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JEE Mains · Physics · STD 12 - 2. Electric potential and capacitance
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
- A \(3.2\; \mu \mathrm{F}\)
- B \(8.4\; \mu \mathrm{F}\)
- C \(1.6\; \mu \mathrm{F}\)
- D \(4.2\; \mu \mathrm{F}\)
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…
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