EE 151 Applied Electricity - KNUST Telecomm. Engineering

KNUST Telecomm. Engineering EE 151 APPLIED ELECTRICITY 2018.09.03 Ing. Dr. Abdul-Rahman Ahmed  KNUST Telecomm....

KNUST Telecomm. Engineering EE 151 APPLIED ELECTRICITY 2018.09.03 Ing. Dr. Abdul-Rahman Ahmed  KNUST Telecomm. EE 151: APPLIED ELECTRICITY Engineering Course Structure: 3 2 3 Course Instructor: Ing. Dr. Abdul-Rahman Ahmed [email protected] 0508 – 351 - 438 Course Materials: https://www.rfmicrowaveknust.com Teaching Assistants: Akambole Isaac Ageebase Alberta Oduraa Quartey 0506257842 0547886101 [email protected] [email protected] 2 Website QR Code KNUST Telecomm. Engineering 3  Course Objective KNUST Telecomm. Engineering  Appreciate basic principles of Electrical Circuits  Be equipped with tools required to analyze  electric and  magnetic circuits  Develop knowledge and fundamental concepts associated with Electrical/ Electronics Engineering 4  Course Outline KNUST Telecomm. Engineering  Unit 1: Circuits and Network Theorems:  Kirchhoff's laws, Thevenin’s Theorem, Norton's Theorem, Superposition Theorem, Reciprocity Theorem and Delta- Star Transformation.  Unit 2: Alternating current circuits  Determination of Average and RMS values, Harmonics, Phasors, impedance,current and power in ac circuits  Unit 3: Three-phase circuits  Connection of three-phase windings, three phase loads, power in three-phase circuits, solving three-phase circuit problems  Unit 4: Magnetic circuits – Components and 5 terminologies, solving magnetic circuit problems  Course Calendar KNUST Telecomm. Engineering Week Week Beginning Topic 1 3rd September Network Theorems 2 10th September Network Theorems 3 17th September AC Circuits 4 24th September AC Circuits 5 1st October 3-Phase Circuits 6 8th October 3-Phase Circuits 7 15th October Magnetic Circuits 8 22nd October Revision/Tutorial 9 29th October Revision/Tutorial 10 5th November Mid-Semester Exams 11 19th November Magnetic Circuits 12 26th November Revision/Tutorial 13 3rd December End of Semester Exams (to be confirmed) 6 Recommended Texts KNUST Telecomm. Engineering 7  Course Delivery Method KNUST Telecomm. Engineering Lectures Tutorials Circuit Design Simulations Laboratory work 8  Classroom Manners KNUST Telecomm. Engineering No phone usage No eating No noise-making No lateness No gossiping No sleeping Etc… 9  ASSESSMENT KNUST Telecomm. Engineering Quizzes Mid-semester examination 30 % End of semester examination 70% Total 100% 10  UNIT 1: CIRCUIT AND KNUST Telecomm. NETWORK THEOREMS Engineering Definition of a circuit An interconnection of elements forming a closed path along which current can flow. R + V(t) L - C Elements of an electric circuit Active elements: Energy producing elements eg. Batteries, Generators, solar cells etc. Passive elements: Energy using elements eg. Resistors, inductors, capacitors 11  Circuit Terminologies KNUST Telecomm. Engineering Node (Junction) A point where currents split or come together [ points c, d, e and f] Path Any connection b c where e flows [eg gbc, be, etc] current a d f h 12  Circuit Terminologies KNUST Telecomm. Engineering Branch A connection (path) between two nodes [eg. cd, cb, ad, df] Loop/Mesh a closed path b c of a circuit [ eg.e cghdc] g a d f h 13  Circuit Terminologies KNUST Telecomm. Engineering Short-circuit A branch of theoretically zero resistance.  It diverts to itself all currents that would have flown in adjacent branches (branches hooked to the same node); except branches with sources. a b R1 c R3 d R1 a R2 R4 V R2 V R3 h g f e b 14  Circuit Terminologies KNUST Telecomm. Engineering Short-circuit cont. Self assessment Which of the resistors in the circuit below have R1 been short-circuited? Ans: R2 and R3 V R2 R3 15  Circuit Terminologies KNUST Telecomm. Engineering  Open circuit – A branch of theoretically infinite resistance. It prevents current from flowing in its branch R2 R3 R5 V R1 R4 16  Resistors in Series KNUST Telecomm. Engineering Resistors are in series when the same current flows through them. There is NO JUNCTION between them. R1 R2 R1 R2 V V R3 Fig. 1 Fig. 2 In Fig. 1 : R1 and R2 are in series In Fig. 2: None of the resistors are in series 17  Resistors In Series KNUST Telecomm. Engineering Self assessment 1 Which of the following resistors are in series? R2 R5 R6 R3 R1 R4 ANS: R1&R2, R3 & R4 and R5 & R6 18  Resistors in Series KNUST Telecomm. Engineering Total (effective) resistance of series resistors total resistance RT for resistors R1, R2, R3, ….., RN which are in series if given by: N RT R1  R2  R3    RN  R n1 n 19  Resistors in Parallel KNUST Telecomm. Engineering Resistors are said to be in parallel when the voltage across them is the same. R1//R2 R1//R2 R1//R2 R1//R3 R3//R2 20  Resistors in Parallel KNUST Telecomm. Engineering  Colloquially, TWO resistors are in parallel if it is possible to move from one to the other without passing through another element. i.e., the two can be circled without passing through another circuit element. R1 V R2 21  Resistors in Parallel KNUST Telecomm. Engineering Self assessment Which of the resistors in the circuit below are in parallel? 1.6Ω 3Ω 6Ω 4Ω 4 // 6 22  Resistors in Parallel KNUST Telecomm. Engineering Total resistance When resistors R1 and R2 are in parallel, the total resistance RT is given by: 1 1 1 R1 R2    RT  RT R1 R2 R1  R2 23  Effective Resistance of a KNUST Telecomm. Circuit Engineering Effective circuit resistance is found by identifying and putting together series and or parallel resistors Eg 1. Find the total resistance of the circuit below. 2 3 11 RT 2 // 3  1  1   23 5 24  Effective Resistance of a KNUST Telecomm. Circuit Engineering Find the effective resistance of the circuit 2 2 RT 2 // 2 1   1 2 22 25  Effective Resistance of a KNUST Telecomm. Circuit Engineering Self Assessment 1 Find the total resistance of the circuit below. 2Ω 3Ω 6Ω 4Ω V  4 6  22 66 RT 4 // 6   2 // 3   2 // 3  // 3   46  5 37 26  Effective Resistance of a KNUST Telecomm. Circuit Engineering Self Assessment 2 Find the total resistance of the circuit below. 1  3 1 3 RT 1 // 1  1 // 1 // 1   1 // 1 // 1  //   2  2 2 8 27  Current Division Rule KNUST Telecomm. Engineering The current division rule is applied to share current between parallel branches. Consider I the circuits below R1 R2 I1 I2 RT  R1  R2 V R1 R2 R1 R2 V  IRT  I R1  R2 R1 R2 I V R1  R2 R1 R2 1 R2 I1   I  I R1 R1 R1  R2 R1 R1  R2 28  Current Division Rule KNUST Telecomm. Engineering Similarly, I I1 I2 V R1 R2 R1 I2 I R1  R2 29  Current Division Rule KNUST Telecomm. Engineering In summary I I1 I2 V R1 R2 R2 R1 I1  I I 2 I R1  R2 R1  R2 30  Current Division Rule KNUST Telecomm. Engineering Consider the circuit below I I1 I2 V R1 R2 R2 R1 I1  I I 2  I R1  R2 R1  R2 31  Current Division Rule KNUST Telecomm. Engineering Example 1 Find the values of I1 and I2 in the circuit below. 10A I1 I2 R2 2 V 3Ω 2Ω I1  I 10 4 A R1  R2 23 R1 3 I 2  I  10  6 A R1  R2 23 32  Current Division Rule KNUST Telecomm. Engineering Example 2 Find the value of I1 in the circuit below. 10A I1 V 3Ω 2Ω 2Ω 10A I1 V 3Ω 1Ω 1 I1  10 2.5 A 1 3 33  Voltage Drop KNUST Telecomm. Engineering Any time a voltage drives current through a resistor, some of the voltage drops across the resistor. The magnitude I of V1 the drop V2 is the product of the resistance and current R1 R2 V V2 V  V1 34  Voltage Drop KNUST Telecomm. Engineering Example  Find the values of I and R in the circuit below. I 4V R 3 10V Solution Voltage across 3Ω resistor = 10 – 4 = 6V Current in 3Ω resistor = I = 6/3 = 2A 35 Resistance R = 4V/I = 4/2 = 2Ω  Revision Exercise KNUST Telecomm. Engineering Find the value of I in the circuit below. I 3 3.5 4 1 2 5 7 25V I I 3 6  4 1 2 4 1 5 25V 25V 36  Revision Exercise KNUST Telecomm. Engineering Thus 12  13 1 25V 25 V 25 RT   IT   13 A 13 RT 25 13 37  Revision Exercise KNUST Telecomm. Engineering I 6  4 1 5 25V 6 I  5 13  3 A 6 4 5 38  Group Assignment 1 KNUST Telecomm. Engineering Find the value of the current in all resistors of the circuit below using total resistance and voltage drop principles. DO NOT use current division rule. 3.5 3 2 1 4 5 7 25V 39 Kirchhoff's Current Law(KCL) KNUST Telecomm. Engineering  The Law The sum of currents entering a node equals the sum of currents leaving the node. i5 i6 i1 i4 i3 i2 Sum of currents entering i1  i3  i5 Sum of currents Leaving i2  i4  i6 i1  i3  i5 i2  i4  i6 40 Kirchhoff's Current Law(KCL) KNUST Telecomm. Engineering  Example Find the value of i in the figure below. i 2 3 4 7 Solution i  2  3 4  7 i  5 11 i 6 41 Kirchhoff's Current Law(KCL) KNUST Telecomm. Engineering  Self assessment Find the value of i in the figure below. i 2 3 4 7 ANS i 2 42  Kirchhoff's Voltage Law(KVL) KNUST Telecomm. Engineering  The law  The algebraic sum of the voltages in a loop (closed path) equals zero. Alternatively, in a loop, the algebraic sum of voltage sources equals the algebraic R R R sum of voltage a I drops.b I 1 1 c 2 I d 2 4 4 I3 I5 R3 R5 V1 V2 h g f e V1=I1R1+I3R3 Loop abgha V1-V2=I1R1+I2R2-I4R4 Loop adeha 43 Kirchhoff's Voltage Law(KVL) KNUST Telecomm. Engineering R1 R2 R4 a I1 b I2 c I4 d I3 I5 R3 R5 V1 V2 h g f e Loop cbgfc 0 = -I2R2 + I3R3 +I5R5 Loop acfha V1=I1R1 + I2R2 – I5R5 44  Kirchhoff's Voltage Law(KVL) KNUST Telecomm. Engineering Example 1 Find the current in all parts of the circuit below. 2Ω 4Ω b c e I1 I2 32V 8Ω 20V I3 a d f Applying KVL to loop bcdab 32 – 2I1 – 8I3 = 0 32 = 2I1 + 8I3 (1) Applying KVL to loop ecdfe20 – 4I2 – 8I3 = 0 20 = 4I2 + 8I3 (2) 45  Kirchhoff's Voltage Law(KVL) KNUST Telecomm. Engineering 2Ω 4Ω b c e I1 I2 32V 8Ω 20V I3 a d f  Applying KCL to nodeI c:  I  I2 (3) 3 1 Solving the equations simultaneously yields I1 = 4A, I2 = -1A and I3 = 3A 46  Kirchhoff's Voltage Law(KVL) KNUST Telecomm. Engineering Example 2 Find the currents in all parts of the circuit below. 2Ω c 5Ω e 2Ω g b I1 I1-I2 I2 I2-I3 I3 7V 3Ω 2Ω 10 V I2 a d f h Solution 2Ω Apply KVL to loop cefdc 5I2 + 2 ( I2 – I3 ) + 2I2 - 3 ( I1 – I2 ) = 0 47  Kirchhoff's Voltage Law(KVL) KNUST Telecomm. Engineering 0 = -3I1 + 12I2 - 2I3 (1) 2Ω c 5Ω e 2Ω g b I1 I1-I2 I2 I2-I3 I3 7V 3Ω 2Ω 10 V I2 a d f h 2Ω 7 = 2I1 + 3(I1 – I2) Apply KVL to loop abcda: 7 = 5I1 – 3I2 (2) 10 = -2 ( I2 – I3 ) + 2I3 Apply KVL to loop ghfeg: 10 = -2I2 + 4I3 (3) 48  KNUST Kirchhoff's Voltage Law(KVL) Telecomm. Engineering Solving the three equations: 0 = -3I1 + 12I2 - 2I3 (1) 7 = 5I1 – 3I2 (2) 10 = -2I2 + 4I3 (3) Simultaneously, I1 = 2.0A, I2 = 1.0A and I3 = 3.0A 49  Thevenin’s Theorem KNUST Telecomm. Engineering Theorem: Any linear circuit connected between two terminal can be replaced by a Thevenin’s voltage(VTH) in series with a Thevenin's resistance (RTH). VTH is the open-circuit voltage across the two terminals RTH is 12 the Ω resistance seen from theRTH two terminals when all A B A sources have been deactivated 6Ω VTH 42 V 35 V B 50  Thevenin’s Theorem KNUST Telecomm. Engineering  To find the current through a resistor in a circuit, the following steps are taken: 1. Remove the resistor from the circuit and mark the two terminals. 12 Ω 3 A B 42 V 35 V 6Ω 51  KNUST Thevenin’s Theorem Telecomm. Engineering 2. Find the open-circuit voltage (VTH) across the two terminals by applying KVL. Treat VTH as a source 12 Ω A B VTH 42 V 35 V 6Ω 52  Thevenin’s Theorem KNUST Telecomm. Engineering 3. Recall the circuit created before step 2 and deactivate all sources. Short-circuit voltage sources and Open-circuit 12 current Ω sources. A B 42 V 35 V 6Ω 53  KNUST Thevenin’s Theorem Telecomm. Engineering 4. Find the total resistance of the circuit resulting from step 3 as seen from12 theΩtwo terminals A B 6Ω 54  Thevenin’s Theorem KNUST Telecomm. Engineering 5. Reproduce the Thevenin’s equivalent circuit and connect the resistor whose current is to be found. RTH A VTH 3 B 55  Thevenin’s Theorem KNUST Telecomm. Engineering 6. Calculate the current in the circuit in step 5. This is RTHfor. the current being sought A VTH VTH 3 i RTH  R B 56  Thevenin’s Theorem KNUST Telecomm. Engineering Example 1 Using Thevenin’s theorem, determine the current in the 12 Ω 3Ω 3-Ω resistor of the circuit below. 42 V 35 V 6Ω 12 Ω A B Solution VTH 6Ω Steps 1 & 42 V 35 V 2 I 57  Thevenin’s Theorem KNUST Telecomm. Engineering 12 Ω A B VTH 6Ω 42 V 35 V I Applying KVL to loop 35 + VTH = 6I dcbed: Applying KVL to loop 42 (12  6) I fabef: 7 I A 3 58  KNUST Thevenin’s Theorem Telecomm. Engineering Substituting for I in equation 1 7 35  VTH 6( ) 3 Steps 3 & 4 V TH  21 V 12 Ω RTH A B 12 6 RTH 12 // 6  4 12  6 6Ω 59  Thevenin’s Theorem KNUST Telecomm. Engineering Steps 5 & 6 RTH = 4Ω A I3 VTH= -21V 3Ω B VTH 21 I3 =   3 A RTH  3 43 60  Thevenin’s Theorem KNUST Telecomm. Engineering Example 2 Find the current in the 510-Ω Ω resistor of the 10 Ω 12 Ω circuit below using Thevenin’s theorem. 4V 15 Ω 8Ω 6V 5Ω VTH 12 Ω Solution A B 4V I1 15 Ω I2 8Ω 6V Steps 1 & 2 61  Thevenin’s Theorem KNUST Telecomm. Engineering 5Ω VTH 12 Ω A B 4V I1 15 Ω I2 8Ω 6V Applying KVL to loop cbgfc:VTH = 15I1 - 8I2 1 4 = (5+15)I1 Applying KVL to loop abgha: I1  A 5 Applying KVL to loop dcfed:6 (12  8) I 2 3 I2  A 10 62  Thevenin’s Theorem KNUST Telecomm. Engineering Substituting for I1 and I2 in equation 1 yields:  1  3 3 VTH 15   8   V  5  10  5 5Ω RTH 12 Ω A B Steps 3 & 4 15 Ω 8Ω 171 RTH (5 //15)  (12 // 8)   20 63  Thevenin’s Theorem KNUST Telecomm. Engineering Steps 5 & 6 RTH A I VTH 10 Ω B VTH 3 3 20 I    = 0.032 A RTH  10  171  5 371 5  10   20  64  Group Assignment 2 KNUST Telecomm. Engineering Use Thevenin’s theorem to find the current in the 5Ω resistor of the circuit below. 2Ω c 5Ω e 2Ω g b 7V 3Ω 2Ω 10 V a d f h 2Ω 65  Norton’s Theorem KNUST Telecomm. Engineering Theorem: Any linear circuit connected between two terminals can be replaced by a Norton’s current(IN) in parallel with a Norton's resistance (RN). RN is the resistance seen from the two terminals when is the allshort-circuit sources have current between the(Rtwo been deactivated terminals N N = RTH) 12 Ω A B A IN 6Ω RN 42 V 35 V B 66  Norton’s Theorem KNUST Telecomm. Engineering  To find the current through a resistor in a circuit, the following steps are taken: 1. Remove the resistor from the circuit and mark the two terminals. 12 Ω 3 A B 42 V 35 V 6Ω 67  Norton’s Theorem KNUST Telecomm. Engineering 2. Find the short-circuit current (IN) through the two terminals by applying KVL. 12 Ω IN A B 42 V 35 V 6Ω 68  Norton’s Theorem KNUST Telecomm. Engineering 3. Recall the circuit created before step 2 and deactivate all sources. Short-circuit voltage sources and Open-circuit 12 current Ω sources. A B 42 V 35 V 6Ω 69  KNUST Norton’s Theorem Telecomm. Engineering 4. Find the total resistance of the circuit resulting from step 3 as seen from12 theΩtwo terminals A B 6Ω 70  KNUST Norton’s Theorem Telecomm. Engineering 5. Reproduce the Norton’s equivalent circuit and connect the resistor whose current is to be found. A IN 3 RN B 71  Norton’s Theorem KNUST Telecomm. Engineering 6. Calculate the current in the circuit in step 5. This is the current being sought for. I A IN 3 RN RN i I N RN  3 B 72  Norton’s Theorem KNUST Telecomm. Engineering Example 1 Using Norton’s theorem, determine the current in the 3-Ω resistor of the circuit below. 12 Ω 3Ω 42 V 35 V 6Ω I+IN 12 Ω IN B A Solution I 6 Ω Steps 1 & 2 42 V 35 V 73  KNUST Telecomm. Norton’s Theorem Engineering 12 Ω IN B A I 6Ω 42 V 35 V 42 = 12(I+IN) Applying KVL to loop abefa: + 6I 42 = 18I + 12IN Applying KVL to loop cbedc:35 = 6I 35 I  A 6 74  KNUST Norton’s Theorem Telecomm. Engineering Substituting for I in equation421:18 35   12 I N  6   21 IN  A 4 Steps 3 & 4 12 Ω RN A B 12 6 6Ω RN 12 // 6  4 12  6 75  KNUST Norton’s Theorem Telecomm. Engineering Steps 5 & 6 I3 A  21 4 A 4 3 B 4  21 I3    3 A 43 4 76  KNUST Norton’s Theorem Telecomm. Engineering Example 2 Determine the current in theAload resistor RL R1 R2 R3 R4 RL E1 E2 E3 E4 B A Solution I1 I2 I3 I4 R1 R2 R3 R4 IN E1 E2 E3 E4 B 77  KNUST Norton’s Theorem Telecomm. Engineering A I1 I2 I3 I4 R1 R2 R3 R4 IN E1 E2 E3 E4 B Solution Applying KCL I N  I1  I 2  I 3  I 4 E1 E2 E3 E4     R1 R2 R3 R4 78  Norton’s Theorem KNUST Telecomm. Engineering Finding RN A R1 R2 R3 R4 RN B 1 1 1 1 1     R N R1 R2 R3 R4 79  Norton’s Theorem KNUST Telecomm. Engineering Finding IL A IL IN RN RL B RN IL  I N RN  RL 80  Superposition Theorem KNUST Telecomm. Engineering The Theorem The current through(or the voltage across) any element in a multiple-source linear circuit can be found by taking the algebraic sum of the current through(or the voltage across) that element due to each individual source acting alone 81  KNUST Superposition Theorem Telecomm. Engineering The Theorem I 12Ω 6Ω 3Ω 42V 35V IA 12Ω IB 12Ω 6Ω 6Ω 3Ω + 3Ω 42V 35V 82  Superposition Theorem KNUST Telecomm. Engineering With the 35V battery acting alone, IB 12Ω 6Ω 3Ω 35V RT 12 // 6   3 7 35 IT  5 A 7 I B IT 5 A 83  Superposition Theorem KNUST Telecomm. Engineering With both batteries acting, IA IB I 12Ω 12Ω 6Ω 3Ω 12Ω 6Ω + = 3Ω 6Ω 3Ω 42V 35V 35V 42V I I B  I A 5  2 3 A 84  KNUST Reciprocity Theorem Telecomm. Engineering The Theorem An ideal ammeter and ideal voltage source when inserted in two different branches of a linear network can be interchanged without changing the reading of the ammeter R1 R2 R1 R2 + V(t) R3 A = A R3 + V(t) - - 85  Reciprocity Theorem KNUST Telecomm. Engineering Example 1 Jointly use superposition and reciprocity theorems to find the current supplied by the 35V battery of the circuit below. 12Ω I 6Ω 3Ω 42V 35V Solution 12Ω IA With the 42V battery acting 6Ω 3Ω alone, 42V 86 Reciprocity Theorem KNUST Telecomm. Engineering IA 12Ω 6Ω 3Ω 42V RT 3 // 6   12 14 42 IT  3 A 14 6 IA  3 2 A 63 87  Reciprocity Theorem KNUST Telecomm. Engineering The second circuit to be solved is: B I IA 12Ω 3Ω 12Ω 6Ω 6Ω 3Ω is comparable to 35V 42V Applying the reciprocity theorem IA IA 12Ω 12Ω 6Ω 3Ω 6Ω 3Ω 42V = 42V 88 Reciprocity Theorem KNUST Telecomm. Engineering I I B  I A 5  2 3 A 89  Reciprocity Theorem KNUST Telecomm. Engineering Use Norton’s theorem to find the current in the 2Ω resistor connected between e and f in the circuit below. 2Ω c 5Ω e 2Ω g b 7V 3Ω 2Ω 10 V a d f h 2Ω Submission date: God willing a week today Submission time: Before lecture starts Where to submit: Class room 90  Delta-star Transformation KNUST Telecomm. Engineering The transformation is employed in situations where neither series nor parallel arrangements can be identified. An arrangement of three(3) resistors where one terminal of a resistor is connected to another resistor and the A other terminalA to a different resistor is a delta A arrangement. R2 R3 R2 R3 B C B R1 C R1 91  Delta-star Transformation KNUST Telecomm. Engineering An arrangement of three(3) resistors where all resistors have a common point of connection through one terminal of each resistor while the remaining terminals are unconnected or connected to elements other than the two other resistors is a star(wye) arrangement. Ra Ra Rc Rb Rb Rc 92  Delta-star Transformation KNUST Telecomm. Engineering A delta arrangement can be changed to star and vice versa using the following relations: A R2 R3 Ra  R1  R2  R3 Ra R1R3 R2 R3 Rb  R1  R2  R3 Rc Rb R1R2 Rc  B C R1  R2  R3 R1 93  Star-Delta Transformation KNUST Telecomm. Engineering A Rb Rc R1 Rb  Rc  Ra Ra R2 R3 Ra Rc Rc Rb R2 Ra  Rc  Rb B R1 C Ra Rb R3 Ra  Rb  Rc When all values are the same, delta values are 3 times star values 94  Delta-star Transformation KNUST Telecomm. Engineering Example 1 Determine the voltage V0 across the 6Ω resistor of the 6Ω circuit below 6 2Ω 2Ω Ra a 2 2 6 10 V 2Ω 6Ω Vo Solution 10 R 2 Rb c b Ra  Rb  Rc 2 3 6 95  Delta-star Transformation KNUST Telecomm. Engineering Example 1 Determine the voltage V0 across the 6Ω resistor of the circuit below 6Ω 6Ω 10V 6Ω 6Ω 6Ω 3 Ω 3 a V0  10V 5V 33 6 Ω 3 Ω 10V Vo b 96  Group Assignment 4 KNUST Telecomm. Engineering Use Norton’s theorem to find the current in the 3Ω resistor connected between points A and B of the A circuit below. 3Ω 2.4Ω 3Ω 3Ω 5Ω B 3Ω 10v Submission date: God willing a week today Submission time: Before lecture starts Where to submit: Electrical Engineering office 97