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Answer:

As the circuit has been closed for a long time,the circuit would have attained steady state conditions,when the switch is opened  at t=0.

Therefore the current in the inductor and 16 Ohm resistor at t= 0    is    16/16 = 1 Amp

When the switch is opened at t= 0, the 16 Volt battery gets disconnected and 1 amp current through the inductor circulates through both the 12 Ohms and 16 ohms resistor as well.

Therefore ,the  initial voltage that will appear across the inductor immediately after t=0 

will be = E0=     1*(16+12) = 28 Volts 

Time constant  =   Total resistance of the circuit/Inductance  =  (16+12)/(8*10^-3 )=  3.5*10^3

Voltage Vt  across the inductor at any time t , after  opening of the switch = Eo *e^(-Rt/L)=

28*e^ (-3.5*10^3t)

Voltage across the inductor at t=0 is     28*e^(-3.5*10^3)*0 =  28*e^0 = 28 Volts

Vt after t = 3 milliseconds = 28*e^{(-3.5*10^3)* (3*10^-3) = 28*e^(-10.5)=

 28*0.0000275= 0.00077 Volts

Vt after t = 5 seconds  =  28*e^{(-3.5*10^3)* (5*10^-3) = 28*(e^-17.5)=

= 28*0.0000000251=  0.0000007 volts

 

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