University of Illinois Fall 2016ECE210 / ECE211 - Homework 01 SolutionDue: Wednesday, August 31 at 6:00 p.m.1. Determine i in the following circuit.Solution:The voltage drop across the resister is 6V. By Ohm’s law, the current across the resister isI =6V2Ω= 3AApplying KCL, we havei = 4A − 3A = 1A.2. (a) In the following circuit, determine all of the unknown element and node voltages.(b) What is the voltage drop in the circuit from the reference to node 4?(c) What is the voltage rise from node 2 to node 3?(d) What is the voltage drop from node 1 to the reference?Solution:(a)v1=13A · 2Ω =23V.Page 1 of 5v2=23V + 2V =83V.ib=v21Ω=83A.v3= v2− 4Ω · (−3)A =443V.v4= 0 − 2Ω · 3A = −6V.va= v1=23V.vb= v2=83V.vc= v2− v3= −12V.vd= v3− v4=623V.ve= 0 − v4= 6V.(b) Voltage drop from reference to node 4: Vdrop= ve= 6V.(c) Voltage rise from node 2 to node 3: Vrise= v3− v2= 12V.(d) Voltage drop from node 1 to reference: Vdrop= v1=23V.3. In the following circuit, one of the independent current sources is injecting energy into the circuit, while theother one is absorbing energy. Identify the source that is injecting the energy absorbed in the circuit andconfirm that the sum of all absorbed powers equals to zero.Solution:By KCL, the current across the resister (from the top to the bottom) is: I = 3A − 1A = 2A.Voltage drop across the resister: V = 2A · 1Ω = 2V.Power of the left source: Plef t= 2V · (−3)A = −6W.Power of the right source: Pright= 2V · 1A = 2W.Power absorbed by resister: Pr= 2A · 2V = 4W.Hence, the left current source is injecting energy.The sum of all absorbed powers is Plef t+ Pright+ Pr= 0.4. Calculate the absorbed power for each element in the following circuit and determine which elements injectenergy into the circuit.Page 2 of 5Solution:By KCL, the current across the resister (from the top to the bottom) is: I = 3A − (−1A) = 4A.Voltage drop across the resister: V = 4A · 1Ω = 4V.Power of the left source: Plef t= 4V · (−3)A = −12W.Power of the right source: Pright= 4V · (−1A) = −4W.Power absorbed by resister: Pr= 4A · 4V = 16W.Hence, both current sources are injecting energy.5. In the circuit given, determine vxand calculate the absorbed power for each circuit element. Which elementis injecting the energy absorbed in the circuit?Solution: Applying KCL to node n.6A = 2vx+vx2ΩSolving the above equation, we getvx=125VPower of the independent source: Plef t= vx· (−6)A = −725W.Power of the dependent source: Pright= vx· 2 · vx=28825W.Power of the resister Pr=vx2Ω· vx=7225W.Hence, the independent source is injecting energy.Page 3 of 56. Some of the following circuits violate KVL/KCL and/or basic definitions of two-terminal elements given inSection 1.3. Identify these ill-specified circuits and explain the problem in each case.Solution:(a) Violates KVL.(b) Correct.(c) Violates KVL.(d) Violates KCL.(e) Correct.7. (a) Let A = 3 − j3. Express A in exponential form.(b) Let B = −1 − j1. Express B in exponential form.(c) Determine the magnitudes of A + B and A − B.(d) Express AB and A/B in rectangular form.Solution:(a) |A| =√32+ 32= 3√2.6A = tan−1(−33) = −π4. A = 3√2e−jπ4(b) |B| =p(−1)2+ (−1)2=√2.6B = π + tan−1(1) =5π4. B =√2ej5π4(c) |A + B| = |3 − j3 + (−1 − j1)| = |2 − j4| =√22+ 42= 2√5|A − B| = |3 − j3 − (−1 − j1)| = |4 − j2| =√42+ 22= 2√5(d) AB = (3 − j3)(−1 − j1) = −6A/B =3−j3−1−j1=(−1+j1)(3−j3)(−1+j1)(−1−j1)=j62= j3Page 4 of 58. (a) Determine the rectangular forms of 7ejπ4, 7e−jπ4, 5ej3π4, 5e−j3π4.(b) Simplify P = 2ej5π4− 2e−j5π4, Q = 8e−jπ4− 8ejπ4, and R =ej3π4e−jπ4Solution:(a) 7ejπ4= 7 cosπ4+ j7 sinπ4=7√22+ j7√227e−jπ4= 7 cos(−π4) + j7 sin(−π4) =7√22− j7√225ej3π4= 5 cos3π4+ j5 sin3π4= −5√22+ j5√225e−j3π4= 5 cos(−3π4) + j5 sin(−3π4) = −5√22− j5√22(b) P = 2(ej5π4− e−j5π4) = 2(cos5π4+ j sin5π4− (cos(−5π4) + j sin(−5π4))) = 4 · j sin5π4= −j2√2Q = 8e−jπ4− 8ejπ4= 8 · −j√2 = −j8√2R = ej(3π4−(−π4))= ejπ= −1Page 5 of
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