Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

Two identical capacitors $C_1$ and $C_2$ of equal capacitance are connected as shown in the circuit. Terminals a and b of the key k are connected to charge capacitor $C_1$ using a battery of emf $V$ volt. Now disconnecting a and b terminals, terminals b and c are connected. Due to this, what will be the percentage loss of energy?

75%

50%

25%

Step-by-Step Solution

Initial State: Capacitor $C_1$ is charged to voltage $V$ . The energy stored is $U_i = \frac{1}{2}CV^2$ . Capacitor $C_2$ is uncharged, so its potential is 0.
Redistribution: When $C_1$ is connected to the identical uncharged capacitor $C_2$ (by connecting terminals b and c), charge flows until they reach a common potential $V'$ . Since capacitance is conserved and identical ( $C_1=C_2=C$ ), the total capacitance is $2C$ . By conservation of charge ( $Q = CV$ ), the common potential is $V' = \frac{Q_{total}}{C_{total}} = \frac{CV}{C+C} = \frac{V}{2}$ .
Final Energy: The total energy stored in the parallel combination is $U_f = \frac{1}{2} C_{total} (V')^2 = \frac{1}{2} (2C) (\frac{V}{2})^2 = C \frac{V^2}{4} = \frac{1}{4}CV^2$ .
Energy Loss: The loss in energy is $\Delta U = U_i - U_f = \frac{1}{2}CV^2 - \frac{1}{4}CV^2 = \frac{1}{4}CV^2$ .
Percentage Loss: $\frac{\Delta U}{U_i} \times 100 = \frac{\frac{1}{4}CV^2}{\frac{1}{2}CV^2} \times 100 = 50\%$ . This result is consistent with NCERT Example 2.10, which states that when a capacitor shares its charge with an identical uncharged capacitor, the final energy is half the initial energy, implying a 50% loss as heat and radiation [NCERT Class 12, Physics Part I, Sec 2.15, Example 2.10].

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