: it doesn't make sense for being zero voltage and drew some current. for H bridge when zero voltage = zero current, here another point of view is the following |dw:1444544265619:dw| node analysis \[\frac{ b-v }{ R } + \frac{ b-v }{ R } +\frac{ b-a }{ 0 }=\frac{ b-v }{ R } + \frac{ b-v }{ R }+\frac{ 0 }{ 0 }\] then 0/0 = 2i !!

  9. anonymous
    • one year ago
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    -2i*

  10. radar
    • one year ago
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    Note that the voltage V(b-a) is zero as indicated by the voltmeter (see original circuit with the measuring devices), and the current in the conductor is 2 ma as indicated by the current meter in that conductor (the combining of the two 1 ma loops. Note that in most instances, for current measurements, the circuit is opened at the point where the current meter is inserted, and reconnected after the measurement. Also the voltage between two points on a conducting wire is for all practical purposes ( on a good conductor is zero;|dw:1444570598941:dw|

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