Analog design interview/screening test question || Amit Bar

[@dushyantsinhdabhi4772]

please keep the video of yours on the right side bottom; so that options are visible

[@SAhellenLily]

After 3sec switch is closed, L open, Diode off VL=L(diL/dt) iL(3) initial value 1/L*v*(t-0)=0.3A And then t>3 sec swith is open, L short, Diode oN iL(t) =0.3+(1*(-9))/L*t...(1) 0=0.3+(-9/30*10^-6)*3*10^-6*t..(1) -0.3=-0.3t,t=1usec... Answer