Sirindhorn International Institute of Technology Thammasat University at Rangsit

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1 Sirindhorn Interntionl Institute of Technology Thmmst University t Rngsit School of Informtion, Computer nd Communiction Technology COURSE : ECS 204 Bsic Electricl Engineering L INSTRUCTOR : Asst. Prof. Dr. Prpun Suksompong (prpun@siit.tu.c.th) WEB SITE : EXPERIMENT : 02 Network Theorems I: Thevenin & Norton Theorems. I. OBJECTIVES 1. To verify Thevenin s theorem for resistive circuits. 2. To verify Norton s theorem for resistive circuits. 3. To lern how to construct current source from the power supply. 4. To ecome fmilir with potentiometer. II. BASIC INFORMATION Let s consider liner circuit whose two terminls nd re connected to n ritrry lod. Thevenin s nd Norton s theorems ssert tht the circuit cn thus e replced y either Thevenin or Norton equivlent circuit, which cts like the originl circuit cross the lod connected to the two terminls. Thevenin nd Norton theorems re very useful in circuit nlysis for simplifying prts of complicted circuits. For resistive circuits, the Thevenin equivlent circuit (shown in Figure 2-1) simply consists of Thevenin voltge source V TH in series with Thevenin resistnce R TH, while the Norton equivlent circuit (shown in Figure 2-2) consists of Norton current source I N in prllel with Norton resistnce, which is the sme s R TH. V TH cn e determined from the open-circuit voltge cross terminls -, i.e., the voltge cross the two terminls when the lod is disconnected. R TH is the equivlent resistnce of the circuit with respect to terminls - fter dectivting ll independent sources in the circuit nd disconnecting the lod. I N cn

2 e determined from the short-circuit current t terminls -, i.e., the current flowing through the short-circuit connecting terminls -. Reminders: (1) A voltge source is dectivted when it gives 0 V. In which cse, it ecomes short connection. (2) A current source is dectivted when it gives 0 A. In which cse, it ecomes n open connection. (3) We my use the terms turn off or disle insted of dectivte. They do not necessrily men powering off the power supply. II.1. Thevenin s theorem Thevenin s theorem provides method for simplifying circuit to stndrd equivlent form. As shown in Figure 2-1, the Thevenin equivlent circuit consists of ) Thevenin equivlent voltge source (VTH) in series with ) Thevenin equivlent resistnce (RTH). VTH nd RTH cn e found, when RLod is disconnected from nodes nd. The Thevenin voltge VTH is defined s the open-circuit voltge etween nodes nd. RTH is the totl resistnce ppering etween nd when ll sources re dectivted. This series comintion of VTH nd RTH is equivlent to the originl circuit in the sense tht if we connect the sme lod cross terminls - of ech circuit, we will get the sme voltge nd current t the terminls of the lod. This equivlence holds for ll possile vlues of lod resistnce. R TH V TH R Lod Figure 2-1: Thevenin equivlent circuit. 2

3 II.2. Norton s theorem The Norton equivlent circuit consists of n independent current source (IN) in prllel with n equivlent resistnce (RN), rrnged s shown in Figure 2-2. We cn simply derive this equivlent circuit from the Thevenin equivlent circuit y using the source trnsformtion concept. Thus, the Norton current ctully equls the short-circuit current t the terminls -, nd the Norton resistnce is identicl to the Thevenin resistnce. I N R N = R TH R Lod Figure 2-2: Norton equivlent circuit. Exmple: Figure 2-3- shows n exmple of circuit whose lod resistor is disconnected to nodes nd, creting n open circuit, so tht the Thevenin voltge VTH cn e mesured etween nodes nd (e.g. using voltmeter). In ddition, Figure 2-3- illustrtes the sme circuit, with the lod resistor disconnected, llowing mesurement of the Thevenin resistnce RTH (e.g. using n ohmmether cross nodes nd ) when ll voltge sources re replced with short circuit nd ll current sources re replced with n open circuit. Figure 2-3-c shows n exmple of circuit whose lod resistor is disconnected from nodes nd nd then short circuit is creted etween the two nodes so tht Norton current IN cn e mesured (e.g. using n mpmeter). In ddition, the Norton equivlent resistnce RN cn e otined in similr mnner s RTH shown in Figure Figure 2-3-d illustrtes the Thevenin nd Norton equivlent circuits derived in this exmple. 5 ohms 120 V 20 ohms 3 A 4 ohms + V TH =108V _ Figure 2-3-: A circuit used for illustrting the Thevenin s nd Norton s theorems. 3

4 5 ohms 20 ohms 4 ohms R TH = R N = 8 ohms Figure 2-3-: The circuit with the voltge nd the current sources dectivted to find RTH. 5 ohms 4 ohms 120 V 20 ohms 3 A I N = 13.5 A Figure 2-3-c: The circuit with the short-circuit etween - for finding the Norton current. 8 ohms R Lod 108 V 13.5 A 8 ohms R Lod Figure 2-3-d-: The Thevenin nd Norton equivlent circuits. Note tht VTH = IN RTH. III. MATERIALS REQUIRED - DC power supplies - Multi-meters - Resistors (1/4-W): 470-, 1.2-k, 10-k, two of 330-, nd potentiometer (vrile resistor). 4

5 IV. PROCEDURE Prt A: Thevenin equivlent circuit 1. Let R1= 330, R2 = 470, R3 = 330, nd RL = 1.2 k. Use DMM to mesure the resistnce of ech resistor, nd record the vlues in Tle Turn on the power supply, nd mesure its output voltge VPS. Adjust VPS to 12 V. Record the mesured vlue of VPS in Tle Connect the circuit in Figure 2-4. A R ohms B V ps 12V R ohms C R3 330 ohms R L 1200 ohms + V TH _ A D E Figure 2-4: The circuit for verifying Thevenin s nd Norton s theorems. The current IL through the lod is mesured y the mmeter A. 4. Mesure IL (the current through R L ), nd record this vlue in Tle 2-1, under the column Originl circuit. 5. Disconnect nd remove RL from the circuit, nd mesure the voltge cross node C-E. This is V TH. Record the vlue in Tle 2-1 under the V TH Mesured column. 6. Turn off the power supply, nd disconnect it from the circuit. 7. Short A-E y connecting wire cross node A nd E. 8. With RL still disconnected, mesure the resistnce cross nodes C nd E. This is R TH. Record the vlue in Tle 2-1 under the R TH mesured column. 5

6 9. Now we will uild the Thevenin equivlent circuit. The circuit shown in Figure 2-5 is our implementtion of Figure 2-1. To do this, djust the power supply so tht VPS = VTH, which hs een previously mesured. Connect DMM in the resistnce mesurement mode cross the potentiometer, nd djust its resistnce until the vlue of RTH is otined. Record the vlues of VTH (cross the power supple) nd RTH (cross the potentiometer) in Tle Connect the circuit s in Figure 2-5. This is the Thevenin equivlent circuit of the circuit in Figure 2-4. V PS =V TH R = R TH R L 1200 ohms A Figure 2-5: Thevenin equivlent circuit. The current IL through the lod is mesured y the mmeter A. 11. Mesure IL, nd record the vlue in Tle 2-1 under the Thevenin equivlent circuit, mesured column. Turn off the power supply. 12. Use the vlues of VPS, R1, R2, nd R3 to clculte VTH for the circuit in Figure 2-4. Record your result in Tle 2-1 under VTH clculted. 13. Clculte RTH in Figure 2-4 using the vlues of R1, R2, nd R3. Record your result in Tle 2-1 under RTH, clculted. 14. Use the clculted vlues of VTH nd RTH to clculte IL. Record the result under IL clculted. 6

7 Prt B: Norton equivlent circuit 1. From the vlue of IL nd RTH in Prt A, copy vlues of IL nd RTH in Tle 2-2, under "IL mesured, Originl circuit" nd "RN clculted," respectively. 2. Turn on the power supply, nd djust it ck to VPS = 12V. Record the mesured vlue of VPS in Tle 2-2. Connect the circuit in Figure 2-4 gin. To mesure IN, short circuit cross RL (from node C to node D), nd mesure the shortcircuit current. 1 Think out why RL does not hve to e removed. Record the vlue in Tle 2-2 under the column IN Mesured. 3. Now we will uild the Norton equivlent circuit. The circuit shown in Figure 2-6 is our implementtion of Figure 2-2. To do this, first djust the potentiometer until the resistnce vlue is equl to the vlue of RTH, mesured in Tle 2-1 recorded from step A.9. (It should lredy e t this vlue.) This will e your RN. Record the ctul (remesured) resistnce vlue cross the potentiometer in Tle 2-2 under the column RN Mesured. Now, for the current source, set the output of the power supply to its lowest vlue, i.e., 0 V. Connect the circuit shown in Figure 2-6. Meter A1 will e used to mesure the Norton current source IN in step 4. Meter A2 will e used in step B.5 to mesure the lod current IL. Note tht with one DMM, use it s Meter A1 here nd then use it s Meter A2 in step 5. A1 A2 I N V PS R = R N (R N = R TH ) R L 1200 ohms Figure 2-6: Norton equivlent circuit. Note tht in step B.2 the DMM is used s mmeter A1 to mesure the mount of current flowing out of the power supply. Then, it is used s mmeter A2 in step B.5. 1 Note tht this is the sme s putting n mmeter cross R L in Figure 2-4. The mmeter itself cts s the short circuit cross R L nd it lso displys the vlue of the short-circuit current. Alterntively, to mesure I N, we cn simply replce R L with n mmeter. 7

8 4. Turn on the power supply nd slowly increse the output of the supply until the current mesured y DMM A1 is equl to the vlue of IN, which hs een previously mesured. Record the vlue of IN tht you get from DMM A1 in Tle 2-2. Also, Record the power supply voltge VPS tht chieves this vlue of IN in Tle Record the lod current IL mesured y DMM A2 in Tle 2-2 under the Norton equivlent circuit, mesured column. Turn off the power. 6. Clculte the vlue of IN from the circuit of Figure 2-4, nd record your result in Tle 2-2 under IN clculted. 7. Clculte y using the clculted vlues of I N nd R N. Record your result in Tle 2-2 under, clculted. Remrk: Since the current source is not ville in the l, this prt of the experiment is modified to suit our ojective y djusting VPS in order to otin the current IN. Tips: From now on, when you re sked to use current source, replce it y power supply (voltge source) nd connect the rest of the circuit. Then, djust the vlue of the output voltge of the power supply so tht the required mount of current psses through it. Cution: Becuse power supply is not true current source, when you mke ny chnge to the circuit connection, the vlue of the current tht pss through the voltge source my chnge. You will need to redjust the voltge vlue of the power supply (voltge source) so tht the required mount of current psses through it every time tht you mke ny chnge to the circuit. Cution 2: DO NOT connect the DMM in its mmeter mode directly to the power supply. The mount of current coming out of the power supply when there is nothing connected to it except the DMM is meningless. 8

9 Tle 2-1: Thevenin equivlent circuit Mesured: R1 = R2 = R3 = RL = VPS = (A.2) V TH R TH Mesured Clculted Mesured Clculted Originl circuit Mesured Thevenin equivlent circuit A.5 A.12 A.8 A.13 A.4 A.11 A.14 Clculted A.9 A.9 TA Signture: Tle 2-2: Norton equivlent circuit VPS = (B.2) VPS = (B.4) I N R N Mesured Clculted Mesured Clculted Originl Circuit Mesured Norton equivlent Circuit Clculted B.2 B.6 B.3 B.1 B.1 B.5 B.7 B.4 9 TA Signture:

10 V. QUESTIONS 1. Consider the circuit given in Figure 2-4. If R 3 is replced y resistor, find the following vlues. V TH = V R TH = = ma If R L is lso replced y 120- resistor, determine the voltge nd current t R L. 2. Wht re the dvntges of using Thevenin s theorem nd Norton s theorem in solving complicted liner circuits? Show some exmples to support your nswers. 3. In step 2 of prt B, we tried to mesure IN. Why cn we leve RL in the circuit? 10