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elica85

  • 4 years ago

what is the current thru the 2ohm resistor in the figure? what power is dissipated by the 2ohm resistor in the figure?

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  1. elica85
    • 4 years ago
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  2. elica85
    • 4 years ago
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    ignore the current direction if it's wrong

  3. suyash011
    • 4 years ago
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    |dw:1332192864001:dw| Now using Kirchoff's Current Law (x-15)/4 + (x-0)/2 + (x+12-0)/4 = 0 x-15 + 2x + x+12 = 0 4x = 3 x = 3/4 I1 = (15-0.75)/4 = 14.25/4 = 3.5625 Amperes I2 = 0.375 Amperes I3 = 3.1875 Amperes

  4. elica85
    • 4 years ago
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    the loop law is summation of potential difference=0...V=-IR so what's the first line of equation?

  5. radar
    • 4 years ago
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    Or you could Thevenize it. Mentally remove the 2 Ohm resistor, compute the voltage that would be present in the simple series ckt with the two sources and the two 4 Ohm resistors.|dw:1332269821486:dw| you will come up with a -1.5 volts. That is the Thevenin voltage. Now mentally short the two sources and look into the circuit (where the 2 Ohms was. You will now see the 2 4 Ohm resistors in parallel giving you the Thevinen resistance of 2 Ohm. This is what you now have:|dw:1332270017523:dw| Connecting the 2 Ohm resistor back in, you can easily calculate the current through it: 1.5/4=.375 A or 375 ma.

  6. radar
    • 4 years ago
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    The power can now be calculated:\[P=i ^{2}R=(.375)^{2}2=.28125 watts\]

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