(a) immediately after the switch is closed (after being open a long time)... ...

Question

(a) immediately after the switch is closed (after being open a long time)...
...the current through the inductor
IL =   mA

...the current through R2
I2 =    mA


(b) a long time after the switch has been closed...
...the current through the inductor
IL =   mA

...the current through R2
I2 =   mA


(c) immediately after the switch is open (after being closed a long time)...
...the current through the inductor
IL =   mA

...the current through R2
I2 =   mA


(d) a time 1.095e-04 s after the switch is open....
...the current through the inductor
IL =   mA

...the current through R2
I2 =   mA

Explanation / Answer

a) when the switch is closed initailly inductor acts as opne ckt.

so, IL = 0

I2 = E/(R1+R2)

= 10/(490 + 400)

= 1.12*10^-2 A

= 11.2 mA

b) after a long time inductor acts as short ckt.

so, IL = E/R1

= 10/490

= 2.04*10^-2 A

= 20.3 mA

I2 = 0

c) IL = 2.04*10^-2 A

= 20.4 mA

I2 = 2.04*10^-2 A

= 20.4 mA

d)

Time constant of the ckt, T = L/R2 = 0.073/400

= 1.825*10^-4 s

so,

at t = 1.095*10^-4 s

I = Imax*e^(-t/T)

= 2.04*10^-2*e^(-1.095/1.825)

= 1.12*10^-2 A

= 11.2 mA

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