JEE PYQ: Capacitance Question 42
Question 42 - 2019 (12 Jan Shift 2)
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:
(1) $\frac{1}{8}\frac{Q^2}{C}$
(2) $\frac{3}{8}\frac{Q^2}{C}$
(3) $\frac{5}{8}\frac{Q^2}{C}$
(4) $\frac{3}{4}\frac{Q^2}{C}$
Show Answer
Answer: (2)
Solution
Energy stored initially: $U_i = \frac{1}{2}\frac{Q^2}{C} = \frac{Q^2}{2C}$. When switch is moved to B, charge redistributes between C and 3C. $U_f = \frac{Q^2}{2 \times 4C} = \frac{Q^2}{8C}$. Energy dissipated $= \frac{Q^2}{2C} - \frac{Q^2}{8C} = \frac{3Q^2}{8C}$.
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