Date i3 2 KCL 12 73 10 KVLI 12 13 13 12 13 SolutionKCL

Date i3 /2 KCL 12 + 73 10 KVLI 12 + 13 13- 12 = 13 =

KCL ==> i1 +i2 -i3 = 0 -------( 1)

KVL 1 ==>

2y -3y = x i1 +2x i1 -x i2

-y = 3x i1 - x i2 --------( 2)

KVL 2==>

3y = 3x i3 + x i3 +2x i3 +x i2

= 6x i3 + x i2 --------( 3)

From equation (1 ) , i3 = i1 + i2

Substitute i3 in equation( 3) you get ,

3y = 6x(i1+i2)+ x i2

= 6x i1 +6x i2 + x i2

= 6x i1 +7x i2 -------------( 4)

-y = 3x i1 - x i2 -----------( 2)

equation(2) X 2 ==>

-2y = 6x i1 -2x i2 ----------( 5)

Equation(4) - equation(5) ==>

3y +2y = 6x i1 +7x i2 -6x i1 +2x i2

5y = 9x i2

i2 = 5y /9x

Substitute this in equation (2 ) you get ,

-y = 3x i1 - x (5y/9x)

-y = 3x i1 - (5/9) y

3x i1 = -y +(5/9) y

= (-4/9) y

i1 = (-4/27) y/x

Substitute i1 and i2 in equation( 1) you get

i3 = (-4/27) y/x +[ 5y /9x]

= [(-4 +15) /27 ] (y/x)

= (11/27)(y/x)