JEE Mains · Physics · STD 12 - 2. Electric potential and capacitance
In the figure shown, after the switch \(‘S’\) is turned from position \(‘A’\) to position \(‘B’\), the energy dissipated in the circuit in terms of capacitance \(‘C’\) and total charge \(‘Q’\) is
- A \(\frac{1}{8}\frac{{{Q^2}}}{C}\)
- B \(\frac{3}{8}\frac{{{Q^2}}}{C}\)
- C \(\frac{5}{8}\frac{{{Q^2}}}{C}\)
- D \(\frac{3}{4}\frac{{{Q^2}}}{C}\)
Answer & Solution
Correct Answer
(B) \(\frac{3}{8}\frac{{{Q^2}}}{C}\)
Step-by-step Solution
Detailed explanation
Before switch is shifted: Energy stored, \(\mathrm{U}_{1}=\frac{1}{2} \mathrm{CE}^{2}\) and charge stored, \(Q=C E\) After switch is shifted: \(V_{M}-V_{N}=\frac{Q}{4 C}=\frac{C E}{4 C}=\frac{E}{4}\) Energy stored,…
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