Exams › JEE Advanced › Physics
Two capacitors A and B, each of capacitance C, are connected in series in the top branch. A switch S connects the junction between them to the negative terminal of a battery of emf E in the central branch. The battery E is in the bottom branch. Initially S is open for a long time, then S is closed. Match List-I with List-II. List-I: (P) Charge flowed through battery when S is closed (Q) Work done by battery (R) Charge on capacitor A after S is closed (S) Heat developed in the system List-II: (1) CE²/2 (2) CE/2 (3) CE²/4 (4) CE
- P -> 2; Q -> 2; R -> 4; S -> 3
- P -> 2; Q -> 2; R -> 4; S -> 1
- P -> 1; Q -> 2; R -> 4; S -> 3
- P -> 2; Q -> 1; R -> 4; S -> 3
Correct answer: P -> 2; Q -> 1; R -> 4; S -> 3
Solution
S open: capacitors in series, each with charge Q_i = CE/2 and energy U_i = C(E/2)²/2 each, total U_i = CE²/4. S closed: each capacitor is directly across E with charge Q_f = CE each, total energy U_f = CE². Charge through battery = CE - CE/2 = CE/2 (P -> 2). Work by battery = E * CE/2 = CE²/2 (Q -> 1). Charge on A = CE (R -> 4). Heat = Work - delta_U = CE²/2 - (CE² - CE²/4) = CE²/2 - 3CE²/4 = -CE²/4? That gives negative heat. Let me recheck initial state: S open with battery connected, capacitors in series: total capacitance C/2, charge on each = (C/2)*E = CE/2. Final: S closed means midpoint grounded? If midpoint connected to negative terminal, then A is across E alone (charge CE) and B might be bypassed. Energy stored initially = 2*(1/2)*(C)*(E/2)² = CE²/4. Finally for A alone: (1/2)CE². Charge increase through battery = CE - CE/2 = CE/2. Work = CE²/2. Heat = CE²/2 - (CE²/2 - CE²/4) = CE²/4. So P->2, Q->1, R->4, S->3.
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