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(D) At $t = 0$ (just after the switch is closed),the inductor acts as an open circuit (infinite resistance). Circuit $(1)$: $i_1 = 0$. Circuit $(2)$:

Vedclass · Accepted Answer

$i_2 = i_3 > i_1$ $(i_1 
eq 0)$; $i_2 > i_3 > i_1$

Vedclass · Answer

(D) At $t = 0$ (just after the switch is closed),the inductor acts as an open circuit (infinite resistance). Circuit $(1)$: $i_1 = 0$. Circuit $(2)$: The inductor branch is open,so the current flows through the resistor in parallel with the battery. $i_2 = E/R$. Circuit $(3)$: The inductor branch is open,so the current flows through the resistor in parallel with the battery. $i_3 = E/R$. Thus,$i_2 = i_3 > i_1$ (where $i_1 = 0$). At $t = \infty$ (a long time later),the inductor acts as a short circuit (zero resistance). Circuit $(1)$: $i_1 = E/(2R)$. Circuit $(2)$: The inductor branch is a short circuit,bypassing the resistor in parallel with it. The total resistance is $R$. $i_2 = E/R$. Circuit $(3)$: The inductor branch is a short circuit,bypassing the resistor in series with it. The total resistance is $R$. $i_3 = E/R$. Wait,re-evaluating circuit $(3)$ at $t = \infty$: The resistor is in parallel with the inductor. The inductor shorts the parallel resistor. The total resistance is $R$. $i_3 = E/R$. Comparing the circuits at $t = \infty$: $i_1 = E/(2R)$,$i_2 = E/R$,$i_3 = E/R$. Thus,$i_2 = i_3 > i_1$.

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.

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