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elica85
Group Title
what is the current thru the 2ohm resistor in the figure? what power is dissipated by the 2ohm resistor in the figure?
 2 years ago
 2 years ago
elica85 Group Title
what is the current thru the 2ohm resistor in the figure? what power is dissipated by the 2ohm resistor in the figure?
 2 years ago
 2 years ago

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elica85 Group TitleBest ResponseYou've already chosen the best response.0
ignore the current direction if it's wrong
 2 years ago

suyash011 Group TitleBest ResponseYou've already chosen the best response.0
dw:1332192864001:dw Now using Kirchoff's Current Law (x15)/4 + (x0)/2 + (x+120)/4 = 0 x15 + 2x + x+12 = 0 4x = 3 x = 3/4 I1 = (150.75)/4 = 14.25/4 = 3.5625 Amperes I2 = 0.375 Amperes I3 = 3.1875 Amperes
 2 years ago

elica85 Group TitleBest ResponseYou've already chosen the best response.0
the loop law is summation of potential difference=0...V=IR so what's the first line of equation?
 2 years ago

radar Group TitleBest ResponseYou've already chosen the best response.1
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.
 2 years ago

radar Group TitleBest ResponseYou've already chosen the best response.1
The power can now be calculated:\[P=i ^{2}R=(.375)^{2}2=.28125 watts\]
 2 years ago
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