## elica85 3 years ago what is the current thru the 2ohm resistor in the figure? what power is dissipated by the 2ohm resistor in the figure?

1. elica85

2. elica85

ignore the current direction if it's wrong

3. suyash011

|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

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

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.