JEE Mains · Physics · STD 12 - 2. Electric potential and capacitance
Three capacitors of capacitances \(25 \mu \mathrm{F}, 30 \mu \mathrm{F}\) and \(45 \mu \mathrm{F}\) are connected in parallel to a supply of \(100\) \(V\). Energy stored in the above combination is \(\mathrm{E}\). When these capacitors are connected in series to the same supply, the stored energy is \(\frac{9}{\mathrm{x}} \mathrm{E}\). The value of \(x\) is _______.
- A \(85\)
- B \(86\)
- C \(87\)
- D \(88\)
Answer & Solution
Correct Answer
(B) \(86\)
Step-by-step Solution
Detailed explanation
In parallel combination : Potential difference is same across all \(\text { Energy }=\frac{1}{2}\left(\mathrm{C}_1+\mathrm{C}_2+\mathrm{C}_3\right) \mathrm{V}^2\) \(=\frac{1}{2}(25+30+45) \times(100)^2 \times 10^{-6}=0.5=\mathrm{E}\) In series combination: Charge is same on all.…
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