A cell supplies a current of `0.9A` through a `2Omega` resistor and a current of `0.3A` through a `7 Omega` resistor. Calculate the internal resistance of the cell .

Explanation

Given : Cell supplies a circuit `I_(1)=0.9A`
Cell supplies to resistor `R_(1)=2`
Cell supplies to circuit `I_(2)=0.3A`
Cell supplies to resistor `R_(2)=7Omega`
To find :
Internal resistor of the cell `r=?`
formula :
`I_(1)=(xi)/(R_(1)+r)`
`xi=I_(1)=(R_(1)+r) " "...(1)`
`I_(2)=(xi)/(R_(2)+r)`
`xi=I_(2)(R_(2)+r) " "...(2)`
From (1) and (2)
`I_(1)(R_(1)+r)=I_(2)(R_(2)+r)`
`r=(I_(1)R_(1)-I_(2)R_(2))/(I_(2)-I_(1))=(0.9xx2-0.3xx7)/(0.3-0.9)`
`=(1.8-2.1)/(-0.6)=(0.3)/(0.6)=(1)/(2)=0.5Omega`
`r=0.5Omega`