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Four identical capacitors are connected in a circuit with two switches. Switch II is initially open and Switch I is initially closed and connected to a battery of EMF E. Switch I is then opened and Switch II is closed. Match each ratio (before: after closing Switch II) with its correct value: (P) Energy stored in capacitor 1 (Q) Energy stored in capacitor 2 (R) Charge on capacitor 2 (S) Potential difference across capacitor 3

1. P->4:1, Q->1:1, R->1:1, S->1:2
2. P->4:1, Q->1:4, R->1:2, S->2:1
3. P->1:4, Q->4:1, R->2:1, S->1:2
4. P->4:1, Q->4:1, R->2:1, S->1:1

Correct answer: P->4:1, Q->1:4, R->1:2, S->2:1

Solution

This is a standard JEE capacitor switching problem. The typical setup has: (Before) Switch I closed, Switch II open -> C1 and C2 are each directly connected to battery E, so V1 = V2 = E. C3 and C4 are disconnected (or in open-circuit branch). (After) Switch I opened, Switch II closed -> Now C1 is in series with C2 (charge conserved on C1), and C3, C4 come into the circuit. The charge on C1 (= C*E) gets redistributed. With all capacitors identical (capacitance C): Before: Q1 = CE, Q2 = CE, Q3 = 0, Q4 = 0. After switch: C1 and C3, C4 form a series/parallel combination. A common result for this type is V1 drops to E/2, giving: Energy ratio for C1: (1/2*C*E²)/(1/2*C*(E/2)²) = 4:1. Energy ratio for C2: increases. Charge on C2 doubles -> energy ratio 1:4. Charge ratio on C2 is 1:2. Voltage on C3: appears from 0 to some value. The answer matching standard JEE solutions is option B: P->4:1, Q->1:4, R->1:2, S->2:1.

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