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

Case (1)
`{:("Resistance"),("of a resistor"):}} = 2 Omega`
Current I = 0.9 A
Case (2)
Resistance `R = 7 Omega`
Current I = 0.3 A
`E = I (R+r) = IR + Ir `
`:." r = E/I -R `
`:." From two cases we have, "`
` E = /0.9 -2 = E / 0.3 - 7`
` E/0.9 - E/0.3 = -7 + 2`
`( E - 3 E)/0.9 = -5`
` - (2E)/0.9 = -5`
` :." E" = 4.5 /2 V `
`{:("Internal"),("resistance r"):}} = E/I = R`
` I = 0.3, R = 7 Omega`
`r = 4.5/(2 xx 0.3 ) -7`
` = 4.5 /0.6 -7`
` = 7.5 - 7 = 0.5 Omega`