Multiple choice

Assume that the switch S is in position 1 for a long time and thrown to position 2 at t = 0.

At t = 0+, the current i1 is

  1. $\dfrac{-V}{2R}$
  2. $\dfrac{-V}{R}$
  3. $\dfrac{-V}{4R}$

  4. zero
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A Correct answer
Explanation

When the switch is in position 1 for a long time, the inductor reaches steady state with current I = V/R flowing through it (inductor acts as short circuit in DC steady state). At t=0+, the inductor current cannot change instantaneously, so it remains V/R. In position 2, this current flows through 2R, creating a voltage of -V/2R across the inductor to oppose the change.

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