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Three condensers of capacities $\mathrm{C}_{1}, \mathrm{C}_{2}, \mathrm{C}_{3}$ are connected in series with a source of
e.m.f. $V$. The potentials across the three condensers are in the ratio of
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e.m.f. $V$. The potentials across the three condensers are in the ratio of
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Verified Answer
The correct answer is:
$\frac{1}{\mathrm{C}_{1}}: \frac{1}{\mathrm{C}_{2}}: \frac{1}{\mathrm{C}_{3}}$
When capacitors are connected in series, the charge is the same on all the capacitors.
$$
\begin{array}{l}
\mathrm{Q}=\mathrm{C}_{1} \mathrm{~V}_{1}=\mathrm{C}_{2} \mathrm{~V}_{2}=\mathrm{C}_{7} \mathrm{~V}_{3} \\
\text { Or } \frac{V_{1}}{\frac{1}{C_{1}}}=\frac{V_{2}}{\frac{1}{C_{2}}}=\frac{V_{3}}{\frac{1}{C_{3}}}
\end{array}
$$
$$
\begin{array}{l}
\mathrm{Q}=\mathrm{C}_{1} \mathrm{~V}_{1}=\mathrm{C}_{2} \mathrm{~V}_{2}=\mathrm{C}_{7} \mathrm{~V}_{3} \\
\text { Or } \frac{V_{1}}{\frac{1}{C_{1}}}=\frac{V_{2}}{\frac{1}{C_{2}}}=\frac{V_{3}}{\frac{1}{C_{3}}}
\end{array}
$$
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