Find the value of `i_(1)/I_(2)`in figure if where `R=10Omega`

Explanation

(a) `0.1 (i_1)+ 1(i_1) - 6 + 1(i_1)-6 = 0`
`rArr 0.1 (i_1) + 1(i_1) + 1(i_1) = 12`
` rArr i_1 = (12/2.1)`
ABCDA
` 0.1(i_2) + 1i-6 =0 `
` rArr 0.1(i_2) + i = 6`
ADEFA,
` i-6+6-(i_2 -i)1 = 0 `
` rArr i-i_2+i=0`
` rArr 2i-i_2=0`
` rArr 2[6-0.1(i_2)]-i_2 = 0 `
` rArr i_2 = 10`
So, ` i_1/i_2 = 0.57`
(b)` 1(i_1) + 1.(i_1)-6 +(i_1)-6=0`
` 3i_1 = 12 `
` rArr i_1 = 4 `
DCFED
` i_2 + i - 6 = 0 `
` rArr i_2 + i = 6 `
ABCDA,
` i_2 + (i_2 -i) -6 = 0 `
` rArr i_2 + i_2 -i =6 `
` rArr 2i_2 - i =6 `
` rArr 2i_2 -i = 6 `
` rArr i=-2`
`i_2 -2 =6`
` rArr i_2 = 8 `
`i_1/i_2 = 1/2 `
(c) ` 10 i_1 + 1 i_1 - 6 + 1 i_1 - 6 = 0 `
` rArr 12i_1 = 12`
` rArr i_1 =1`
` rArr 10^i_2 - 1i - 6 =0 `
` rArr 10i_2 -i =6`
`rArr (10i_2 + (i_2 -i) 1 -6 = 0) `
` rArr 10i_2 - i +i_2 = 6`
` rArr 11 i_2 - i =6 `
` i_2 = 0.57`
` So, i_1/i_2 = 1.75 ` .
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