How does one do these types of problems? Problem 1. The key is moved to position "B" at t=0. Find i(t).202matBiTOVProblem 2.453my64(t) [v]I,12 F↑(lt)2112i(t) = ? ; t =03 Hilt) = ? tzuV (0)=4V(10)=2A2H

Since you have posted multiple questions, we will provide the solution only to the first question as per Q&A guidelines. Please repost the remaining questions separately.1.The given circuit isWe need to determine the expression of i(t) for t≥0.At t =0-, the switch has been connected to position 'A'.The circuit will reach steady state.The capacitor will be open and the inductor will be short.i.eBy writing KVL equation in the loop,⇒−10+2i(0−)+3i(0−)=0 ;⇒i(0−)=2 A.The voltage v(0−) is measuring across a short circuit. ⇒v(0−)=0 V.But the voltage across the capacitor and the current through the inductor can't change instantaneously.⇒v(0+)=v(0−)=0 V.⇒i(0+)=i(0−)=2 A.For t > 0, the switch has been connected to position 'B'.i.eThe current through the resistor is⇒iR(t)=v(t)R ;⇒iR(t)=v(t)3............................................................................(1)The current through the capacitor is⇒iC(t)=Cdv(t)dt ;⇒iC(t)=112dv(t)dt ....................................................................(2)The voltage across the inductor is⇒v(t)=Ldi(t)dt ;⇒v(t)=3di(t)dt ...........................................................................(3)By writing KCL equation at node A,⇒iR(t)+i(t)+iC(t)=0 ;⇒v(t)3+i(t)+112dv(t)dt=0 ;⇒3di(t)dt3+i(t)+112d[3di(t)dt]dt=0 ;⇒d2i(t)dt2+4di(t)dt+4(t)=0 .................................................(4)The homogeneous equation is⇒D2+4D+4=0 ;⇒D=−2,−2.The homogeneous solution is⇒iH(t)=e−2t[A+Bt],A.The particular solution is⇒iP(t)=0f(0)=00+0+4=0,A.The total solution is⇒i(t)=iP(t)+iH(t) ;⇒i(t)=0+e−2t[A+Bt], A for t≥0.⇒i(t)=e−2t[A+Bt], A for t≥0.But i(0+)=2 A.⇒i(0+)=e−2(0)[A+B(0)]=2 ;⇒A=2.From equation (3),⇒v(t)=3di(t)dt ;⇒v(t)=3d{e−2t[A+Bt]}dt ;⇒v(t)=3{−2e−2t[A+Bt]+e−2t[0+B]}⇒v(t)=e−2t[−6A+3B−6Bt], V for t≥0.But v(0+)=0 V.⇒v(0+)=e−2(0)[−6A+3B−6B(0)]=0 ;⇒−6A+3B=0 ;⇒−6(2)+3B=0 ;⇒B=4.The expression of the current i(t) is⇒i(t)=e−2t[2+4t], A for t≥0.1.The expression of the current i(t) is⇒i(t)=e−2t[2+4t], A for t≥0.