University of Illinois Fall 2022ECE210 / ECE211 -Homework 1 - SolutionsDue: Tuesday, August 30 @ 11.59pm.Problems:1.Sign acknowledging you will abide by this course's and the University's Academic Integritypolicies or face sanctions for not doing so. These policies include, among others, plagiarism:representing the words or ideas of others as your own; cheating: using unauthorized materialsand/or information. The possible sanctions include, among others, point reduction, reducedletter grades and an F in the course. If your solution upload does not include your signature,your homework will NOT be graded, resulting in a zero.sign:2. Some of the following circuits violate KVL/KCL and/or basic denitions of two-terminal elements given inSection 1.3 of the textbook. For each one of them indicate if it is ill-specied or not, and explain your reasoning.Solution:(a) Violates KVL : the outer loop doesn't sum to zero!(b) Applying KVL clockwise starting at the bottom left corner:−2V+ v3A+ (1 Ω)(3A) = 0 ⇒ v3A= −1V.Circuit is correct.(c) Violates KCL : currents entering any node don't sum to zero!(d) This circuit violates KVL, top node cannot be both 6 V and 3 V!(e) Applying KVL counter-clockwise on outer loop, starting at top left node:2V− (2Ω)(2A) −v2A= 0 ⇒ v2A= −2V.KCL at the top left node:0 = i2V+2V1Ω+ 2A⇒ i2V= −4A.Circuit is correct.3. In the following circuit, determineiand the absorbed power at each source.Solution:©2022 Juan Alvarez. All rights reserved. Redistributing without permission is prohibited.Page 1 of 5Applying KCL at the upper node yieldsi +6 V1 Ω= 4 Ai = −2 A.To determine the absorbed power of any element it is important to follow the SRS sign convention.The absorbed power at the6 Vvoltage source isP6V= (6 V) (i) = (6 V) (−2 A) = −12 W,which is to say that the6 Vvoltage source is delivering a power of12 W.The absorbed power at the4 Acurrent source isP4A= (6 V) (−4 A) = −24 W,which is to say that the4 Acurrent source is delivering a power of24 W. Notice that we used−4A becausethe current is entering the source through its negative terminal.4. For the following circuit, determinei1andv1.Solution:Applying KCL at the upper node yields5 A = i1+12 V6 Ωi1= 3 A.To determine thev1,apply a KVL on the left-side loop counter-clockwise starting at the top left node:v1− 12 + (6 Ω)(i1) = 0v1− 12 + (6 Ω)(3) = −6 V.5. In the following circuit determinevxand calculate the absorbed power of each circuit element. Which elementis injecting the energy absorbed in the circuit?Solution:The KCL equation at the top nodevxcan be written as7 =vx3+ 2 vx,©2022 Juan Alvarez. All rights reserved. Redistributing without permission is prohibited.Page 2 of 5which yieldsvx= 3V.Absorbed powers:P3Ω=v2x3 Ω=323= 3W (absorbs energy)P7A= v7A×(−7A) = 3V×(−7A) = −21W (injects energy). Notice that we used−7A because the currententers the source through the negative terminal.P2vx= v2vx× (2vx) = vx× 2vx= 2v2x= 2(3)2= 18W (absorbs energy)The independent current source is injecting the energy. Notice that the21 Wdelivered by the independentsource matches the sum of the absorbed powers of the resistor and the dependent source.6. In the following circuit determine:(a) all the unknown element and node voltagesva, vc, vd, ve, vf, v1, v2, v3(b) the voltage drop in the circuit from the reference to node 4(c) the voltage rise from node 2 to node 3(d) the voltage drop from node 1 to the referenceSolution:(a) The KCL at nodev2can be written as1A+ 3A=v21 Ωv2= 4V.Node Voltages:v1: move from ground to node 1 through node 2:v1= v2− 6 V = 4 − 6 V =⇒ v1= −2V.v2:v2= 4V.v3: move from ground to node 3 through node 2:v3= v2+ 3A× 4Ω = 4 + 12 V=⇒ v3= 16V.v4: move from ground to node 4 and recall that moving from positive to negative terminal loses energy:v4= −(3A) × 2Ω = −6 V.Element voltages:va= v1− 0 = −2Vvc= v2− 0 = 4Vvd= v2− v3= 4V− 16V= −12Vve= v3− v4= 16V− (−6V) = 22Vvf= 0 − v4= 6V.(b) Voltage drop from the reference to node 4:0 − v4= vf= 6V.(c) Voltage rise from node 2 to node 3:v3− v2= −vd= 12V.(d) Voltage drop from node 1 to the reference:v1− 0 = va= −2V.©2022 Juan Alvarez. All rights reserved. Redistributing without permission is prohibited.Page 3 of 57. LetA =√3 − j√3andB = −1 − j√3.(a) ExpressAin exponential form.(b) ExpressBin exponential form.(c) Determine the magnitudes ofA + BandA − B.(d) ExpressABandA/Bin rectangular form.Solution:i.A =√3 − j√3,|A| =q(√3)2+ (−√3)2=√6,∠A = arctan−√3√3= −π4∴ A =√6e−jπ4ii.B = −1 − j√3,|B| =q(−1)2+ (−√3)2= 2,∠B = ±π + arctan−√3−1= −π +π3= −23πrecall±πmust be added because Re{B} < 0∴ B = 2e−j23πiii.|A + B| =√3 − j√3 − 1 − j√3=q√3 − 12+ (−2√3)2=p3 − 2√3 + 1 + 12 =p16 − 2√3|A − B| =√3 − j√3 + 1 + j√3=√3 + 1= 1 +√3iv.AB = (√6e−jπ4)(2e−j23π) = 2√6e−j1112π= 2√6cos−1112π+ j sin−1112π = −√3(1 +√3) −√3(√3 − 1)j ≈ −4.732 − 1.268j.A/B = (√6e−jπ4)/(2e−j23π) =√62ej512π=√62cos512π+ j sin512π =14√3(√3 − 1) +14√3(1 +√3)j ≈ 0.317 + 1.183j.8. Determine the rectangular forms of(a)√2ejπ4(b)√2e−jπ4(c)√2ej3π4(d)√2e−j3π4(e)2ejπ6(f)2e−jπ6Solution:i.√2ejπ4=√2cosπ4+ j sinπ4 = 1 + j.ii.√2e−jπ4=√2cos−π4+ j sin−π4 = 1 − j.iii.√2ej3π4=√2cos3π4+ j sin3π4 = −1 + j.iv.√2e−j3π4=√2cos−3π4+ j sin−3π4 = −1 − j.v.2ejπ6= 2cosπ6+ j sinπ6 =√3 + j.vi.2e−jπ6= 2cos−π6+ j sin−π6 =√3 − j.9. Simplify(a)P = 2ej5π4− 2e−j5π4(b)Q = 8e−jπ4− 8ejπ4(c)R =ej3π4e−jπ4.Solution:i.P = 2ej5π4− 2e−j5π4= j4 sin5π4= −j2√2ii.Q = 8e−jπ4− 8ejπ4= −j16 sinπ4= −j8√2iii.R =ej3π4e−jπ4= ejπ= −1©2022 Juan Alvarez. All rights reserved. Redistributing without permission is prohibited.Page 4 of 510. This last problem will introduce the basic python programming concepts that will be useful to you in thiscourse, and others. You will use an IPython notebook via PrairieLearn with unlimited attempts. Make surethat you complete the PrairieLearn question before this homework's due date to get credit for it. You donot need to submit an aswer for this