In the given circuit, charges Q₂ on the 2 capacitor changes as C is varied from 1 to 3. Q₂ as a function of 'C' is given properly by: (figures are drawn schematically and are not to scale) (a) (b) (c) (d)

Answer: (d). 1 and 2 in parallel give 3, in series with C: Q = 3C/C+3E. The 2 takes 2/3 of it: Q₂ = (2CE)/(C+3). (dQ₂)/(dC) = (6E)/((C+3)²) > 0 and decreasing, so Q₂ rises with a decreasing slope (concave down). So the answer is (d).

In the given circuit, charges Q₂ on the 2 capacitor changes as C is… | JEE Main 2015

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Capacitors in DC Circuits & RC

In the given circuit, charges $Q_2$ on the $2\ \mu\text{F}$ capacitor changes as $C$ is varied from $1\ \mu\text{F}$ to $3\ \mu\text{F}.\ Q_2$ as a function of $'C'$ is given properly by:

$\text{(figures are drawn schematically and are not to scale)}$

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Answer: (d)

$1\ \mu\text{F}$ and $2\ \mu\text{F}$ in parallel give $3\ \mu\text{F}$ , in series with $C$ : $Q = \dfrac{3C}{C+3}E$ .
The $2\ \mu\text{F}$ takes $\dfrac{2}{3}$ of it: $Q_2 = \dfrac{2CE}{C+3}$ .
$\dfrac{dQ_2}{dC} = \dfrac{6E}{(C+3)^2} > 0$ and decreasing, so $Q_2$ rises with a decreasing slope (concave down).
So the answer is (d).

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