In the figure shown after switch|CAPACITANCE| JEE-MAIN 2019

JEE-MAIN 2019

12-01-2019 S1

Question

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

Select the correct option:

$\frac{3}{8} \frac{\mathrm{Q}^2}{\mathrm{C}}$

$\frac{1}{8} \frac{\mathrm{Q}^2}{\mathrm{C}}$

$\frac{5}{8} \frac{\mathrm{Q}^2}{\mathrm{C}}$

$\frac{3}{4} \frac{\mathrm{Q}^2}{\mathrm{C}}$

✓ Correct! Well done.

✗ Incorrect. Try again or view the solution.

Solution

Question Tags

JEE Main

Physics

Easy

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