JEE Mains · Physics · STD 12 - 2. Electric potential and capacitance
Consider the combination of \(2\) capacitors \(C _{1}\) and \(C _{2},\) with \(C _{2}> C _{1},\) when connected in parallel, the equivalent capacitance is \(\frac{15}{4}\) time the equivalent capacitance of the same connected in series. Calculate the ratio of capacitors, \(\frac{ C _{2}}{ C _{1}}\)
- A \(\frac{15}{11}\)
- B \(\frac{111}{80}\)
- C \(\frac{29}{15}\)
- D None of these
Answer & Solution
Correct Answer
(D) None of these
Step-by-step Solution
Detailed explanation
When connected in parallel \(C _{ eq }= C _{1}+ C _{2}\) When in series \(C _{ eq }^{\prime}=\frac{ C _{1} C _{2}}{ C _{1}+ C _{2}}\) \(C _{1}+ C _{2}=\frac{15}{4}\left(\frac{ C _{1} C _{2}}{ C _{1}+ C _{2}}\right)\) \(4\left( C _{1}+ C _{2}\right)^{2}=15 C _{1} C _{2}\)…
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