JEE Main 2015 · 4 Apr · Shift 1 · Physics Q16

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PhysicsSingle correct+4 / -1Question 16 of 90

In the given circuit, charge $Q_2$ on the 2μF capacitor changes as C is varied from 1μF to 3μF. $Q_2$ as a function of 'C' is given properly by : (figures are drawn schematically and are not to scale)

Capacitor C in series with a parallel pair of 1 μF and 2 μF capacitors, across a battery E.

A straight line rising with C

Graph of Q₂ (charge) against C from 1 μF to 3 μF: a rising straight line.

A rising curve, concave down (flattening)

Graph of Q₂ (charge) against C from 1 μF to 3 μF: a rising curve that flattens out (concave down).

A rising curve, concave up (steepening)

Graph of Q₂ (charge) against C from 1 μF to 3 μF: a rising curve that steepens (concave up).

A falling curve

Graph of Q₂ (charge) against C from 1 μF to 3 μF: a decreasing curve.

Answer and worked solution

Official answer: B

The 1 μF and 2 μF capacitors in parallel make 3 μF, in series with C:
$Q=\frac{3C}{C+3}E,\qquad Q_2=\frac23Q=\frac{2CE}{C+3}.$
This increases with C, with $\frac{dQ_2}{dC}=\frac{6E}{(C+3)^2}$ decreasing: a rising curve that flattens out, graph (2).

Topics:Series and parallel combinations

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JEE Main 2015 · 4 Apr · Shift 1, Physics question 16