University of Illinois Fall 2016ECE210 / ECE211 - Homework 02Due:Wednesday,September7at6:00p.m.1. (a) Write nodal equations for the circuit below, assuming that v0=ground.Solveforv1and v2.(b) Write nodal equations for the same circuit, but this time, assu me that v1=ground.Solveforv0and v2.(c) How are your answers to parts (a) and (b) related?Solution:(a) Taking v0as reference (ground), apply KCL to node v1:v18+v1 v23=611v1 8v2= 144.Similarly, apply KCL to node v2:v2 v13+v22+2 = 05v2 2v1= 12.Solve for v1and v2,wegetv1= 16Vv2=4V.(b) Taking v1as reference (ground) instead, apply KCL to node v0:v08+v0 v22+6 = 25v0 4v2+ 32 = 0.Similarly, apply KCL to node v2:v23+v2 v02+2 = 05v2 3v0= 12.Solve for v0and v2,wegetv0= 16Vv2= 12V(c) Observation: The relative relationship betwe en v0, v1,andv2are remained in question (a) and (b),regardless of reference point.Numerically:v1 v0= 16  0 = 0 + 16 = 16Vv2 v1=4 16 = 12  0=12V.Page 1 of 62. The circuit shown below is called a bridge or ladder circuit; variations of this circuit are used in many signalprocessing applications. If R1is a variable resistor, and if all of the other resistors are constant, it is possible tovary the “output voltage” v1 v2across a wide range of values by adjusting R1.(a) Use nodal analysis to nd v1and v2in the following circuit, given that R1=2k⌦, R2=8k⌦, R3=4k⌦,R4=4k⌦,andR5=4k⌦.Hint:youdonotneedtoknowthecurrentthroughthesource.(b) Assume that R2through R5are as given in part (a), but R1is variable. What value of R1results in v1= v2?Solution:(a) Pick the “-” end of voltage source as reference (ground). Apply KCL to node v1:v1 2R1+v1R2+v1 v2R5=0Similarly, apply KCL to node v2:v2 2R3+v2R4+v2 v1R5=0Given that R1=2k⌦, R2=8k⌦, R3=4k⌦, R4=4k⌦,andR5=4k⌦,solveforv1and v2,wegetv1=2819Vv2=2219V.(b) Using the equations in part (a), substitute the value of R2through R5only and let v1= v2,weget:v1 2R1+v18=0v1 24+v14=0Solve for v1and R1:R1=8k⌦v1=1V3. Find ixusing nodal analysis.Solution:Page 2 of 6Apply KCL (node analysis) to the 2 lo op s. Let the voltage at the node above the depend ent current sourcebe vx.Wecangenerate2equations:vx2+vx 62=4ixix=vx2Solve for ixand vx,wegetix= 1.5Avx= 3V.4. Find ixusing lo op analysisSolution:Apply KVL (loop analysis) to the 2 loops. We can gene rate 2 equations:3ix+(2) + (3ix) + 2(ix iy)=05iy+ 2(iy ix)+3ix+(5) = 0Solve for ixand iy,wegetix=1.5Aiy=0.5A.5. In the following circuit, find the open-circuit voltage and the short-circuit current between nodes a and b:Solution:To find the open-circuit voltage, we must observe that no current may flow through the rightmost resistor.Therefore, vocis equal to the voltage drop across the current source. Writing a KCL equation at the nodeabove the current source gives usi +2i =3i =0,which implies i =0. This means that the current through the other resistor is also 0, which means thatvoc= 4V.Page 3 of 6To find the short-circuit current, we first draw a short b etween nodes a and b. Constructing a KVL equationaround the outer loop yields4=1i +1isc.Constructing a KCL equation at the node above the current source yieldsisc= i +2i =3i.Solving the system yieldsi =1A =) isc= 3A.6. Determine the Thevenin equivalent of the following network between nodes a and b:Solution:To find the open-circuit voltage, we construct a KCL equation at the node above the 4A source, yieldingix= ir+4.Clearly the dependent source current must run through the top resistor, soir= 2vx.The current through the left resistor can be computed by Ohm’s law, givingix= vx/1=vx.Page 4 of 6Thus we havevx=4 2vx3vx=4vx=43V.The open-circuit voltage is clearly seen to be the sum of the voltages across the resistors. Thus, we havevoc= vx+1ir=4/3  2vx=4/3  8/3= 43V.To find the short-circuit current, we construct a KCL equation at node a,yieldingisc= ir 2vx.We may then note that the voltage across both the resistors is the same since the y are in parallel. Thusir= vx/1=vxandisc= ir 2ir= ir.Note that the depende nt source does not contribute in this equivalent circuit, since it takes as much currentfrom the node as it supplies as shown in the diagram below.Thus it may be ignored for computational purposes. This yields the following equivalent diagram.The remaining current from the 4A source is equally split between the 1⌦ resistors. Thereforeir= 2A = iscisc= 2A.Page 5 of 6The Thevenin resistance is then given by RTh=vocisc=4/32=23, and the Thevenin equivalent is given below.Page 6 of