22Fall_hw_sol_coverHomework 2 Solutions Word DocumentUniversity of Illinois Fall 2022ECE210 / ECE211 -Homework 2 -SolutionsDue: Wednesday, September 7 @ 11.59pm.(notice the Wednesday deadline due to Monday holiday)Homework Policy:HWs are due on Tuesdays at 11.59pm and submissions will be made via Gradescope. You can join the coursethere with the entry code 8NEND8, use your UIN as your Student ID #. You can nd instructions on howto scan/upload your HW on the course website, including the Gradescope's app.Late assignments will bededucted12.5% per late hour, so after 8 hours, you will get zero, so please markyour calendar with the deadlines to avoid losing points. Don't wait until the last minute to submit it andthen run into internet issues.In order to account for sickness, travel or internet issues, the lowest homework grade will be dropped for 211students and the lowest two homework grades will be dropped for 210 students.You will be expected toprovide detailed explanations of your solutionsin order to obtain credit inyour homeworks. Graders should not have to guess or make assumptions about why you are using a certainequation, or how you came up with an equation, etc. Conversely, solutions lacking full explanations willreceive zero credit even when the answer provided may be correct.Do notsolve the HW directly in this pdf because there is not enough space for it.Make sure you box your answers and match problem parts accurately in Gradescope, or you will bededucted5% of the corresponding problem part. You can nd instructions on how to do that on the course websiteMake sure that your HW is neat enough to read. Graders has the exibility todeductup to 20% for lack ofneatness.Please report angles in radians and in the range(−π, π]Failure to do so could result in loss of some credit.Homeworks constitute an essential component of your learning experience in the course and prepare you foryour exams in eective ways. Investing time to do your homeworks with care will pay o when you are takingyour exams.Solutions will be posted in Canvas 8 hours after the corresponding deadline.Regrades: You will receive an email from Gradescope so you can log in and see your graded homework. Ifafter looking at the posted solutions, you feel there was an inaccuracy in the grading of your homework, youcan request a regrade within Gradescope itself. Make sure you submit regrade requests before 11.59pm ofthe Tuesday after your graded homework is made available via Gradescope. Regrades willnot be acceptedafter that date.©2022 Juan Alvarez. All rights reserved. Redistributing without permission is prohibited.Page 1 of 11) Signature 2) Survey. Solution: Find the coefficients K1 and K2 using the superposition method (textbook pp. 48-52). Then, solve for y, the current through the 20 Ω resistor. Let 𝑓𝑓1= 20𝑉𝑉 source and 𝑓𝑓2= 4𝐴𝐴 source. Analyze the circuit using the source suppression method to find 𝐾𝐾1, 𝐾𝐾2. Equation Description 1𝑉𝑉 −𝑉𝑉𝑎𝑎5Ω=𝑉𝑉𝑎𝑎10Ω+𝑉𝑉𝑎𝑎−𝑉𝑉𝑏𝑏10Ω 2𝑉𝑉 −2𝑉𝑉𝑎𝑎= 𝑉𝑉𝑎𝑎+ 𝑉𝑉𝑎𝑎−𝑉𝑉𝑏𝑏 𝟐𝟐𝟐𝟐 = 𝟒𝟒𝟐𝟐𝒂𝒂−𝟐𝟐𝒃𝒃 KCL at node 𝑉𝑉𝑎𝑎: 𝑉𝑉𝑎𝑎−𝑉𝑉𝑏𝑏10Ω=𝑉𝑉𝑏𝑏20Ω 2𝑉𝑉𝑎𝑎−2𝑉𝑉𝑏𝑏= 𝑉𝑉𝑏𝑏 2𝑉𝑉𝑎𝑎= 3𝑉𝑉𝑏𝑏↔ 𝟐𝟐𝒂𝒂=𝟑𝟑𝟐𝟐𝟐𝟐𝒃𝒃 KCL at node 𝑉𝑉𝑏𝑏: 2𝑉𝑉 = 4 �32𝑉𝑉𝑏𝑏�−𝑉𝑉𝑏𝑏 2𝑉𝑉 = 5𝑉𝑉𝑏𝑏 𝟐𝟐𝒃𝒃= 𝟎𝟎. 𝟒𝟒𝟐𝟐 Use substitution to solve for Vb 𝑦𝑦1=𝑉𝑉𝑏𝑏20Ω=0.4𝑉𝑉20Ω 𝒚𝒚𝟏𝟏= 𝟎𝟎. 𝟎𝟎𝟐𝟐 𝑨𝑨 Ohm’s Law 𝑦𝑦1= 𝐾𝐾1𝑓𝑓1 𝐾𝐾1=𝑦𝑦1𝑓𝑓1=0.02𝐴𝐴1𝑉𝑉 𝑲𝑲𝟏𝟏= 𝟎𝟎. 𝟎𝟎𝟐𝟐 𝛀𝛀−𝟏𝟏 Superposition equation for 𝑦𝑦1(3) Continued: Now, perform the same analysis of the circuit with the voltage source suppressed (short circuit) and the current source set to 1A: Equation Explanation 0 −𝑉𝑉𝑎𝑎5Ω=𝑉𝑉𝑎𝑎10Ω+𝑉𝑉𝑎𝑎−𝑉𝑉𝑏𝑏10Ω −2𝑉𝑉𝑎𝑎= 𝑉𝑉𝑎𝑎+ 𝑉𝑉𝑎𝑎−𝑉𝑉𝑏𝑏 𝟒𝟒𝟐𝟐𝒂𝒂= 𝟐𝟐𝒃𝒃 𝒐𝒐𝒐𝒐 𝟐𝟐𝒂𝒂=𝟐𝟐𝒃𝒃𝟒𝟒 KCL at node Va 𝑉𝑉𝑎𝑎−𝑉𝑉𝑏𝑏10Ω+ 1𝐴𝐴 =𝑉𝑉𝑏𝑏20Ω 2𝑉𝑉𝑎𝑎−2𝑉𝑉𝑏𝑏+ 2𝐴𝐴 = 𝑉𝑉𝑏𝑏 𝟐𝟐𝑨𝑨 = 𝟑𝟑𝟐𝟐𝒃𝒃−𝟐𝟐𝟐𝟐𝒂𝒂 KCL at node Vb 2𝐴𝐴 = 3𝑉𝑉𝑏𝑏−2 �𝑉𝑉𝑏𝑏4� 2𝐴𝐴 = 2.5𝑉𝑉𝑏𝑏 𝟐𝟐𝒃𝒃= 𝟎𝟎. 𝟖𝟖𝟐𝟐 Solve the system of 2 equations for Vb 𝑦𝑦2=𝑉𝑉𝑏𝑏20Ω= 0.4𝐴𝐴 Ohm’s Law 𝑦𝑦2= 𝐾𝐾2𝑓𝑓2 𝐾𝐾2=𝑦𝑦2𝑓𝑓2=0.4𝐴𝐴1𝐴𝐴 𝑲𝑲𝟐𝟐= 𝟎𝟎. 𝟒𝟒 Superposition equation for 𝑦𝑦2 𝑦𝑦 = 𝐾𝐾1𝑓𝑓1+ 𝐾𝐾2𝑓𝑓2 𝑦𝑦 = 0.02 Ω−1∗20𝑉𝑉 + 0.4 ∗4𝐴𝐴 𝒚𝒚 = 𝟐𝟐𝑨𝑨 Full superposition equation Answer: The current flowing through the 20Ω resistor = 2A4) Consider the following circuit. Use the loop current method to obtain the loop currents, i1, i2, i3. Equation Explanation −𝒊𝒊𝟏𝟏+ 𝒊𝒊𝟐𝟐= 𝟐𝟐𝑨𝑨 Loop current relation 1Ω(𝑖𝑖2−𝑖𝑖3)−4𝑉𝑉 −2Ω(𝑖𝑖3) = 0 1Ω∗𝑖𝑖2−1Ω∗𝑖𝑖3−2Ω∗𝑖𝑖3= 4𝑉𝑉 𝒊𝒊𝟐𝟐−𝟑𝟑𝒊𝒊𝟑𝟑= 𝟒𝟒𝑨𝑨 KVL for 𝑖𝑖3 loop 4𝑉𝑉 −2Ω∗𝑖𝑖1−1Ω∗𝑖𝑖2−4𝑉𝑉 −2Ω∗𝑖𝑖3= 0𝑉𝑉 𝟐𝟐𝒊𝒊𝟏𝟏+ 𝒊𝒊𝟐𝟐+ 𝟐𝟐𝒊𝒊𝟑𝟑= 𝟎𝟎𝑨𝑨 KVL for outermost loop Solve the above system equations using your method of choice (substituting variables, matrix methods such as row reduction or Cramer’s rule, eliminating equations, etc.) Answer: 𝒊𝒊𝟏𝟏= −𝟐𝟐𝟏𝟏𝟏𝟏𝑨𝑨 𝒊𝒊𝟐𝟐=𝟐𝟐𝟎𝟎𝟏𝟏𝟏𝟏𝑨𝑨 𝒊𝒊𝟑𝟑= −𝟖𝟖𝟏𝟏𝟏𝟏 𝑨𝑨5) Assume that node v0 is ground. Write nodal equation for V1 and V2 (Node Voltage Method). Solve for V1 and V2. Equation Explanation 6𝐴𝐴+𝑉𝑉2−𝑉𝑉18Ω=𝑉𝑉13Ω 48𝑉𝑉+