In steady state,a capacitor of capacitance `2 mu F` is charged to `4 mu C` , as shown in figure. If the internal resistance of the cell is `0 . 5 Omega` , then the emf of the cell is

Options

A. 4V
B. 5 V
C. 2 . 5 V
D. 2 V

Answer

C

Explanation

`V_("cap") = (Q)/(C) = (4 xx 10^(-6))/(2 xx10^(-6))`
= 2v
Now `V_("cap") = v ("across" 2Omega" resistor")`
`:. I = (V)/(R) = (2)/(2) = IA`
Now, using the relation
E = V+ Ir
`= 2 + 1 xx 0 . 5`
= 25 V