University of Illinois Fall 2022ECE210 / ECE211 -Homework 4 -SolutionsDue: Tuesday, September 20, 2022 at 11:59 PM CST1. Signature.2. The circuit shown below is in DC steady state before the switch ips att = 0.(a) FindvL(0−)andiL(0−).(b) FindvL(t)andiL(t)fort > 0.Solution:In the DC steady state, before closing the switch, the inductor acts like a short, such thatvL(0−) = 0V. Becausethe switch is open att = 0−, KCL at thevLnode requires that the currentiL(0−1)be the same as that throughthe3Ωresistor on the top loop. By Ohm's Law, therefore we have thatiL(0−) =12V3Ω= 4AAfter closing the switch, we have the following circuit:In this case, the6Ωresistor that is in parallel with the voltage source does not aect the time constant ofthe inductor in the RL circuit, since it supports the same voltage as the source across it. The time constantis aected by the voltage drop across the two resistors connected in parallel, which have equivalent resistance3Ω||6Ω = 2Ω, such thatτ = L/R = (6H)/(2Ω) = 3s. The general solution for the current across the resistor isthus:iL(t) = K1e−t/τ+ K2Whent → ∞, the inductor acts like a short, therefore, by Ohm's LawiL(t → ∞) =12V2Ω= 6A = K2Page 1 of 4Applying the initial condition yieldsiL(0+) = iL(0−) = 4A = K1+ K2such thatK2= −2A. Therefore,iL(t > 0) = 6 − 2e−t3AUsing thev-irelation for the inductorvL(t) = Lddti(t), we ndvL(t > 0) = 6H ×ddt(6 − 2e−t3)A = 6V × (23e−t3) = 4e−t33. Considering the circuit shown below, assume the switch has been in position A for a long time (long enough toreach steady state). It moves to position B at timet = 0s and then back to position A at timet = 2s. FindvaluesLandRsuch thati(t) = 1 − e−1A att = 2s, andi(t) = (1 − e−1)e−1A att = 6s.Solution:Beforet = 0, the circuit is in steady state, and, without any sources, there is no current owing through theinductor, such thati(0−) = 0A. Because current through an inductor must be continuous in time,i(0+) = 0Aalso. After the switch moves to position B, we have anRLcircuit with time constantτ = L/(1Ω). The generalequation for the current through the inductor is:i(t) = K1e−tτ+ K2If the switch were to stay in position B for a long time, the inductor would act as a short, and the current owingthrough it would be 1A, which means thatK2= 1A. Applying the initial condition, we can then solve forK1:i(0+) = 0A = K1+ K2= K1+ 1Asuch thatK1= −1A. Therefore the current through the inductor in the interval0 < t < 2s isi(t) = 1 − e−t/LWhen the switch moves back to postiionAatt = 2s, the current through the inductor again must be continuousin time, i.e.,i(2+) = i(2−) = 1 − e−2/LNow the general equation for the current has a new time constant:τ = L/Rand, starting att = 2s, it becomesi(t) = K3e−(t−2)/τ+ K4With no souce in the new circuit, we havei(t → ∞) = 0, and consequently,K4= 0A. Applying the initialcondition, we can solve forK3:i(2+) = i(2−) = (1 − e−2/L)e−(t−2)L/RThe condition of the problem says thati(t) = 1 − e−1A att = 2s. Therefore, the inductance of the inductormust beL = 2H. The other condition of the problem says thati(t) = (1 − e−1)e−1A att = 6s. Therefore,−(t − 2)L/R|t=6s= −1Page 2 of 4For this to be true, the resistanceRmust beR =L4=24= 0.5Ω4. Assume that the switch in the circuit below has been closed for a long time and opens att = 0s. Find and sketchvR(t),vC(t), andiC(t).Solution:In the steady-state, before the switch opens (i.e., att < 0), there is no current owing through the fully chargedcapacitor, such thatiC(0−) = 0. The voltage right before the switch closes (i.e., att = 0−) can be obtainedfrom Ohm's Law asvR(0−) = vC(0−) = vC(0+) = (10A)(4Ω) = 40VAfter the switch opens, we notice that the Thevenin equivalent resistance that the capacitor sees between itsterminals is:RT= 4 + 8 + 8 = 20ΩThe general equation for thisRCcircuit with time constantτ = RC = 20s is:vC(t) = K1e−t/20+ K2Whent → ∞, there will be no current owing through the capacitor, and it will be fully charged with the samevoltage across the4Ωand8Ωresistors in series:vC(t → ∞) = (10A)(4Ω + 8Ω) = 120Vwhich yieldsK2= 120V. Finally, applying the initial condition:vC(0+) = 40V = K1+ K2yieldsK2=-80 V. Consequently, the voltage across the capacitor fort > 0becomesvC(t) = 120 − 80e−t/20VWe can ndiC(t)using thev-irelationiC(t) = CddtvC= 4e−t/20AFrom KCL at the top node, we can obtainvR(t)4Ω+ iC(t) = 10Awhich yields the voltage across the resistorvR(t) = 40 − 4iC(t) = 40 − 16e−t/20VFor sketchingvR(t),vC(t), andiC(t), it is useful to know some approximations ofe−x, e.g.:e−1≈ 0.37ande−5≈ 0.01, which have the following practical applications:Page 3 of 4After one time constant(t = τ), a capacitor has either discharged down to 37% of its initial charge (whendischarging) or has reached 63% of its nal charge (when charging)Att = 5τ, a capacitor is either almost fully charged (when charging) or almost completely discharged (whendischarging)5. Assuming linear operation and making use of the ideal op-amp approximations, determinev0(t)att=1 ms inthe following circuit. Assume thatvc(0) = −1V.Solution:Assuming linear operation, there is no current owing into the op-amp through either terminal (+or−), so wecan isolate theRCcircuit with time constantτ = RC = (1kΩ)(1µF) = 1ms, giving the general solution for thevoltage across the capacitor asvC(t) = K1e−t/τ+ K2Whent → ∞, all the charge in the capacitor should have dissipated, thereforevC(t → ∞) = 0 = K2. Applyingthe initial condition yieldsvC(0+) = vC(0−) = −1V = K1+ K2such thatK1= −1V. ThereforevC(t > 0) = −e−t/(1ms)VWriting a KCL equation in the(−)terminal of the op-amp, wherev−≈ v+= vC(t), we have1V − vC(t)1kΩ=vC(t) − v0(t)1kΩwhich simplies tov0(t = 1ms) = −2e−1− 1 ≈ (−2 × 0.37) − 1 = −1.74VPage 4 of