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

Engineering

Physics

Parallel Plate Capacitor

Question

C₁ = C₂

C₁ < C₂

C₁ > C₂

C₁ and C₂ cannot be compared

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According to the graph we can say that the potential difference across the capacitor C₁ is more than that across C₂

Since charge Q is same i.e,

Q = C₁V₁ = C₂V₂

\Rightarrow \frac{C_{1}}{C_{2}} = \frac{V_{2}}{V_{1}}

⟹ C₁ < C₂(V₁ > V₂)

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