Find the value of `i_(1)//i_(2)`in figure if
(a)`R=0.1(Omega)`
(b)`R=1(Omega)`
(c)`R=10(Omega)`.
Note from your answers that in order to get more current from a combination if two batteries they should be joined in parallel if the external resistance is small and in series if the external resistance is large as compared to the internal resistances.

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 ` .
,
,
,
,
,