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
- A \(1: 1: 1\)
- B \(\mathrm{C}_1: \mathrm{C}_2: \mathrm{C}_3\)
- C \(\mathrm{C}_1^2: \mathrm{C}_2^2: \mathrm{C}_3^2\)
- 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 same on all the capacitors.
\(\begin{aligned}
& \mathrm{Q}=\mathrm{C}_1 \mathrm{~V}_1=\mathrm{C}_2 \mathrm{~V}_2=\mathrm{C}_3 \mathrm{~V}_3 \\
& \Rightarrow \mathrm{~V}_1: \mathrm{V}_2: \mathrm{V}_3=\frac{1}{\mathrm{C}_1}: \frac{1}{\mathrm{C}_2}: \frac{1}{\mathrm{C}_3}
\end{aligned}\)
\(\begin{aligned}
& \mathrm{Q}=\mathrm{C}_1 \mathrm{~V}_1=\mathrm{C}_2 \mathrm{~V}_2=\mathrm{C}_3 \mathrm{~V}_3 \\
& \Rightarrow \mathrm{~V}_1: \mathrm{V}_2: \mathrm{V}_3=\frac{1}{\mathrm{C}_1}: \frac{1}{\mathrm{C}_2}: \frac{1}{\mathrm{C}_3}
\end{aligned}\)
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