MHT CET · Physics · Capacitance
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
- A \(\mathrm{C}_{1}: \mathrm{C}_{2}: \mathrm{C}_{3}\)
- B \(C_{1}^{2}: C_{2}^{2}: C_{3}^{2}\)
- C \(1: 1: 1\)
- D \(\frac{1}{\mathrm{C}_{1}}: \frac{1}{\mathrm{C}_{2}}: \frac{1}{\mathrm{C}_{3}}\)
Answer & Solution
Correct Answer
(D) \(\frac{1}{\mathrm{C}_{1}}: \frac{1}{\mathrm{C}_{2}}: \frac{1}{\mathrm{C}_{3}}\)
Step-by-step Solution
Detailed explanation
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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