emf of battery is 2.2 V. When 5`Omega` resisitor i connected across the battery, its termina voltage is 1.8 V. In tern al resistance of battery .... `Omega`.

Options

A. `(10)/(9)`
B. `(9)/(10)`
C. `(9)/(5)`
D. `(5)/(9)`

Answer

A

Explanation

`(10)/(9)`
`epsilon = V + I r `
`therefore r = (epsilon - V)/(I)`
= `(2.2 - 1.8)/((V)/(R)) = (0.4)/((1.8)/(5)) = (0.4 xx 5)/(1.8) `
`therefore r = (20)/(18)`
= `(10)/(9) Omega`