. The switch in the circuit shown below has been open for a long time before closingat t = 0.25 V+50 Ω100 Ω2017/300 Ω200 Ωa) Find io(0), iL(0¯), and vi(0-)b) Find i.(0*), iL(0*), and v1(0*)c) Find io(co), iL(00), and VL(00)d) Write the expression for i(t), for t≥ 0.e) Write the expression for io(t), for t≥ 0+.f) Write the expression for vL(t), for t≥ 0¹.t=050 mH+VL

Since you have posted a question with multiple sub-parts, we will provide the solution only to the first three sub-parts as per Q&A guidelines. Please repost the remaining sub-parts separately.The given circuit isWe need to determine a. io(0−), iL(0−) and vL(0−).b. io(0+), iL(0+) and vL(0+).c. io(∞), iL(∞) and vL(∞).At t = 0-, the switch has been opened.The circuit will reach steady state. i.e the inductor will be short circuit.i.eLet the voltages at the nodes A and B be VA and VB respectively.Let us consider the node B as reference. i.e grounded.⇒VB=0 V.From the above circuit,⇒io(0−)=0 A. [ ∵ open circuit ]⇒vL(0−)=0 V. [ ∵ It is measuring across a short circuit ]By writing nodal equation at node A,⇒VA−2550+VA−0300+VA−0100+200=0 ;⇒6VA−150+VA+VA=0 ;⇒VA=18.75 V.The current iL(0−) is given by⇒iL(0−)=VA−0100+200 ;⇒iL(0−)=18.75−0100+200 ;⇒iL(0−)=62.5 mA.At t = 0+, the switch has been closed.The inductor will be replaced by an open circuit in parallel with a current source of value iL(0−).i.eLet the voltages at the nodes A, B and C be VA, VB and VC respectively.Let us consider the node B as reference. i.e grounded.⇒VB=VC=0 V.The current through the inductor can't change instantaneously.⇒iL(0+)=iL(0−)=62.5 mA.By writing nodal equation at node A,⇒VA−2550+VA−0300+VA−0100=0 ;⇒6VA−150+VA+3VA=0 ;⇒VA=15 V.By writing nodal equation at node C,⇒0−VA100+io(0+)+(62.5×10−3)=0 ;⇒0−15100+io(0+)+(62.5×10−3)=0 ;⇒io(0+)=87.5 mA.By writing KVL equation in rightmost loop,⇒200iL(0+)+vL(0+)=0 ;⇒200(62.5×10−3)+vL(0+)=0 ;⇒vL(0+)=−12.5 V.As t→∞, the circuit will reach steady state. i.e the inductor will be short circuit.i.eFrom the above circuit,⇒vL(∞)=0 V. [ ∵ It is measuring across a short circuit ]Here, the current iL(∞) is the current flowing through 200 Ω resistor. But 200 Ω resistor is connected in parallel with a short circuit. Hence, no current will flows through 200 Ω resistor.⇒iL(∞)=0 A.i.eLet the voltages at the nodes A and B be VA and VB respectively.Let us consider the node B as reference. i.e grounded.⇒VB=0 V.By writing nodal equation at node A,⇒VA−2550+VA−0300+VA−0100=0 ;⇒6VA−150+VA+3VA=0 ;⇒VA=15 V.The current io(∞) is ⇒io(∞)=VA−0100 ;⇒io(∞)=15−0100 ;⇒io(∞)=150 mA.a. ⇒io(0−)=0 A.⇒iL(0−)=62.5 mA.⇒vL(0−)=0 V.b.⇒io(0+)=87.5 mA.⇒iL(0+)=62.5 mA.⇒vL(0+)=−12.5 V.c.⇒io(∞)=150 mA.⇒iL(∞)=0 A.⇒vL(∞)=0 V.