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A circuit contains an ideal battery of EMF E. Elements X and Y are unknown. The switch S is initially open and is closed at t = 0. Match Column-1, Column-2, and Column-3: Column-1: (I) X = R, Y = C; (II) X = R, Y = R; (III) X = R, Y = parallel R and C; (IV) X = R, Y = battery E (same polarity). Column-2: (i) entire energy from batteries is dissipated as heat; (ii) total charge drawn from battery after long time is CE/2; (iii) total heat produced is less than energy supplied by battery; (iv) current I = 0 just after closing and after long time. Column-3 shows I vs t or q vs t or H vs t graphs (P, Q, R, S). Of the four situations in Column-1, which has the maximum current just after the switch is closed?

1. (A) I, iii, P
2. (B) IV, i, Q
3. (C) II, ii, R
4. (D) III, iv, S

Correct answer: (C) II, ii, R

Solution

At t = 0+ just after switch closes: a capacitor is uncharged so it acts as a short circuit (V=0). Case I: Y=C acts as short, current path is E through X=R only: I₀ = E/R. Case II: Y=R, two resistors in series (or relevant configuration): I₀ = E/(2R). Case III: Y = R||C, capacitor shorts the parallel R, Y has zero impedance at t=0+, I₀ = E/R. Case IV: Y = battery E in same polarity, so total EMF = 2E, only X=R limits current: I₀ = 2E/R. Maximum current at t=0+ is in Case IV = 2E/R. Looking at options: (B) mentions IV, i, Q. Case IV: both batteries supply energy E to current through R; all energy is dissipated as heat (no capacitor to store energy), so Column-2 match is (i). Answer is (B) IV, i, Q.

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