The figure shows three circuits with identical batteries, inductors, and resistors. Rank the circuits according to the current through the battery (i) just after the switch is closed and (ii) a long time later, greatest first.
[i₁, i₂ and i₃ are respective current in circuit (1), (2) and, (3)]

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$(i) i_{2} > i_{3} > i_{1} (i_{1} = 0)$ $(ii) i_{2} > i_{3} > i_{1}$

$(i) i_{2} < i_{3} > i_{1} (i_{1} \neq 0)$ $(ii) i_{2} > i_{3} > i_{1}$

$(i) i_{2} = i_{3} = i_{1} (i_{1} = 0)$ $(ii) i_{2} > i_{3} > i_{1}$

$(i) i_{2} = i_{3} > i_{1} (i_{1} \neq 0)$ $(ii) i_{2} > i_{3} > i_{1}$

answer is A.

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Detailed Solution

Just after the switch is closed maximum emf is induced in the inductor which opposes the flow of current and the current in the inductor becomes zero . So $i_{1} = 0, i_{2} = \frac{E}{R} a n d i_{3} = \frac{E}{2 R} . H e n c e i_{2} > i_{3} > i_{1} . A l o n g t i m e a f t e r t h e k e y i s c l o s e d, t h e i n d u c t o r b e h a v e s l i k e a r e s i s \tan c e l e s s w i r e T h e r e f o r e, i_{1} = \frac{E}{2 R}, i_{2} = \frac{E}{\frac{R}{2}} = \frac{2 E}{R} a n d i_{3} = \frac{E}{R} . S o i_{2} > i_{3} > i_{1} .$