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:

loop C:
50V - 3(I1 - I2 - I3) + 2(I3) =0
3(I1) -3(I2) -5(I3) = 50 V

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+

 

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.)

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+

 

I guess you got 10A through the 10 Ohms resistor because you reversed V2 in your simulation.