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 aect the time constant ofthe inductor in the RL circuit, since it supports the same voltage as the source across it. The time constantis aected 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 4After 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 simplies tov0(t = 1ms) = −2e−1− 1 ≈ (−2 × 0.37) − 1 = −1.74VPage 4 of
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