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(C) In a series combination of capacitors,the charge $Q$ on each capacitor is the same.The potential difference across a capacitor is given by $V = \f

Vedclass · Accepted Answer

${C_1} < {C_2}$

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(C) In a series combination of capacitors,the charge $Q$ on each capacitor is the same.The potential difference across a capacitor is given by $V = \frac{Q}{C}$.From the graph,the potential drop across the first capacitor $(C_1)$ is greater than the potential drop across the second capacitor $(C_2)$.Let $V_1$ be the potential difference across $C_1$ and $V_2$ be the potential difference across $C_2$.Since $V_1 > V_2$ and $Q = C_1 V_1 = C_2 V_2$,we have $\frac{C_1}{C_2} = \frac{V_2}{V_1}$.Since $V_1 > V_2$,it follows that $\frac{V_2}{V_1} < 1$,which implies $\frac{C_1}{C_2} < 1$,or ${C_1} < {C_2}$.

Figure $(a)$ shows two capacitors connected in series and joined to a battery. The graph in figure $(b)$ shows the variation in potential as one moves from left to right on the branch containing the capacitors. If the graph represents the potential across the capacitors,then:

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