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anonymous

  • one year ago

I have the circuit that is shown in the following picture http://i.imgur.com/mcNe4b4.jpg. Currently I'm not sure that I understand two things about it. Firstly how do I know the polarity over the resistance. Secondly they ask what the power is for each circuit element. I found that the current over the resistance is 2A and thus Pr = 20 W, Pv=20 W. Thus the current source have to be -40W, but given the relation that P=v(t)i(t) I get -(2*10)=-20 W.

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  1. anonymous
    • one year ago
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  2. anonymous
    • one year ago
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    For the first question is it, perhaps because the current is moving clock-wise and thus there is a voltage drop over the resistance which determines the voltage polarity?

  3. radar
    • one year ago
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    Is the element on the left a current source, or just an ammeter in the circuit measuring 2 amps?

  4. radar
    • one year ago
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    I am thinking, that it is an ammeter, and the current is electron flow rather than conventional current direction. The power in the resistor is as you say 20 watts (I^2R) and the power provided by the battery is also 20 watts (EI) and their is negligible voltage drop across the ammeter and the power used by the ammeter can be ignored.

  5. radar
    • one year ago
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    |dw:1437765877388:dw|

  6. anonymous
    • one year ago
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    Thank you for replying, the element on the left is a current source and the answer is supposed to be that it provides 40W (Pc=-40 W) while both the resistor and voltage source absorbs 20 W each.

  7. radar
    • one year ago
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    That complicates the matter. It is strange that the current source produces the same current as would be allowed by the voltage source and the resistor. The only way I can see that, is the current source is 4 amps in the opposite direction that the current is produced by the battery.???? Very confusing. That would mean there is 20 volts across the current source thus there would be 40 watts generated by that source with 20 watts being absorbed by the resistor and 20 watts absorbed by the battery. ??? Not for sure how that would work.

  8. radar
    • one year ago
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    |dw:1437769374761:dw|

  9. radar
    • one year ago
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    |dw:1437769695576:dw|

  10. anonymous
    • one year ago
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    Thanks, I think I get it now. The voltage over an ideal current source is whatever is necessary in order to produce the amount of power (watts) that is connected to it as a load. So in this case the load is 40 W (20 W +20 W) and thus, as you illustrated the voltage across the ideal current source is 20 V i.e. the current source provides 40 W.

  11. radar
    • one year ago
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    I am thinking that is what is happening, a little more complicated than I first suspected. I liked it better if it had not of been a current source, but just a ammeter lol.

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