University of Illinois Sp 2016ECE210 / ECE211 - Homework 06Solutions1. In the following circuit determine the node-voltage phasors V1, V2, and V3and express them in polar form.Solution:V1can be determined directly from the circuit,V1= 2 V.Applying KCL at V2givesV21+ 1∠90o| {z }j+V2− V1−j= 0.Likewise, the KCL equation at V3isV3j2+V3− V12= 1∠90o| {z }j.From the first KCL equation, we obtainV2+ j + j(V2− 2) = 0V2=j1 + j=ejπ2√2ejπ4V2=√22ejπ4V.From the second KCL equation, we have−jV3+ V3− 2 = j2V3=2 + j21 − j= 2j V.2. In the circuit shown for Problem 1, determine the loop-current phasors I1, I2, and I3and express them in polarform.Solution:In order to reduce the number of unknowns, we are going to express I3in terms of I2. From a KCL at node V3we notice thatI3= 1∠90o+ I2= j + I2.Applying KVL in the super-loop around the outermost circuit2 − 2I2− j2 (j + I2) = 0.Now we write a KVL equation around the loop done by current I12 − (−j) (I1− I2) − 1 (I1− (j + I2)) = 0.Page 1 of 7From the first KVL equation we obtainI2=√2e−jπ4A.this current in the second KVL equation yields to2 + jI1− I1+ j + (1 − j)I2= 0I1=2 − j1 − j=√5e−j arctan12√2e−jπ4=r52ejπ4−arctan12A.≈ 1.581ej0.32175Finally, from the first equation we obtainI3= 1 A.3. Use the phasor method to determine v1(t) in the following circuit:+-+-v2(t)2cos(4t)V4ix(t)ix(t)1H116Fv1(t)2ΩSolution:The phasor equivalent of this circuit is shown below.+-+-2j 4Ix4Ixj 4V1+ 4IxV12 VWriting the KCL equation for the super-nodeV1− (2)2+V1j4+V1+ 4Ix−j4= 0,but, from the V − I relation for the inductor’s impedance we haveIx=V1j4,which will substitute Ixin the KCL equation for the super-nodeV1− 22− jV14+ jV1+ 4V1j44= 02(V1− 2) − jV1+ jV1+ V1= 0V1=43V.Therefore the steady-state voltage v1(t) can be express asv1(t) =43cos(4t)V.4. Use the following network to complete (a) and (b):(a) Calculate the equivalent impedance of the following network for (i) ω = 5 krad/s, (ii) ω = 25 krad/s, and(iii) ω = 125 krad/s.(b) Assuming a cosine voltage input to the network with a fixed amplitude and variable frequency ω, at whichvalue of ω is the amplitude of the capacitor voltage maximized? At the same frequency, what will be theamplitude of the resistor current?Page 2 of 7+23V23⌦ab3 HW64 Problem 4.14(a) Calculate the equivalent impedance of the following network for (i) ! =5krad/2, (ii) ! = 25krad/s, and (iii) ! = 125 krad/s.(b) Assuming a cosine voltage input to the network, with a fixed amplitudeandvariable freque nc y !, at which value of ! is the amplitude of th e capacitorvoltage maximiz e d? At the same frequency, what will be the amplitude of t h eresistor current?2µF50⌦0.8mHSolution:Convert the circuit into f r eq u en cy domain:1j!CRj!L3Solution:(a) We begin by converting the circuit into the frequency domain:+23V23⌦ab3 HW64 Problem 4.14(a) Calculate the equivalent impedance of the following network for (i) ! =5krad/2, (ii) ! = 25krad/s, and (iii) ! = 125 krad/s.(b) Assuming a cosine voltage input to the network, with a fixed amplitudeandvariable freque nc y !, at which value of ! is the amplitude of th e capacitorvoltage maximiz e d? At the same frequency, what will be the amplitude of t h eresistor current?2µF50⌦0.8mHSolution:Convert the circuit into f r eq u en cy domain:1j!CRj!L3Parallel combining the capacitor and resistor yieldsParallel combine the inductor and capacitor into the equivalent load impedanceZL(!) given byZL(!)=1j!Cj!L1j!C+ j!L(1)=j!L1  !2LC(2)=1Cj!1LC !2(3)ZL(!)R4, where the equivalent load impedance ZL(ω) is given byZL(ω) =1jωCjωL1jωC+ jωL=jωL1 − ω2LC=1Cjω1LC− ω2=1jCωω2−1LCFrom there, we simply series combine to get the total equivalent impedenceZeq(ω) = R + ZL(ω)= R +1jCωω2−1LC= R −jCωω2−1LC= 50 − j(5 · 105)ωω2− 625 · 106Ω= 50 − j(5 · 105)ωω2− (25 · 103)2ΩPage 3 of 7Evaluating at these at ω ∈ {5000, 25000, 125, 000} yieldsω Zeq(ω)5,000 50 + j256≈ 50 + j4.166625,000 50 − j∞ = ∞125,000 50 − j256≈ 50 − j4.1666(b) Recognizing the form of the circuit as a voltage divider with complex load impedence, we may write thetransfer functionH(ω) =Vout(ω)Vin(ω)=ZL(ω)R + ZL(ω)=1jRCωω2+1jRCω −1LCWe want to find the frequency ωmwhere the |H(ω)| is maximized. Noting |H(ω)| is maximized when|H(ω)|2is maximized, we haveωm= arg maxω|H(ω)|= arg maxω|H(ω)|2. Evaluating |H(ω)|2yields|H(ω)|2= H∗(ω)H(ω)=ω2ω2+ R2C2ω2−1LC2Differentiating with respect to ω yields∂∂ω(ω2ω2+ R2C2ω2−1LC2)=2ωhω2+ R2C2ω2−1LC2i− ω22ω + R2C2ω2−1LC(2ω) hω2+ R2C2ω2−1LC2i2Setting the derivative to 0 and multiplying both sides by the denominator yields0 =2ωm"ω2m+ R2C2ω2m−1LC2#− ω2m2ωm+ 2R2C2ω2m−1LC(2ωm)="ω2m+ R2C2ω2m−1LC2#− ω2m− 2ω2mR2C2ω2m−1LC= R2C2ω2m−1LC2− R2C22ω2mω2m−1LC=ω4m−2LCω2+1L2C2− 2ω4m+2LCω2m= −ω4m+1L2C2∴ ωm=1√LC= 25, 000radsTo find the current through the resistor at this frequency, we observe that the current is the same as thecurrent through the load ZL(ω). We should be able to determine the voltage across the load from thetransfer function H(ω), which was previously derived. Thus finding the current is simply a matter of usingOhm’s law. At the resonance frequency, the load impedance is evaluated to beZLω =1√LC=1jC1√LC1LC−1LC= ∞Page 4 of 7Since the impedence is infinite, there can be no current flowing through the load, which thus meansIrω =1√LC= 0. (1)5. Use the following network to answer (a) through (d):+-+-VVsIs1j 11j 3j 1R1R2(a) Determine the phasor V when Is= 0.(b) Determine the phasor V when Vs= 0.(c) Determine V when Vs= 4 V and Is= −2 A, and calculate the average p ower absorbed in the resistors.(d) What is the Thevenin equivalent and the available average power of the network when Vs= 4 V andIs= −2 A?Solution:(a) When Is= 0, there is no current flowing through the inductor or the capacitor or the resistor R1, thereforeV = Vs.(b) When Vs= 0 the only current flowing through the inductor is Is, consequentlyV = Is(j1) = jIs.(c) By superposition,V = Vs+ jIs= (4 − j2)V = 2√5e−j0.4636V.For the resistor R1, the average absorbed power is ,PR1= 0W,since no current is flowing through it. For the other