A cell supplies a current of 0.9 A through a `1 Omega` resistor and a current of 0.3 A through a `2Omega` resistor. Calculate the internal resistance of the cell.

Explanation

Current from the cell, `I_(1)=0.9A`
Resistor, `R_(1)=2Omega`
Current from the cell, `I_(2)=0.3A`
Resistor, `R_(2)=7Omega`
Internal resistance of the cell, `r=?`
Current in the circuit, `I_(1)=(xi)/(r+R_(1))`
`xi=I_(1)(r+R_(1)) " "...(1)`
Current in the circuit, `I_(2)=(xi)/(r+R_(2))`
`xi=I_(2)(r+R_(2))" "...(2)`
Equating equation (1) and (2),
`I_(1)r+I_(1)R_(1)=I_(2)R_(2)+I_(2)r`
`(I_(1)-I_(2))r=I_(2)R_(2)-I_(1)R_(1)`
`r=(I_(2)R_(2)-I_(1)R_(1))/(I_(1)-I_(2))=((0.3xx7)-(0.9xx2))/(0.9-0.3)=(2.1-1.8)/(0.6)=(0.3)/(0.6)`
`r=0.5Omega`