University of Illinois Fall 2022ECE210 / ECE211 -Homework 3 -SolutionsDue: Tuesday, September 13 @ 11.59 pm.1. Signature.2. Determine the Thevenin and Norton equivalent circuits of the following network between nodesaandb, anddetermine the available power of the network.Solutions:To nd the open circuit voltageVT, we setup the KCL equation at the center top node6 − VT3= −2i = −26 − VT3.Solving forVTwe haveVT= 6V.To nd the short circuit currentIN, we short nodesaandb, and setup the KCL equation at the center topnode6 − Va3= −2i +Va3= −26 − Va3+Va3.Solving forVawe haveVa=92V⇒ IN=Va3=32A.FromVTandIN, the equivalent resistance isReq=VTIN= 4 Ω.The Thevenin and Norton circuits are plotted below.The available power of the circuit isPa=V2T4R=94W.©2022 Juan Alvarez. All rights reserved. Redistributing without permission is prohibited.Page 1 of 63. Consider the circuit shown.(a) Assuming ideal op-amp approximations, determinevoas a function ofvs.(b) Assuming ideal op-amp approximations, and letvs= 2V,Rs= 1k Ω, andRL= 10k Ω. Determine thevalues ofisandiL.(c) The values ofisandiLare not equal and would appear to invalidate Kirchho's Current Law as appliedto the super-node indicated by the blue rectangle in the gure below. Where is the missing current?Explain what is missing from our model.Solutions:(a) Using the ideal op-amp approximation, we haveis= 0, v+= v−= vs= vo.(b) Givenvs= 2V,Rs= 1k Ω, andRL= 10k Ω, from (a) we haveis= 0,vo= iL· 10k Ω = 2V⇒ iL= 0.2mA.(c) Current can enter and exit the KCL at the supernode through 4 branches, not just 2. The gure showstwo branches entering/exiting the supernode: one throughRs,with currentis, and another throughRL,with currentiL.The extra two branches are formed by the external power supplies, which are notgenerally shown on the op-amp symbol.4. Consider the circuit shown, and assume ideal op-amp approximations.©2022 Juan Alvarez. All rights reserved. Redistributing without permission is prohibited.Page 2 of 6(a) Determinev2in terms ofv1.(b) Determine the values ofi1andi2whenv1= 0.5V.(c) Determine the power injected by (the input voltage source)v1= 0.5V.(d) Determine the power dissipated by the ve resistors whenv1= 0.5V.Solutions:(a) Using ideal op-amp approximation, we havev+= v−= 0V.Next, we establish the KCL equation at thev−node, and top center node (we label it to have voltagevx)v15k=0 − vx10k,0 − vx10k=vx2k+vx− v24k.From above, we ndv2= −345v1.(b) Givenv1= 0.5V, we havei1=v15k= 0.1mA.andi2=v2− vx4k+v26k.Calculatingv2= −3.4V, vx= −1V,we havei2=−2.44k−3.46k= −76mA.(c) The power absorbed by the voltage source isP = −v1· i1= −0.05mW(or the power injected is 0.05 mW).©2022 Juan Alvarez. All rights reserved. Redistributing without permission is prohibited.Page 3 of 6(d) The power dissipated by the resistors areP5k= i21· 5k = (0.1mA)2· 5k = 0.05mWP10k= (0.1mA)2· 10k = 0.1mWP2k= (0.5mA)2· 2k = 0.5mWP4k= (0.6mA)2· 4k = 1.44mWP6k= (0.567mA)2· 6k = 1.93mW5. Consider the circuit shown, and assume ideal op-amp approximations.(a) Determine the value ofvoas a function ofv1.(b) Letv1= 3V. A design aw is identied during production and your boss asks that you redesign thecircuit such thatvo= 2v1= 6V. For ease of repairing the circuits already produced, she wants you tochange only one resistor. Furthermore, in the interest of power conservation, she asks that your changeprovides the correct output with the least power absorbed by the combination of four resistors.Solutions:(a) We setup the KCL equations at nodesv+andv−v1− v+2k=v+1k,v−10k=vo− v−10k.Using the ideal op-amp approximation, we havev+= v−= v, such thatvo=23v1.(b) The total power absorbed by four resistors can be calculated asPR=v21R1+ R2+v2oR3+ R4sinceR1andR2are in series driven byv1, andR3andR4are in series driven byvo. In order to reducethe power consumption by the resistors, we prefer to increase the resistance of one of the resistors. Usingv1= 3V andvo= 6V, the KCL equations in (a) can be written as3 − vR1=vR2,vR3=6 − vR4.Since we can only change one resister, we can choose to change either (i)R1orR2, or (ii)R3orR4.i. KeepingR3andR4unchanged, we havev10k=6 − v10k⇒ v = 3V,©2022 Juan Alvarez. All rights reserved. Redistributing without permission is prohibited.Page 4 of 6from which we have3 − 3R1=3R2⇒ R2→ ∞.That is, replacingR2with an open circuit. In this case, the total power consumption by the resistorswould bePR=v2oR3+ R4=3620k= 1.8mW.ii. KeepingR1andR2unchanged, we have3 − v2k=v1k⇒ v = 1V,from which we have1R3. =6 − 1R4⇒ R4= 50k Ω.In this case, the total power consumption by the resistors would bePR=93k+3660k=3620k= 3.6mW.The correction that also results in the least power dissipated by the resistors is, therefore, to merelyremoveR2.6. Consider the circuit shown and apply ideal op-amp approximations. Determine the node voltagesv1,v2, andvogiven thatvs1= 6V andvs2= 4V.Solutions:We work on the top op-amp (#1) and the bottom op-amp (#2) separately forv1andv2. After that, we solveforvousing the two voltages.Op-amp #1:With op-amp approximation, we havev+= v−= vs1= 6V.Setup KCL equation at thev−node, we havev−15k=v1− v−5k,from which we obtainv1= 8V.Op-amp #2:With op-amp approximation, we havev+= v−= vs2= 4V.Sincev2shares the same node asv−, we havev2= 4V.©2022 Juan Alvarez. All rights reserved. Redistributing without permission is prohibited.Page 5 of 6Obtainedv1andv2,vocan be obtained by setting up KCL equation at nodevov1− vo10k+v2− vo10k=vo10k,from which we havevo= 4V.©2022 Juan Alvarez. All rights reserved. Redistributing without permission is prohibited.Page 6 of