A resistance of `4 Omega ` is connected across a cell Then it is replaced by another resistance of `1 Omega` it is found the power dissipeted in resistance in both the is `16` walt Then

Options

A. internal resistance of the cell is `2 Omega`
B. emf of the celll is `12V`
C. maximum power the can be dissipated in the external resistance is `18` walt
D. short circuit current from the cell is infinite

Answer

a,b,c

Explanation

`((epsilon)/(4+r))^(2) xx 4 = ((epsilon)/(1+r))^(2) xx1`
`r = 2 Omega`
Current in circuit,`I = (epsilon)/(4+2)`
`16 = ((epsilon)/(4+2))^(2) xx 4or epsilon = 12 V`
power dissipated is maximum when exertnal resistance is equal to internal resistance i.e.
`R = r = 2 Omega`
`P_(max) = ((12)/(2+2))^(2) xx 2 = 18W`