Show that a series RL circuit is a low pass filter if the output is taken across the resistor. Calculate the cut-off frequency (f_c) if L = 2 mH and R = 10 kΩ.

For t<0, the switch is closed.

Capacitor acts as open to DC.

Voltage across the capacitor,

     v(0^-)= {(12||4)×V}÷{(12||4)+6} = 8VFor t=0, the switch is opened,Voltage across the Capacitor cannot change instantaneously,So, v(0^-)= v(0) = 8VAt t>0, The capacitor is discharging.R_th = (12||4) = 3 ohmC= 1/6 FTime constant, Π = R_th×C = 0.5 So, v(t) = v(0)e^-t/Π  = 8e^-t/0.5 =8e^-2t V (Ans.)