Consider the following circuit, wherethe voltage source v,(t) = u(t) implements a step functionthe dependent current source operates with B = 1n-1Additionally, there is no energy stored in the circuit for t <0 seconds.Bvo(t)1H1Ω100아Į i(t)Vg(t)1F:vat)+a) Determine I(s) [Hint: use mesh-current method]b) Compute the values lim i(t) and lim i(t) [Hint: use the initial- and final-value theorems]t→0c) Determine if the results obtained in (b) agree with the known circuit behavior (Hint: when t → 0the circuit's voltages and currents approach constant values; hence, you can analyze the circuitbehavior under DC conditions to double-check the result obtained in b)).

Given the circuit as shown below: vg(t)=u(t) β=1 a) We need to determine I(s). b) We need to compute the values limt→0i(t) and limt→∞ i(t). c) We need to compare the values of i(t) by using normal expression of i(t) with that using initial value and final value theorem. a) At t=0- : i(t)=0 The Laplace equivalent circuit is as shown below: Applying KCL at Va : Va1s+Va-1s1+Va-Vb1=0Va(s+2)=1s+VbVa=1(s+2)1s+Vb ....(1) Applying KCL at node Vb : Vb1+Vb-Vas+βVa=0Vb1+Vb-Vas+Va=0Vb1+1s=Va1s-1Vb1+1s=1(s+2)1s+Vb1s-1Vb1+1s-1(s+2)1s-1=1s(s+2)1s-1Vbs(s+2)+(s+2)-(1-s)s(s+2)=1-ss2(s+2)Vb=1-ss(s+2)+(s+2)-(1-s) =1-ss(s2+2s+1) =1-ss(s+1)I(s)=Vb1 =1-ss(s+1) =1s-2s+1I(s)↔ℒ-1i(t)=(1-2e-t)u(t) Answer thus, I(s)=(1-2e-t)u(t)b) limt→0i(t)=lims→∞sI(s) =lims→∞1-s1+s =-1limt→∞i(t)=lims→0sI(s) =lims→01-s1+s =1 Answer thus, limt→0i(t)=-1 and limt→∞i(t)=1c) i(t)=(1-2e-t)u(t)i(t=∞)=1i(t=0)=-1 Thus with normal method the values of i(t) is equal to that using initial value and final value theorem.