Super Node - Engineering Electrical Circuits - Lecture Slides, Slides of Electrical Circuit Analysis

Download Super Node - Engineering Electrical Circuits - Lecture Slides and more Slides Electrical Circuit Analysis in PDF only on Docsity! Super Node/Mesh Thevénin/Norton Docsity.com OutLine • ReIterate NODE & LOOP/MESH Analysis by way of Examples • SuperNode Method • SuperMesh Technique • Loops vs Meshes; describe difference • Loop & Node Compared • Introduction to Thevenin & Norton theorems Docsity.com Docsity.com KCC AT NODE & PAL ZAR HO 2yv-2¥A = Va-Ve + 2V4-2¥o 2N= s¥a Ve — 2Vo C1) KCL AT No0Ee B 2ymA = Va-/2ZV 4 ¥e-¥a iV —Ve ZAL 2A TAL AN <p l2V 4 Va-Va +2Va -2¥z LON= -VA +¥Y¥e -2V, C2) Docsity.com keh AT NODE Ve /Ah4e (AL (ALL Ve~Vo tAa- Vo = Vo or Os Wa -Ve t3Ve £% 3 GOANX LN 3 BAKMOWNS Save BY GAUSSIAN SLLANIAATIONW ne: I \ | i Docsity.com Ve = C.MVAEV Vs = 216 VY as Bruce Mayer, PE Engineering instructor Chabot College 25555 Hesperian Bivd Hayward, CA 94545 eMail: bmayer@chabotcollege.edu o¥FeEB(2 @ row KVL ON Lye Loop (mA CAL Ty -6V CAC (Le tlayt2 a (Iu+l, ) = 22 Docsity.com - - * _ A zo 3 Ty —3 mA +ety 4253 4+ Te 41% CLlye +2232 =2mA am aly +I3 =l%A Oe Ly = GL +m) /2 (2) BY Le = (Cat ima)/s [ Te = (-¥L3 -[nA) 2 (C1) oO x £3 + BmA 3 20 fEry=gen [E> Tox Ends Thus D2=-l4.m BY LOOP DEFINITION Lo =—Le —435 5° Bo =-2ma-Eve mh) = —-Ye m4 [ro = Fa ma C ReCall Node (KCL) Analysis • Need Only ONE KCL Eqn 0 12126 12322 = − + − + k VV k VV k V  The Remaining Eqns From the Indep Srcs ][6 ][12 3 1 VV VV −= =  Solving The Eqns ][5.1][64 0)()(2 22 12322 VVVV VVVVV =⇒= =−+−+  3 Nodes Plus the Reference. In Principle Need 3 Equations... • But two nodes are connected to GND through voltage sources. Hence those node voltages are KNOWN Docsity.com SuperNode Technique • Consider This Example • Conventional Node Analysis Requires All Currents at a Node  2 eqns, 3 unknowns... Not Good (IS unknown) • Recall: The Current thru the Vsrc is NOT related to the Potential Across it  But Have Ckt V-Src Reln ][621 VVV =−  More Efficient solution: • Enclose The Source, and All Elements In parallel, Inside a Surface. – Call That a SuperNode SUPERNODE SI 0 6 6@ 11 =++− SIk VmAV 0 12 4@ 22 =++− k VmAIV S Docsity.com +−1 V 2V 1sI 2sI 1R 2R 3R SV ][6],[10],[20 4,10 21 321 mAImAIVV kRkRR ssS === Ω=Ω== Find the node voltages And the power supplied By the voltage source 2012 =−VV 010 1010 21 =−+ mA k V k V ][100 10* 21 VVVk =+⇒ ][2021 VVV =+−⇒ ][40100 ][1202 :adding 21 2 VVV VV =−= = To compute the power supplied by the voltage source We must know the current through it: @ node-1 VI = − ++= k VVmA k VIV 4 6 10 211 mA5 BASED ON PASSIVE SIGN CONVENTION THE POWER IS ABSORBED BY THE SOURCE!! mWmAVP 100][5][20 =×= Docsity.com Illustration using Conductances • Write the Node Equations – KCL At v1  At The SuperNode Have V-Constraint • v2 − v3 = vA  KCL Leaving Supernode  Now Have 3 Eqns in 3 Unknowns • Solve Using Normal Techniques    Docsity.com Example • Find Io • Known Node Voltages  Now use KCL at SuperNode to Find V3 VVVV 12 ,6 42 =−=  The SuperNode V-Constraint VVVVVV 12or12 3131 +==−  Mult by 2 kΩ LCD, collect Terms to Find: SUPERNODE 123 +=V Docsity.com Numerical Example – Dep Isrc • Find Io by Nodal Analysis • Notice V-Source Connected to the Reference Node  KCL At Node-2  Sub Ix into KCL Eqn  Mult By 6 kΩ LCD VV 31 =  Then Io 02 63 212 =−+− xIk V k VV  Controlling Variable In Terms of Node Potential k VI x 6 2= 0 6 2 63 2212 =−+ − k V k V k VV VVVV 602 212 =⇒=− mA k VVIO 13 21 −= − = Docsity.com Current Controlled V-Source • Find Io • Supernode Constraint  Controlling Variable in Terms of Node Voltage xkIVV 212 =− k VI x 2 1= 121 22 VVkIV x =⇒=⇒  KCL at SuperNode 0 2 2 2 4 21 =+++− k VmA k VmA  Multiply by LCD of 2 kΩ ][421 VVV =+ 02 21 =+− VV Recall  Then 38 83 22 VVVV =⇒=  So Finally mA k VIO 3 4 2 2 == Docsity.com ReCall MESH Analysis • Use Mesh Analysis  So Vo 1I 2I mAI 4 :1Mesh 1 =−  Sub for I1 to Find I2 0124 2Mesh 21 =+−− kIkI mAmAII 3 4 12 16 12 4 1 2 === ][8333.16 6 2 VmAkV IkV O O =⋅Ω= ⋅Ω= Docsity.com SuperNODE vs SuperMESH • Use superNODE to AVOID a V-source current, IVs, in KCL Eqns 12V 2KΩ 1KΩ 2KΩ2mA IO 4mA IO Vx  Use superMESH to AVOID an I-source, ΔVIs, in KVL Eqns Docsity.com Shared Isrc – General Loop Approach • Strategy – Define Loop Currents That Do NOT Share Current Sources • Even If It Means ABANDONING Meshes • For Convenience, Begin by Using Mesh Currents Until Reaching Shared CURRENT Source as V-across an I-source is NOT Known – At That Point Define a NEW Loop  To Guarantee That The New Loop Gives An Independent Equation, Must Ensure That It Includes Components That Are NOT Part Of Previously Defined Loops 1I 2I 3I Docsity.com General Loop Approach cont. • A Possible Approach – Create a Loop by Avoiding The Current Source  The Eqns for Current Source Loops mAI mAI 4 2 2 1 = =  The Eqns for 3rd Loop (3 Eqns & 3 Unknowns) 0)(1)(2)(21][6 13123233 =−+−+++++− IIkIIIkIIkkIV  The Loop Currents Obtained With This Method Are Different From Those Obtained With A SuperMesh • A SuperMesh used Previously Defined Mesh Currents 3I 1I 2I Docsity.com Dependent Sources • General Approach – Treat The Dep. Source As Though It Were Independent – Add One Equation For The Controlling Variable • Example at Rt.: Mesh Currents Defined by Sources  Mesh-3 by KVL k VI mAI X 2 4 2 1 = = 0)(1)(21 4313 =−+−+− IIkIIkkIx  Mesh-4 by KVL 012)(1)(1 2434 =+−+− VIIkIIk Docsity.com Dependent Sources cont. • The Controlling Variable Eqns  In Matrix Form  Solve by Elimination or Linear-Algebra )(2 13 24 IIkV III x x −= −=  Combine Eqns, Then Divide by 1kΩ mAIII mAIII III mAI 122 823 0 4 432 432 321 1 −=+−− =−+ =−+ =             − =                         −− − − 12 8 0 4 2110 2310 0111 0001 4 3 2 1 I I I I Docsity.com Loop & Node Compared • Consider the Ckt  Find Vo by NODE Analysis • ID Nodes • Make a SuperNode  Vsrc to GND ][VV 44 =  SuperNode Constraint xVVV 221 =− Docsity.com Loop & Node Compared (4) • Now by Loops  Start with 3 Meshes  The Mesh/Loop Eqns • Loop-1: mAI 21 = ( ) 0112 322 =−++− IIkkIVx • Loop-3: 1 I 2I 3I  Add a General Loop to avoid the Isrc 4I mAI 23 −= • Loop-2: – Note I3 = –2 mA • Loop-4: ( ) 04121 4143 =++−+−+ VkIVIIIk x – Note I1 = 2 mA Docsity.com Loop & Node Compared (5) • The Controling Var  As Before Vx = V2  And V2 is related to the net current From Node-2 to GND  By Net Current & Ohm’s Law ( )4311 IIIkVx −−Ω= ( ) 0112 322 =−++− IIkkIVx  SubOut Vx in Loop-2 & Loop-4 Eqns: ( ) 04121 4143 =++−+−+ VkIVIIIk x 1I 2I 3I 4I  After Subbing Find: V84 4 =kI VkIkI 642 42 =+ Docsity.com Loop & Node Compared (6) • Summarize Loops  Using The Loop/Mesh Eqns Find  Recall the GOAL  General Comments • Nodes (KCL) are generally easier if we have VOLTAGE Sources • Loops (kVL) are generally easier if we have CURRENT Sources mAI 21 = mAI 12 −= mAI 23 −= mAI 24 = 2I 21kIVo = Docsity.com Low Dist Pwr Amp cont • To Even STAND A CHANCE to Match the Speakers & Amp We Need to Simplify the Ckt – Consider a Reduced CIRCUIT EQUIVALENT • Replace the OpAmp+BJT Amplifier Ckt with a MUCH Simpler (Linear) Equivalent + - RTH VTH • The Equivalent Ckt in RED “Looks” The Same to the Speakers As Does the Complicated Circuit Docsity.com Thevenin’s Equivalence Theorem LINEAR CIRCUIT May contain independent and dependent sources with their controlling variables PART A LINEAR CIRCUIT May contain independent and dependent sources with their controlling variables PART B a b_ Ov + i LINEAR CIRCUIT PART B a b_ Ov + i − + THR THv PART A  Thevenin Equivalent Circuit for PART A  vTH = Thevenin Equivalent VOLTAGE Source  RTH = Thevenin Equivalent Series (Source) RESISTANCE Docsity.com Norton’s Equivalence Theorem  Norton Equivalent Circuit for PART A  iN = Norton Equivalent CURRENT Source  RN = Norton Equivalent Parallel (Source) RESISTANCE LINEAR CIRCUIT May contain independent and dependent sources with their controlling variables PART A LINEAR CIRCUIT May contain independent and dependent sources with their controlling variables PART B a b_ Ov + i LINEAR CIRCUIT PART B a b_ Ov + i NRNi PART A Docsity.com Interpret Thevenin & Norton  This equivalence can be viewed as a source transformation problem. It shows how to convert a voltage source in series with a resistor into an equivalent current source in parallel with the resistor • SOURCE TRANSFORMATION CAN BE A GOOD TOOL TO REDUCE THE COMPLEXITY OF A CIRCUIT RTH i + _ OvOCv +_ Thevenin SCi THR − + Ov a b i Norton SC OC NTH i vRR == In BOTH Cases Docsity.com Source Transformations • Source transformation is a good tool to reduce complexity in a circuit ...WHEN IT CAN APPLIED – “IDEAL sources” are NOT good models for the REAL behavior of sources • .e.g., A Battery does NOT Supply huge current When Its Terminals are connected across a tiny Resistance as Would an “Ideal” Source  These Models are Equivalent When + - Improved model for voltage source Improved model for current source SV VR SI IR a b a b SS IV RIV RRR = ==  Source X-forms can be used to determine the Thevenin or Norton Equivalent • But There May be More Efficient Methods Docsity.com Example  Solve by Src Xform  In between the terminals we connect a current source and a resistance in parallel  The equivalent current source will have the value 12V/3kΩ  The 3k and the 6k resistors now are in parallel and can be combined  In between the terminals we connect a voltage source in series with the resistor  The equivalent V-source has value 4mA*2kΩ  The new 2k and the 2k resistor become connected in series and can be combined Docsity.com Source Xform Summary • These Models are Equivalent + - Improved model for voltage source Improved model for current source SV VR SI IR a b a b SS SS IV VRI RIV RRR = = ==  Source X-forms can be used to determine the Thevenin or Norton Equivalent  Next Review Several Additional Approaches To Determine Thevenin Or Norton Equivalent Circuits Docsity.com Determine the Thevenin Equiv. • vTH = OPEN CIRCUIT Voltage at A-B if Part-B is Removed and Left UNconnected • iSC = SHORT CIRCUIT Current at A-B if Voltage at A-B is Removed and Replaced with a Wire (a short) • Then by R = V/I SC OC TH i vR = Docsity.com Graphically... N SC OC TH SCabNOCabTH R i vR iIivVv == ==== 1. Determine the Thevenin equivalent source Remove part B and compute the OPEN CIRCUIT voltage abV 2. Determine the SHORT CIRCUIT current Remove part B and compute the SHORT CIRCUIT current abI LINEAR CIRCUIT May contain independent and dependent sources with their controlling variables PART A a b _ 0= + v SCi abI Second circuit problem One circuit problem _ abV + LINEAR CIRCUIT May contain independent and dependent sources with their controlling variables PART A a b _ OCv + 0=i  Then Docsity.com Thevenin Example • Find Vo Using Thevenin’s Theorem • Identify Part-B (the Load)  Break The Circuit At the Part-B Terminals “PART B” V6 Ωk5  DEactivate 12V Source to Find Thevenin Resistance • Produces a SHORT Docsity.com Thevenin Example cont. • Note That RTH Could be Found using ISC  By Series-Parallel R’s  Then by I-Divider SC I Ω=+= kReq 5.7266  Then Itot mAkVItot 6.15.712 =Ω= totI mAmAISC 2.126 66.1 = + =  Finally RTH Ω=== kmAVIVR SCOCTH 52.16 • Same As Before Docsity.com Thevenin Example cont.2 • Finally the Thevenin Equivalent Circuit  And Vo By V-Divider ][1)6( 51 1 VV kk kVO =+ = ][1V Docsity.com Thevenin Example cont • For the open circuit voltage we analyze the circuit at Right (“Part A”)  Use Loop/Mesh Analysis 0)(246 2 211 2 =−++− = IIkkIV mAI mAmAII 3 5 6 26 2 1 = + =  Then VOC [ ] [ ] ][3/3243/20 *2*4 21 VVVV IkIkV OC OC =+= +=  Finally The Equivalent Circuit Docsity.com CALCULATE Vo USING NORTON PART B Ω== kRR THN 3 SCI mAmA k VII NSC 223 12 =−== NI NR k4 k2       + == N N N O IkR RkkIV 6 22 I ( ) ][ 3 4)2( 9 32 VmAkVO =   Ω= COMPUTE Vo USING THEVENIN PART B THV 02 3 12 =+ − mA k VTH kkRTH 43 += + - THR THV k2 − + OV ][ 3 4)6( 72 2 VVVO =+ =612−=⇒ THV Docsity.com DEpendent & INdependent Srcs  Find The Open Circuit Voltage And Short Circuit Current  Solve Two Circuits (Voc & Isc) For Each Thevenin Equivalent  Any and all the techniques may be used; e.g., KCL, KVL, combination series/parallel, node & loop analysis, source superposition, source transformation, homogeneity  Setting To Zero All Sources And Then Combining Resistances To Determine The Thevenin Resistance is, in General, NOT Applicable!! LINEAR CIRCUIT May contain independent and dependent sources with their controlling variables PART A a b_ Ov + i + - THR THV a b OCTH VV = SC OC TH I VR = Docsity.com Illustration • Use Thevenin to Determine Vo • Partition Guidelines – “Part-B” Should be as Simple As Possible – After “Part A” is replaced by the Thevenin equivalent should result in a very simple circuit – The DEpendent srcs and their controlling variables MUST remain together  Use SuperNode to Find Open Ckt Voltage “Part B” 1V  Constraint at SuperNode OCOC VVVV +=⇒=− 1212 11  KCL at SuperNode Ω= =+ + + −−+ k 2 :Where 0 22 12 1 )()12( ' a k V k V k aIV OCOCXOC Docsity.com Illustration cont • The Controlling Variable  Solving 3 Eqns for 3 Unknowns Yields k VI OCX 2 ' = )1/2(4 36 )1/(4 36 kkka VOC + −= + −=  Now Tackle Short Circuit Current 0 2 " == k VI AX AV  At Node-A find  VA=0 → The Dependent Source is a SHORT • Yields Reduced Ckt Ω= k 322||1 Docsity.com Illustration cont • Using the Reduced Ckt  Now Find RTH mA kk VISC 182||1 12 −=−= Ω= − − == k 3 1 mA 18 V 6 SC OC TH I VR  Setting All Sources To Zero And Combining Resistances Will Yield An INCORRECT Value  Finally the Solution ( ) VVV V Rkk kV TH TH 7 186 3333.2 1 11 1 0 0 −=−= ++ = )2( kaRTH = OCV  Note: Some ckts can produce NEGATIVE RTH Docsity.com Note on Example • The Equivalent Resistance CanNOT Be Obtained By DeActivating The Sources And Determining The Resistance Of The Resulting InterConnection Of Resistors – Suggest Trying it → Rth,wrong = 2.5 kΩ • Rth,actual = 0.75 kΩ Req Docsity.com EXAMPLE: Find Vo By Thevenin • Select Partition  Use Meshes to Find VOC “Part B” KVL for V_oc  In The Mesh Eqns mAIVI X 2; 2000 2 ' 1 == ( )       −=−= 2 ' 21 ' 2 44 I k VkIIkV XX mAIIIkkIkIVX 4)(422 12111 ' =⇒−=⇒=  The Controlling Variable  By Dep. Src Constraint  Solve for VOC VVmAkVOC 11342 =+∗=  Now KVL on Entrance Loop ][3*20 1 VIkVOC ++−= Docsity.com Find Vo By Thevenin cont • Now Find ISC • The Mesh Equations  The Controlling Variable mAIVI x 2; 2000 2 " 1 == 0)(23 1 =−+− IIkV SC )(*4 21 " IIkVX −=  Solving for I1 Find Again mAI 41 =  Find ISC by Mesh KVL mA k IkVISC 2 11 2 *23 1 =+= Ω=== k mA V I VR SC OC TH 2)2/11( ][11  Then Thevenin Resistance  Use Thevenin To Find Vo [ ]VV kk kV 433][11 62 6 0 =+ = THV THR 2I scI 1I Docsity.com All Done for Today Source Transformations => the two sowces are Cquivelout cohen Toe n =k. Example. Use Source Franebrm chins Te Kind Thesenin Cquiveledt : Docsity.com Thevenin Example • Find Vo Using Thevenin’s Theorem  in the region shown, could use source transformation twice and reduce that part to a single source with a resistor.  Alternative: apply Thevenin Equivalence to that part (viewed as “Part A”)  Deactivating (Shorting) The 12V Source Yields Ω=+= kRTH 4362  Opening the Loop at the Points Shown Yields ][8][12 63 6 VVVV THOC =+ == Docsity.com Thevenin Example cont. • Then the Original Circuit Becomes After “Theveninizing”  Apply Thevenin Again  For Open Circuit Voltage Use KVL  Result is V-Divider for Vo  Deactivating The 8V & 2mA Sources Gives Ω= kR TH 41 − + 1 THV VVmAkVTH 1682*4 1 =+= VVV 8][16 88 8 0 =+ = Docsity.com Example + - 2SI 3SI 1SV 1SI 1R 2R 3R 4R 1V 2V 3V 4V −+ OV − + R1 = 1k; R2 = 2k, R3 = 1k, R4 = 2k Is1 =2mA, Is2 = 4mA, Is3 = 4mA, Vs1 = 12 V 210 VVV −=  Need Only V1 and V2 to Find Vo  Known Node Potential ][12:@ 133 VVVV S ==  Now KCL at Node 1 0 21 ][2 0:@ 121 4 1 1 21 11 =+ − +− =+ − +− k V k VVmA R V R VVIV S  Find Vo  To Start • Identify & Label All Nodes • Write Node Equations • Examine Ckt to Determine Best Solution Strategy  Notice Docsity.com Example cont. + - 2SI 3SI 1SV 1SI 1R 2R 3R 4R 1V 2V 3V 4V −+ OV − + R1 = 1k; R2 = 2k, R3 = 1k, R4 = 2k Is1 =2mA, Is2 = 4mA, Is3 = 4mA, Vs = 12 V 0 21 12 1 ][4 0:@ 42212 2 42 3 32 1 12 32 = − + − + − +− = − + − + − +− k VV k V k VVmA R VV R VV R VVIV S  At Node 4 0 2 ][4][2 0:@ 24 2 24 214 = − +− = − +− k VVmAmA R VVIIV SS  To Solve the System of Equations Use LCD-multiplication and Gaussian Elimination  At Node 2 Docsity.com Example cont. • The LCDs 0 21 ][2 121 =+−+− k V k VVmA 0 21 12 1 ][4 42212 =−+−+−+− k VV k V k VVmA 0 2 ][4][2 24 =−+− k VVmAmA ][423 21 VVV =− *2kΩ *2kΩ ]3252 421 VVVV =−+− *2kΩ ][442 VVV =+− (1) (2) (3)  Now Add Eqns (2) & (3) To Eliminate V4 ][182][3642 2121 VVVVVV =+−⇒=+− (4)  Now Add Eqns (4) & (1) To Eliminate V2 ][11][222 11 VVVV =⇒= ][5.14][182][11 22 VVVVV =⇒=+− ][5.3][5.14][11210 VVVVVV −=−=−=  BackSub into (4) To Find V2  Find Vo by Difference Eqn Docsity.com Dep V-Source Example • Find Io by Nodal Analysis • Notice V-Source Connected to the Reference Node  SuperNode Constraint  KCL at SuperNode  Mult By 12 kΩ LCD VV 63 = xVVV 221 =− 212 3VVVVx =⇒= 062)6(2 2211 =−+++− VVVV  Controlling Variable in Terms of Node Voltage Docsity.com Dep V-Source Example cont • Simplify the LCD Eqn  By Ohm’s Law mA k V k VIo 8 3 24 9 12 1 = Ω = Ω = VV VV VV VVV 5.4 184 3 and 1833 1 1 12 21 = =∴ = =+ Docsity.com Numerical Example • Select Soln Method – Loop Analysis • 3 meshes • One current source – Nodal Analysis • 3 non-reference nodes • One super node – Both Approaches Seem Comparable → Select LOOP Analysis • Specifically Choose MESHES • Select Mesh Currents  Write Loop Eqns for Meshes 1, 2, 3 by KVL + - + VO _ 6k 4k 2k 2k IS VS IS = 2mA, VS = 6V Determine VO 1I 2I 3I SII =1 0)(2)(4 1232 =−+−+− IIkIIkVS 06)(4)(2 32313 =+−+− kIIIkIIk Docsity.com Find Vo- Compare Mesh vs. Loop  Using MESH Currents  Using LOOP Currents Treat The Dependent Source As One More Voltage Source  Mesh-1 & Mesh-2 0)(422 211 =−++− IIkkIVx 04)(22 121 =+++− kIIIkVx 0)(463 122 =−++− IIkkI 063)(22 221 =+−++− kIIIkVx  Loop-1 & Loop-2 Docsity.com Compare Loop vs. Mesh cont.  Using MESH Currents  Using LOOP Currents  Now Express The Controlling Variable In Terms Of MESH or LOOP Currents 14kIVx =)(4 21 IIkVx −= Solving 3104 042 21 21 =+− =+− kIkI kIkI 386 066 21 21 =+− =+− kIkI kIkI Docsity.com Compare Loop vs. Node cont.  Using MESH Currents  Using LOOP Currents Solutions mAImAI 5.1,3 21 == mAImAI 5.1,5.1 21 == ][96 2 VkIVO == Finally  Notice The Difference Between MESH Current I1 and LOOP Current I1 even Though They Are Associated With The Same Path  The Selection Of LOOP Currents Simplifies Expression for Vx and Computation of Vo Docsity.com Find Vo Using Mesh Analysis • Draw the Mesh Currents  Controlling Variable In Terms Of Loop Currents 1I 2I  Write KVL Mesh Eqns For Mesh-1 & Mesh-2 0)(4212 122 =−++− IIkkI 0)(422 211 =−++− IIkkIkIx 2II x = Docsity.com Find Vo Using Mesh Analysis cont • Substitute & Collect Terms  Solve for I2 1I 2I 1264 066 21 21 =+− =− kIkI kIkI mAIkI 612624 22 =⇒=+− ][122 2 VkIVO ==  Finally Vo 122 1 =kI Docsity.com WhiteBoard Work  Let’s Work This Problem  Find the OutPut Voltage, VO 1KΩ1KΩ 1KΩ 12V VO + - 1KΩ IO2IX IX Docsity.com