An inductor (L=0.03H) and a resistor (R=0.15 KΩ) are connected in series to a battery of 15V EMF. The key K1 has been kept closed for a long time. Then at t=0, K1 is opened and key K2 is closed simultaneously. At t=1ms, the current in the circuit will be :(e⁵≈150) 100mA, 67mA, 6.7mA, 0.67mA

As the inductor will behave as zero resistance after infinite time in LR circuit the current will be
i∞=V/R=15v/150Ω=.1A
As the switch K1 is opened and K2 is closed The Inductor will behave as source and current at any time can be found using KCL
−L(di/dt)−IR=0.. (i)
Integrating the equation (i)
loge(i/i∞)=−t/(L/R) ⇒i=i∞e^−t/(L/R)
In LR decay circuit the current is given as
i=i∞e^−Rt/L=i∞e^−t/τ
at time t=1ms current can be found by substituting
L=.03H
R=150Ω
i=.67mA