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Figure (i) shows two capacitors connected in series and connected by a battery. The graph (ii) shows the variation of potential as one moves from left to right on the branch AB containing the capacitors. Then

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$C_{1} = C_{2}$

$C_{1} < C_{2}$

$C_{1} > C_{2}$

C₁ and C₂ cannot be compared

answer is C.

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Detailed Solution

The charge on both of the capacitors are equal and from the figure we can say potential difference across C₂ is greater than C₁.

$\Rightarrow \frac{q}{C_{2}} > \frac{q}{C_{1}}$

$\Rightarrow C_{1} > C_{2}$

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