Ministry of Higher Education and Scientific Research AlMustansiriyah

Ministry of Higher Education and Scientific Research Al-Mustansiriyah University Faculty of Engineering Computer Engineering Department Electrical Eng. Lab 1' class Assistant Lecturer Hazim alkargole 2017 20189 ﻭﺯﺍﺭةﺎﻟﺘﻌﻠﻴﻢ ﺍﻟﻌﺎﻟﻲ ﻭﺍﻟﺒﺤﺚ ﺍﻟﻌﻠﻤﻲ ﺍﻟﺠﺎﻣﻌﺔ ﺍﻟﻤﺴﺘﻨﺼﺮﻳﺔ ﻛﻠﻴﺔ ﺍﻟﻬﻨﺪﺳﺔ ﻗﺴﻢ ﻫﻨﺪﺳﺔ ﺍﻟﺤﺎﺳﻮﺏ

EXAMPLE : USE THEVENIN AND NORTON TO FIND V 0 12 5 +- 72 Using Thevenin Theorm: 8Ω 20 4 + Vo -

1. FIND VOC : 12 i 1 +- 72 8 5 a 20 b KVL around the upper loop : 12 i 1 8 i 1 5 (i 1 i 2 ) 0 25 i 1 5 i 2 0 KK(1)

KCL around lower loop : 5 (I 2 I 1 ) 20 I 2 72 5 i 1 25 i 2 72 K K (2) i 1 0. 6 A, i 2 3 A Voc 8 i 1 20 i 2 8 (0. 6) 20 (3) Voc 64. 8 V

2. FIND RTH 12 5 12 8 a 20 5 //20 Ω b b RTH = (8+4) // 12 = 12 // 12 = 6 4Ω 8 a

3. RECONNECT THE LOAD : RTH +- VOC 4 Vo Voc 4 R TH 4 (64. 8) 4 6 Vo 25. 92 V + Vo - 44Ω kΩ

USING NORTON THEORM: 1. Find ISC : 12 i 1 +- 72 8 5 i 2 20 KVL around upper loop : 12 i 1 8 (i 1 i 3 ) 5 (i 1 i 2 ) 0 25 i 1 5 i 2 8 i 3 0 K K (1) i 3

KVL around lower loop : 5 (i 2 i 1 ) 20 (i 2 i 3 ) 72 5 i 1 25 i 2 20 i 3 72 K K (2) KVL around right loop : 8 (i 3 i 1 ) 20 (i 3 i 2 ) 0 8 i 1 20 i 2 28 i 3 0 i 1 6 A, i 2 12. 72 A , ISC 10. 8 A K K (3) i 3 10. 8 A

2. Find RTH From before , RTH = 6 3. Reconnect the load a ISC i 1 i 2 + RTH 4Ω Vo - b Vo (4 Ω) i 2 (4 Ω) RTH I SC R TH 4 6 10. 8 25. 92 V Vo 4 6 4

CASE(2) : CIRCUITS CONTAINING ONLY DEPENDENT SOURCES Here there is NO energy source in the circuit. VOC is always zero and ISC is always zero So we can only find RTH Procedure for finding RTH 1. Connect an independent voltage ( or current) source at the terminals , Vx (or Ix) 2. Find the corresponding current ( or voltage) at the terminal , Io ( or Vo) 3. Find RTH = Vx /Io or RTH = Vo/Ix RTH a b

EXAMPLE: FIND THEVENIN EQUIVALENT CIRCUIT 3 kΩ 2 kΩ a 2000 Ix 4 kΩ Ix b 1. Apply voltage source at the terminals (Vx=1 V) 2 kΩ i 2 i 1 4 kΩ Ix Vx = 1 V +- 2000 Ix 3 kΩ V 1

KCL at node V 1 : I 1 I 2 I X 0 2000 I X V 1 1 V 1 IX 0 2 k 3 k where V 1 (4 k) I X 2000 I X 4000 I X 1 4000 I X IX 0 2000 3000 1 4 IX 2 I X IX 0 3 k 3 k 1 4 I X 2 3 3000 I X 0. 1 m A

VX V 1 i 2 3 k VX 4 k I X 1 4 k 0. 1 m A 3 k 3 k i 2 0. 2 m A R TH 1 V VX 5 k Ω i 2 0. 2 m A RTH a b

CASE (3) : CIRCUITS CONTAINING BOTH INDEPENDENT AND DEPENDENT SOURCES Procedure of Thevenin or Norton Theorms: a. Find the open circuit voltage and the terminals , VOC b. Find the short circuit current at the terminals, ISC. c. Compute RTH = VOC/ISC Note : RTH can not be found as in the case of only independent sources d. Construct the Thevenin or Norton circuits RTH a ISC RTH +- Voc a b Thevenin circuit b Norton circuit

EXAMPLE : FIND THEVENIN EQUIVALENT CIRCUIT WITH RESPECT TO THE TERMINALS A, B 20 Ω 160 ix a ix 4 A 80 Ω 60 Ω + Voc 40 Ω - 1. Find VOC : 240 V b 60 Ω i 20 Ω 160 ix Z ix +- i 1 80 Ω 40 Ω + Voc -

KVL around the lift loop : 240 80 i 160 i x 40 i x 0 80 i 200 i x 240 K K (1) KVL around right loop : 80 i 1 40 i x 2 i 1 i x 0 K K (2) KCL at Z: i i 1 i x 0 i 1. 125 A i 1 0. 375 A i x 0. 75 A K K (3)

VOC I X (40 Ω) (0. 75 A) (40 Ω) VOC 30 V 2. Find ISC : 20 Ω 160 ix ix 4 A 60 Ω V 80 Ω Since we have short circuit , 80 // 40 // 0 = 0 40 Ω

ix=0 160 ix source is zero 60 ISC 4 3 A 60 20 20 Ω 60 ΩV 3. Find RTH Current divider VOC 30 V � R TH 10 Ω I SC 3 A RTH Voc a +- 4. ISC 4 A b

Example : Use Thevenin theorem to find the Thevenin equivalent circuit with respect to a, b 2 i x Z 1. Find VOC 8 A a 1 b KCL at node z : 2 i x 8 i 1 0 3 i x i 1 8 i 1 5 +- 40 V ix K K (1)

KVL around outer loop 40 5 I X 1 I 1 0 5 i x i 1 40 K K (2) ix 4 A , i 1 20 A VOC 1 i 1 20 V 2 ix Find ISC : +- 40 V ix a 5 8 A 1 I SC b

2 ix KVL around outer loop : 40 5 i x 0 Z i x 8 A 5 8 A +- KCL at z : 40 V ix I SC 2 i x 8 ISC 3 i x 8 ISC 3 8 8 32 A 3. Find RTH : Thevenin equivalent circuit is Voc a +- VOC 20 R TH 0. 625 Ω I SC 32 RTH b

4. MAXIMUM POWER TRANSFER • A technique in which the load is selected to maximize the power transfer. • This technique is based on the Thevenin equivalent circuit. RTH + i Voc - PL v L i i 2 R L 2 V �OC RL R TH R L + VL -- RL

We wish to select RL to maximize PL: d. PL Take 0 d. R L d. PL (R TH R L ) 2 VOC 2 R L VOC 2 2(R TH R L ) 0 4 d. R L R TH R L 2 Voc (R TH R L ) - 2 R L 0 4 (R TH R L ) R TH R L - 2 R L 0 R TH R L 0 R L R TH If RL = RTH , what is the maximum Power Transfer?

RTH + Voc PL max i R L 2 - 2 � V OC R TH 2 R TH 2 VOC R TH VOC 2 4 R 2 TH 4 R TH 2 VOC � PL max 4 R TH i RL

Example: 8 kΩ 3 m. A 2. 5 k Ω 10 V +- 20 k Ω 4 kΩ +- 10 k Ω 10 V • Find RL for maximum Power Transfer ? • Find the maximum Power transfer to RL ? RL

LET’S FIND THEVENIN EQUIVALENT CIRCUIT. V 1 8 k Ω V 2 i 1 3 m. A 4 kΩ 2. 5 k Ω i 2 10 V +- i 4 + 20 k Ω i 3 +- 10 k Ω 10 V KCL at node V 1 : 3 m A i 1 i 2 0 V 1 V 2 0 3 m A 4 kΩ 8 kΩ Voc -

1 1 1 V 2 3 m V 1 4 k 8 k 0. 375 m V 1 0. 125 m V 2 3 m K K (1) KCL at node V 2: i 2 i 3 i 4 0 V 1 V 2 10 V 2 0 8 k 20 k 12. 5 k 1 1 V 0. 5 m V 1 2 8 k 20 k 12. 5 k K K (2) 0. 125 m V 1 0. 255 m V 2 0. 5 m V 1 10. 34 V V 2 7. 03 V

VOC 10 10 K I 4 V 2 10 10 10 k 7. 03 12. 5 k VOC 4. 375 V To find RTH : 8 kΩ 2. 5 k Ω RTH 4 kΩ 20 k Ω 10 k Ω

R TH 8 K 4 K // 20 K 2. 5 K // 10 K 12 K // 20 K 2. 5 K // 10 K 7. 5 k 2. 5 k // 10 k RTH R TH 5 k Ω 2 V 4. 375 �OC 4 R TH 4 (5 k) 2 PL max 0. 957 m W +- Voc RL = RTH

EXAMPLE : 2000 Ix 3 k ohm 4 k ohm + - 1 k ohm 4 m A Ix 1. Find RL for maximum Power Transfer? 2. Find max. power transfer to RL ? First , find Thevenin equivalent: 2 k ohm RL

2000 Ix' + Ix’ 4 m A 3 k ohm 4 k ohm + - 1 k ohm Voc 2 k ohm -Using source transformation 2000 Ix' 4 k ohm + - 4 k ohm + + 16 V - Ix’ 2 k ohm Voc --

KVL AROUND THE LOOP: -16 + 4 K IX’ - 2 K IX’ + 2 K IX’=0 IX’= 4 MA. Voc= (2 kΩ) Ix’= 8 V. Now, find Isc: - 4 k ohm + 16 V - V 1 4 k ohm + 2000 Ix'' I 1 Ix’’ 2 k ohm Isc

KCL at V 1: I 1 - Ix’’ – Isc = 0 16 (V 1 2 k. Ix'') V 1 0 4 k 2 k 4 k Where V 1=2 k Ix’’ Hence, And 16 V 1 0 4 k 2 k 4 k V 1 Isc 1. 333 m. A 4 k Or V 1=5. 333 V

R TH 8 V VOC 6 kΩ I SC 1. 333 m A PL(max) 2 V 8 64 �OC 4 R TH 4 6 k 24 k PL(max) 8 m. W 3 2