To Whom it may concern, Please excuse the horrible image. I am trying to find I1

Question

To Whom it may concern, Please excuse the horrible image. I am trying to find I1, I2,and I3 from this drawing. I am told that I can use Ohms Law and Ican use kirchoff's rules to solve for this. The numbers in ohmsstarting from the bottom left corner and going around are 13, 7,18, 5, 9, and 12 The voltage starting from the bottom left corner and goingaround are 6, 12, and 10. In the center of the diagram the numbers are 15 volts and 21ohms. by the way, in the top box I forgot to add inside a circulararrow in the clockwise direction. This problem confuses me to death. Any help would beappreciated. Cheers

Explanation / Answer

. 1.   I recommend using loop analysis to solvethis problem. Let's entitle the Loop in the top box L1 andthe lower box L2, both are in a clockwise direction internal to theboxes. . 2.   L1analysis   ;   start at the lowerleft corner and go around the box clockwise, keep the current andresistance on the LHS of the equation and sources on the RHS of theequation 3.   (7+18+5+12)*L1 - (12)*L2 = -12 +15            ;      42*L1- 12*L2 = 3 4.   L2analysis   ;   start at the lowerleft corner and go around the box clockwise, keep the current andresistance on the LHS of the equation and sources on the RHS of theequation 5.   -(12)*L1 + (13+12+9+12)*L2 =6-15-10         ;         12*L1 -46*L2 = 19 . 6.   Solve for L1 and L2 using simultaneousequations 7.   L1 = -426.2E-3 ampere 8.   L2 = -50.3E-3 ampere . 9.   I1 = L2 - L1 = -375.9E-3 ampere 10.   I2 = -L1 = 426.2E-3 ampere 11.   I3 = L2 = -50.3E-3 ampere . 4.   L2analysis   ;   start at the lowerleft corner and go around the box clockwise, keep the current andresistance on the LHS of the equation and sources on the RHS of theequation 5.   -(12)*L1 + (13+12+9+12)*L2 =6-15-10         ;         12*L1 -46*L2 = 19 . 6.   Solve for L1 and L2 using simultaneousequations 7.   L1 = -426.2E-3 ampere 8.   L2 = -50.3E-3 ampere . 9.   I1 = L2 - L1 = -375.9E-3 ampere 10.   I2 = -L1 = 426.2E-3 ampere 11.   I3 = L2 = -50.3E-3 ampere .

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