University of Illinois Spring 2016ECE210 / ECE211 - Homework 01 solutionDue: Wednesday, January 27 @ 6pm.1. Determine R and v in the following circuit.Then, select a node as refer ence and obtain all theremaining node voltages.1V2/3 Vv4ΩR3ΩSolution:Applying Ohm’s law to 3Ω resis tor,I =1 V3 Ω=13ACurrent is the same everywhere in the single loop, so apply Ohm’s law to R:R =2/3 V1/3 A= 2ΩUsing KVL, we havev = −23V − 4 ×13V −1V = −3 VThen taking the negative end of the voltage source as reference, v0= 0, we can determine the node voltagesclockwisely from v0as v1= −23V , v2= −2V and v3= −3V .2. Determine i in the following circuit.Then, calculate the power a bsorbed/injected byeach circuit element. Clearly indicate if the power isabsorbed or injected.2Ω3A 8ViSolution:Applying KCL a t the upper node yieldsi +8 V2 Ω= −3 Ai = −7 A.For finding the absorbed/injected power of any element, it is important to follow the assumed direction andsign:(a) Fo r the source, including independent a nd independent source, power is absorbed by the element ifdirection of current is the same as the direction of voltage drop (from + to - end); Power is injected fromthe element if direction of curre nt is opposite to the direction of voltage drop.(b) For passive device, such as resis tor, power is always absorbed by the element.Therefore, the power absorbed by the 8 V voltage source isp8V= (8 V) (−7 A) = −5 6 W,which is to say that the 8 V voltage source is delivering a power of 56 W .Page 1 of 6The power absorbed by the 3 A current source isp2A= (8 V) (3 A) = 24 W,which is to say that the 2 A current source is absorbing a power of 24 W.The power absorbed by the 2 Ω r e sistor isp4Ω= (8 V) (4 A) = 3 2 W,which is to say that the 2 Ω resistor is absorbing a power of 32 W.You may verify the result by considering the conservation of power: the 56 W delivered by the independentvoltage source matches the sum of the absorbed powers of the resistor and the current source.3. Determine vxin the following circ uit.Then calculate the power absorbed/injected byeach circuit element. Clearly indicate if the poweris absorbed or injected.7A2vx+-3ΩvxSolution:The KCL equation at the top node of vxcan be written as7 +vx3+ 2 vx= 0,which yieldsvx= −3 V.Absorbed/injected powers:p3Ω= vx× ix= (−3 V) ×−3 V3 Ω= 3 W (absorbs energ y)p7A= v7A× 7A = (−3V ) × 7A = −21 W (injects e nergy)p2vx= v2vx× (2vx) = (−3V) × (−6A) = 18 W (absorbs energy)The independent current source is injecting the energy. Notice that the 10 W delivered by the independentsource matches the sum of the absorbed powers of the resistor and the dependent source.4. Determine i and v in the following circuit.3Ωi2Ωv3V 2vSolution:Applying KVL we obtain3 − 3i − 2v − 2i = 0,but from Ohm’s law in the 2 Ω res istor we know thatv = 2i.Inserting this vxin the KVL e quation yields3 − 3i − 2 (2i) − 2i = 03 − 9i = 0i =13A.Therefore,v = 2i =23V.Page 2 of 65. Determine i1and v1in the following circ uit.v15 Ai16 Ω6 Ω12 V2ΩSolution:Applying KCL a t the upper central node, yieldsi1+12 V6 Ω= 5 Ai1= 3 A.Applying KVL in the left loop, we obtainv1+ i1(6 Ω) − 12 V = 0v1= 12 V − (3 A) (6 Ω)v1= −6 V6. Some of the following circuits violate KVL/KCL and/or basic definitions of two-terminal elements given in Section1.3. Identify these ill-specified circuits and explain the problem in each case.1Ω2Ω1A2Aa) b) c)2V1Ω3V8V8V4ΩSolution:(a) Circuit is correct!(b) This circuit violates KVL, to p node cannot be both 2V and 3 V!(c) This circuit violates KCL. Conisde the bottom node that connects both so urces, based on KCL, theoverall current which flows into the node must be 0. Here it cannot be 3A.7. For each one of the following circuits, find Req.1Ω1Ω1Ω2Ω2Ω2ΩReqa) b)1Ω3Ω1Ω4Ω2Ω2Ω4Ω3ΩReqSolution:(a) Simplify the cir cuit as shown below:Thus, Req=2∗22+2= 1ΩPage 3 of 6(b) Simplify the circuit as shown below:Thus, Req=4∗2.44+2.4= 1.5Ω8. Let A = −√3 − j√3 and B = 2e−jπ3.(a) Determine the real and imaginary components of A, and plot them as vectors, as well as A itself, in thecomplex plane.(b) Express A in exponential form, and plot it as a vector in the complex plane, identifying its magnitudeand phase.(c) Determine the magnitude and phase of B, and plot it as a vector in the complex plane, identifying itsmagnitude and phase.(d) Express B in rectangular form, and plot it as a vector in the complex plane, identifying its real andimaginary components.(e) Let X = A + B. Determine its rea l and imaginary components and plot it as a vector in the complexplane, identifying its real and imaginary components.(f) Let Y = A − B. Determine its real and imaginary components and plot it as a vector in the complexplane, identifying its real and imaginary components.(g) Let Z = AB. Determine its magnitude and phase and plot it as a vector in the complex plane, identifyingits magnitude and phase.(h) Let W = A/B. Determine its magnitude and phase and plot it as a vector in the complex plane ,identifying its ma gnitude and phase.Solution:(a) Re(A)=−√3 Im(A)=−√3(b) A=√6e−3π4,mag(A)=√6, phase(A)=−3π4(c) mag(B)=2, phase(B)=−π3(d) B=1 − j√3(e) X = A + B = −√3 − j√3 + 1 − j√3=1 −√3 − j2√3(f) Y = A − B = −√3 − j√3 − 1 + j√3=−1 −√3(g) Z = AB =√6e−3π4∗ 2e−jπ3= 2√6e−j(3π4+π3)= 2√6e−j13π12= 2√6ej11π12(h) W = A.B =√6e−3π4/2e−jπ3=q32e−j(3π4−π3)=q32e−j5π12Page 4 of 69. Determine the rectangular forms of the following complex numbers Z, as well as their real and imaginarycomponents and their magnitude and phase. Also plot them as vectors in the complex plane.(a) Z =√2e−j3π4−√2ej3π4,(b) Z = 3ejπ6+ 3e−jπ6.Solution:(a) Z=-1-j-(-1+j)=-2j Real part=0, imag part=-2, mag=2, phase=−π/2(b) Z =3√32+3j2+3√32−3j2= 3√3Real part=3√3, imag part=0, mag=3√3, phase=010. Let A = j(e−j3− ej3) and B =Re{3ejπ6e−j5}.(a) Express A in terms of a sine function.(b) Express A in terms o f a cosine function.(c) Express B in terms of a cosine function.Page 5 of 6(d) Express B in terms of a sine function.Solution:(a) A = 2 ∗ej3−e−j32j= 2sin(3)(b) A = 2cos(π2− 3)(c) B = Re{3ej(π6−5)} = 3cos(π6− 5)(d) B = sin(π2− (π6− 5)) = sin(π3+ 5)Page 6 of