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JEE Mains · Physics · STD 12 - Electric potential and capacitance
- JEE Mains 2020
- Medium
- MCQ
Effective capacitance of parallel combination of two capacitors \(\mathrm{C}_{1}\) and \(\mathrm{C}_{2}\) is \(10\; \mu \mathrm{F}\). When these capacitors are individually connected to a voltage source of \(1\; \mathrm{V},\) the energy stored in the capacitor \(\mathrm{C}_{2}\) is \(4\) times that of \(\mathrm{C}_{1}\). If these capacitors are connected in series, their effective capacitance will be
Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) \(1.6\; \mu \mathrm{F}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{C}_{1}+\mathrm{C}_{2}=10\)
\(\frac{1}{2} \mathrm{C}_{2} \mathrm{V}^{2}=4 \times \frac{1}{2} \mathrm{C}_{1} \mathrm{V}^{2}\)
\(\therefore \quad \mathrm{C}_{2}=4 \mathrm{C}_{1}\)
\(\therefore \quad \mathrm{C}_{1}=2 \;and\; \mathrm{C}_{2}=8\)
For series combination
\(C_{e q}=\frac{C_{1} C_{2}}{C_{1}+C_{2}}=1.6\)
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