In the givven 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: (figures are drawn schematically and are not to scale)

Options

A.
B.
C.
D.

Answer

B

Explanation

Here `C_(1) = 1 muF, C_(2) = 2 muF, C_(p) = 2+1 = 3 muF`
This combination is in series with `C`
`:. C_(eq) = (3C)/(3 + C)`
Total charge, `q = C_(eq) E = ((3C)/(3 + C)) E`
Charge on `2 muF` capaicatance
`q_(2) = (2)/(3) xx (3 CE)/(3 + C) = (2 CE)/(3 + C) = (2E)/(1 + (3)/(C )) = Q`, say
As `(dQ)/(dC) gt 0. (d^(2) Q)/(dC^(2)) lt 0`.
therefore `Q_(2)` versus `C` graph is shown correctly in Fig.