University of Illinois Spring 2016ECE210 / ECE211 - Homework 03Solutions1. In the following circuit find the open-circuit voltage and the short-circuit current between nodes a to b anddetermine the Thevenin and Norton equivalent of the network between nodes a and b.Solution:To find the open-circuit voltage, apply KCL at the upper node:4 − VT2= 3ix4 − VT2= 34 − VT2VT= 4VTo find the short circuit current, apply KVL at left inner loop:4 = 2ix+ 3iscSince ix= −isc2, the short circuit current is:2−isc2+ 3isc= 4isc= 2ATo solve for RT,RT=VTisc= 2ΩTherefore, the Thevenin equivalent is:Page 1 of 5the Norton equivalent is:2. Determine the Thevenin equivalent of the following network between nodes a and b, and then determine theavailable power of the network:Solution:To find the Thevenin voltage, apply KCL at the upper left node:−Vx+ 4 + (VT− Vx) = 02Vx− VT= 4Next, apply KCL at the upper right node:Vx− VT− 3Vx= 0Vx= −VT2To solve for VT:VT= −2VTo find the Thevenin Resistance, apply test current source between nodes a and b and make the indepedentcurrent source to open.The KCL equations at the upper left node and the upper right node give(Vx− VT) − 3Vx+ 1 = 0−Vx+ (Vx− VT) = 0Solve for these two equations we have:Vx=14VVT=12VTo solve for RT,RT=VTiT=12ΩPage 2 of 5Therefore, the Thevenin equivalent is:The maximum available power is:Pav=V2T4RT= 2W3. Calculate the absorbed power in RLfor:(a) RL= 3kΩ(b) RL= 6kΩ(c) RL= 12kΩSolution:First, we can calculate the voltage across RL.VL= 18RL6k + RLNext, we can calculate the absorbed power in RL.PL=V2LRL= 324RL(RL+ 6k)2Therefore, the absorbed power is(a)PL= 0.012W(b)PL= 0.0135W(c)PL= 0.012WNext, consider the circuit below, which implements a bu er between the source and the load. Assume the circuitbehaves linearly and make use of the ideal op-amp approximations. Calculate the absorbed power in RLfor(d) RL= 3kΩ(e) RL= 6kΩ(f) RL= 12kΩSolution:First, we need to calculate the voltage across RL.Page 3 of 5We will have a relationship between V1and V2which is:V1= V2= 18VWith the voltage and resistance of the load, we can calculate the consumed energy.PL=V22RL(a)PL= 0.108W(b)PL= 0.054W(c)PL= 0.027W4. In the op-amp circuit shown below, determine the voltage vxassuming linear operation.Solution:According to the characteristics of op-amp, V1, V2, and V3become:V1= V2= 8VWith the voltage division rule, we can calculate the V4.V3=V24k(4k + 2k) = 12VTherefore, we can calculate vx.vx= V1− V3= −4V5. In op-amp circuit shown below, determin the voltage Va, VR, and Vo. Assume the circuit behaveds linearly andmake use of the ideal op-amp approximation.Solution:To find the Va, apply KCL at node V1:5 − V13+Va− V16= 0According to the characteristics of op-amp, V1becomes:V1= 0VPage 4 of 5Therefore, we can calculate Va.Va= −10VAccording to the characteristics of op-amp, V2becomes:V2= 0VTherefore, we can calculate VR.VR= 8 − V2= 8VTo find the Vo, apply KCL at node V2:Va− V22+VR8=V2− Vo4Vo= −4Va+ VR2= 16VPage 5 of