http://i.imgur.com/IDrSyCY.png ( THe VOLTAGE AT V1 is same as V2 I mean the positive terminal and negative terminal Is just like V2 I didnt see that when I editin in paint.) please help me deal this problem. I solve this problem but I don't know where I my solution got wrong.. I try multisim to check my answer but in Resistor 10 it measure 10 amps in multisim. I dont know where I got wrong in my solution. Please help me. PS I dont want to use mesh/nodal . I just want to master the Kirchoff's law Thanks to all of you !!!! I came up with my solution. loop A: 100 V - 10(I1) - 5(I2) = 0 10(I1) + 5(I2) = 100 V loop B 50 V - 2(I1 - I2) - 2(I3) + 5(I2) = 0 (* -1 SO THAT I WILL GET POSITIVE VALUE OF I1) 2(I1) - 7(I2) + 2(I3) = 50V loop C: 50V - 3(I1 - I2 - I3) + 2(I3) =0 3(I1) -3(I2) -5(I3) = 50 V USING DETERMINANTS I GET THE VALUE OF D AS: (I1) (I2) (I3) (V) 10 5 0 | 100 2 -7 2 | 50 3 -3 -5 | 50 D= 490 SOLVING FOR I1 USING DETERMINANT D1 = 100 5 0 50 -7 2 50 -3 -5 D1 = 5850 I1 = D1/ D = 5850/ 490 = 11.938 A D2 = 10 100 0 2 50 2 3 50 -5 D2 = -1900 I2 = D2/D = -1900/490 = -3.8775 A D3 = 10 5 100 2 -7 50 3 -3 50 D3 = -250 I3 = D3/D = -250/490 = -0.510 A let me know where did i got wrong . I just want to Solve this using KVL and KCL please help me thanks !!!!!
You made a mistake here: It should be: 50V + 3(I1 - I2 - I3) - 2(I3) =0 OR: -3I1 + 3I2 +5I3 = 50 V And solve that I get: I1 = 485/49 A ≈ 9.9 A I2 = 10/49 A ≈ 0.2 A I3 = 775/49 A ≈ 15.8 A http://www.wolframalpha.com/input/?i=+10*a+5*b+=+100,+2*a-7*b+2*c+=+50,+-3*a+3*b+5*c+=+50+
I didn't read that. In that case, your set of equations is correct. Here is the solution. You got the right answer. http://www.wolframalpha.com/input/?i=+10*a+5*b+=+100,+2*a-7*b+2*c+=+50,+3*a-3*b-5*c+=+50+