A cell of emf `epsilon` and internal resistance `r` is connected across a variable load resistance `R`. It is found that when `R = 4 Omega` the current is `1A ` and when `R` is increased to `9 Omega`, the current reduces to `0.5A`. Find the value of the emf `epsilon` and internal resistance`r`

Answer

`5V , 1 Omega`

Explanation

Current is circuit, `I = (epsilon)/(R + r)`
case (i) `1 =(epsilon)/(4+ r)`……(i)
case (ii) `0.5 =(epsilon)/(9+ r)`……(ii)
Solving (i) and (ii) we get
`r = 1 Omega and epsilon = 5 V`