Three condensers of capacities $\mathrm{C}_{1}, \mathrm{C}_{2}, \mathrm{C}_{3}$ are connected in series with…

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
  1. $\mathrm{C}_{1}: \mathrm{C}_{2}: \mathrm{C}_{3}$
  2. $C_{1}^{2}: C_{2}^{2}: C_{3}^{2}$
  3. $1: 1: 1$
  4. $\frac{1}{\mathrm{C}_{1}}: \frac{1}{\mathrm{C}_{2}}: \frac{1}{\mathrm{C}_{3}}$

Solution

When capacitors are connected in series, the charge is the same on all the capacitors. $\begin{array}{l} \mathrm{Q}=\mathrm{C}_{1} \mathrm{~V}_{1}=\mathrm{C}_{2} \mathrm{~V}_{2}=\mathrm{C}_{7} \mathrm{~V}_{3} \\ \text { Or } \frac{V_{1}}{\frac{1}{C_{1}}}=\frac{V_{2}}{\frac{1}{C_{2}}}=\frac{V_{3}}{\frac{1}{C_{3}}} \end{array}$

Asked in: MHT CET 2020 (13 Oct Shift 1)

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