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JEE Main 2019 (12 Jan, Shift 1) · Charge Sharing & Loss of Energy

Charge Sharing & Loss of Energy question from JEE Main 2019 (12 Jan, Shift 1)

Q31JEE Main 2019 · 12 JAN · SHIFT 1Medium

Charge Sharing & Loss of Energy

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:

Figure

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Answer: (d)

  1. At A, CC is charged to ε\varepsilon: Q=CεQ = C\varepsilon, Ui=Q22CU_i = \frac{Q^2}{2C}.
  2. At B, QQ is shared with 3C3C: Uf=Q22(4C)=Q28CU_f = \frac{Q^2}{2(4C)} = \frac{Q^2}{8C}
  3. ΔU=Q22C−Q28C=38Q2C\Delta U = \frac{Q^2}{2C} - \frac{Q^2}{8C} = \frac{3}{8}\frac{Q^2}{C}
  4. So the answer is (d).
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