In the given circuit, charge `Q_(2)` on the `2muF` capacitor changes as C is varied from `1muF` to `3muF`. `Q_(2)` as a function of 'C' is given properly by : (figure are drawn schematically and are not to scale)

Options

A.
B.
C.
D.

Answer

B

Explanation

Assume negative terminal of the battery as grounded (0 V)
Suppose, potential of point x is V.
From the circuit diagram, we can write
`Q_(C)=Q_(1)+Q_(2)`
or `C(E-V)=1xxV+2xxV or V[C+3]=CE`
or `V=(CE)/(3+C)`
`therefore Q_(2)=C_(2)(V)=(2CE)/(3+C)=(2E)/(1+3//C)`
As `C_(1)` varried from `1muF` to `3muF`, charge increases with decreasing slope.
Note As `C to oo, Q_(2) to 2E=`constant.