An equiring physics student connects a cell to a circuit and measures the current drawn from the cell to `I_1`. When he joins a second identical cell is series with the first, the current becomes `I_2`. When the cells are connected in parallel, the current through the circuit is `I_2`. When the cells are connected are in parallel, the current through the circuit is `I_3`. Show that relation between the current is `3I_3 I_2 =2I_1(I_2 + I_3).`

Explanation

`I_(1)=E/(r+R)...(1)`
`I_(2)=(2E)/(2r+R)....(2)`
`I_(3)=E/((r//2)+R) I_(3)=(2E)/(r+2R)....(3)`
`I_(2)+I_(3)=(2E)/(2r+R)+(2E)/(r+2R)=(2Exx3(r+R))/((2r+r)(r+2R))`
`(I_(2)+I_(3))=3xx(2E)/((2r+R))xx(2E)/((r+2R))xx((r+R))/(2E)`
`(I_(2)+I_(3))=(3xxI_(2)I_(3))/(2I_(1))implies3I_(2)I_(3)=2I_(1)(I_(2)+I_(3))`