The emf of a cell is `epsilon` and its internal resistance is r. its termnals are cannected to a resistance R. The potential difference between the terminals is `1.6V for R = 4 Omega, and 1.8 V for R = 9 Omega. Then,

Options

A. `epsilon =1 V, r = 1 Omega`
B. `epsilon =2V, r = 1 Omega`
C. `epsilon =2 V, r =2Omega`
D. `epsilon =2.5 V, r = 0.5 Omega`

Answer

B

Explanation

Current in the cirut is
`I = (epsilon)/(R +r)`
Potential difference across cell - potential differnce across R
` = iR = (epsilonR)/(R +r)`
Set up two equations with the given data and sovle for `epsilon, r.`