University of Illinois Fall 2022ECE210 / ECE211 -Homework 6 SolutionDue: Tuesday, October 4 @ 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.The current source in the circuit shown below is given byis(t) = 5 cos(200t + π/3) A.(a)What is the linear frequencyfof the current source?Solution:f =ω2π=2002π=100πHz(b)Transform the circuit into a phasor circuit.Solution:The impedance of the capacitor isZC=1jωC=1j200∗100∗10−6= −j50 Ω. Thephasor circuit is shown below.(c)What is the phasor voltageV?Solution:The combined impedance of the resistor and capacitor is given byZtot= R//ZC=2(−j50)2 − j50=625313− j25313=25√2√313e−j tan−1(1/25)Ωand the voltage phasor V can be calculated by Ohm's LawV = IZtot= 5ejπ/325√2√313e−j tan−1(1/25)=125√2√313ej(π/3−tan−1(1/25))V©2022 Juan Alvarez. All rights reserved. Redistributing without permission is prohibited.Page 1 of 3(d)What is the phasor currentIR?Solution:By Ohm's Law,IR=VR=V2=125√626ej(π/3−tan−1(1/25))A(e)What is the phasor currentIC?Solution:By Ohm's Law,Ic=VZC=V−j50=5√626ej(π/3+tan−1(25))A(f)What is the time dependent voltagev(t)?Solution:Converting the phasor voltage V to the time dependent voltagev(t),v(t) = Re{V ej200t} =125√2√313cos(200t + π/3 − tan−1(1/25)) V3.Consider the phasor circuit below.(a)Find the Thevenin equivalent circuit seen between terminals ab.Solution:The combined parallel impedance of the inductor and the capacitor is givenbyZ1= ZL//ZC= (j10)//(−j5) =(j10)(−j5)j10 − j5= −j10 ΩBy using the voltage division between the5 Ωresistor andZ1, we can obtain the TheveninvoltageVT,VT= 10 ·−j105 − j10= 8 − j4 ΩTo nd the Thenenin equivalent impedanceZT, we can supress the voltage source (makeit into a short circuit), and can easily see that the resistor, inductor, and capacitor are inparallel, soZT= 5//j10// − j5 = 4 − j2 ΩThe Thevenin equivalent circuit seen between terminals ab is shown below.©2022 Juan Alvarez. All rights reserved. Redistributing without permission is prohibited.Page 2 of 3(b)What value of impedance should be connected across terminals ab so that the powertransferred from the source to the load is maximum?Solution:The power transferred from the source to the load is maximum when the loadimpedance equals to the complex conjugate of the Thevenin impedance, i.e,ZL= Z∗T= 4 + j2 Ω4.In the following circuit, determine the phasorVand express it in polar form.Solution:Applying KCL at node 1,V2− V1=V − (V2− 2)−j+Vjrearranging this equation givesV2=V + j21 + jApplying KCL at the ground node,V21+Vj= j2Solving for V from those two equations, we haveV = −45− j25=2√55ej(tan−1(1/2)−π)VNote: there is an extra−πin the phase because the phasor V is in the third quadrant in thecomplex plane.©2022 Juan Alvarez. All rights reserved. Redistributing without permission is prohibited.Page 3 of
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