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
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(c) According to graph we can say that potential difference across the capacitor ${C_1}$ is more than that across ${C_2}$. Since charge $Q$ is same i.e., $Q = {C_1}{V_1} = {C_2}{V_2}$
$==>$ $\frac{{{C_1}}}{{{C_2}}} = \frac{{{V_2}}}{{{V_1}}}$ $==>$ ${C_1} < {C_2}$ $({V_1} > {V_2}).$
$==>$ $\frac{{{C_1}}}{{{C_2}}} = \frac{{{V_2}}}{{{V_1}}}$ $==>$ ${C_1} < {C_2}$ $({V_1} > {V_2}).$
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