The switch S in the following circuit is closed after having been open for a ver

Question

The switch S in the following circuit is closed after having been open for a very long time. The current through the battery immediately after the switch is closed is i(t=0) and the value after a very long time t is i(t->infinity). Calculate the ratio i(t=0)/i(t->infinity).

EMF = 12.0V
R_1 = 60.0Ohm
R_2 = 30.0Ohm
L = 20.0 mH

The switch S in the following circuit is closed after having been open for a very long time. The current through the battery immediately after the switch is closed is i(t=0) and the value after a very long time t is i(t - > infinity). Calculate the ratio i(t=0)/i(t - > infinity). EMF = 12.0V R_1 = 60.0Ohm R_2 = 30.0Ohm L = 20.0 mH

Explanation / Answer

Here , at time t = 0

inductor is open circuited

hence,

i(t = 0) = 12/(60 + 30)

i(t = 0) = 0.133 A

now, at time t = infinite, inductor is short curcuited

i(t = infinite) = 12/60

i(t = infinite) =0.2 A

hence ,

the ratio i(t = 0)/i(t = infinite) = 0.133/.2

i(t = 0)/i(t = infinite) = 0.6665

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