Thevenin Norton Equivalencies  GATE Study Material in PDF


 Mervyn Horn
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1 Thevenin Norton Equivalencies  GATE Study Material in PDF In these GATE 2018 Notes, we explain the Thevenin Norton Equivalencies. Thevenin s and Norton s Theorems are two equally valid methods of reducing a complex network down to something simpler to analyse. These methods help a great deal in simplifying complex networks and thereby save a great amount of time. These concepts are important Network Theorems that form the bedrock of Electrical Networks. These free GATE Study Notes will deal with Thevenin Norton Equivalencies in circuits with only independent sources, in circuits with only dependent sources as well as circuits with both dependent and independent sources. You can download these GATE Study Material in PDF. These notes are useful for GATE EE, EC, IES, BARC, DRDO, BSNL and other exams. However, before you move on to Thevenin and Norton Theorems it is advised that you have your basics clear. Recommended Reading Basic Network Theory Concepts Source Transformation & Reciprocity Theorem Kirchhoff s Laws, Node and Mesh Analysis Voltage Division, Current Division, StarDelta Conversion 1 P a g e
2 Thevenin's Theorem Any two terminal bilateral linear DC circuits can be replaced by an equivalent circuit consisting of a voltage source and a series resistor. Norton s Theorem Any two terminal bilateral linear DC circuits can be replaced by an equivalent circuit consisting of a current source and a parallel resistor. Now we will discuss how to solve different kind of problems using these theorems. 1. Problem with only Independent Sources To find Rth : Replace all voltage sources with short circuit and all current sources with open circuit. Find the resistance across the load terminals, that resistance is called Thevenin s resistance (Rth) or Norton s resistance (RN). To find Voc : Calculate the open circuit voltage across load terminals. This open circuit voltage is called Thevenin s voltage (Vth). 2 P a g e
3 To find Isc : Short the load terminals and then calculate the current flowing through it. This current is called Norton current (or) short circuit current (Isc). Example 1: Find ix using Thevenin s and Norton s theorems. Solution: Rth: Rth = RN =1Ω Vth: By Nodal Analysis V th P a g e 2 = 0
4 Vth = 12V Isc: Isc = 12A Thevenin s equivalent: By kvl i x = 12 3 = 4A Norton s equivalent: By current division i x = = 4A 4 P a g e
5 2. Problem with both Dependent and Independent Sources To find Voc: Calculate the open circuit voltage across load terminals. This open circuit voltage is called Thevenin s voltage (Vth). To find Isc: Short the load terminals and then calculate the current flowing through it. This current is called Norton current (or) short circuit current (Isc). To find Rth: Since there are Independent sources in the circuit, we can t find Rth directly. We will calculate Rth using Voc and Isc and it is given by R th = V oc I sc Example 2: Find the value of ix using Thevenin s and Norton s theorems. Solution: Voc: By Nodal Analysis V oc P a g e 2V x = 0
6 V oc 2V x = 10 (1) By KVL 10 + V x + V oc = 0 V oc = 10 V x (2) By (1) & (2) V oc = 10V Isc: By Nodal Analysis V V x + V 1 2 = 0 3V 1 4V x = 20 (3) By KVL, V x = 10 V 1 (4) By (3) & (4) V 1 = 60 7 V I sc = V 1 2 = 30 7 A Rth: R N = R th = V oc I sc = 10 30/7 = 7 3 Ω 6 P a g e
7 Thevenin s equivalent: By KVL i x = = A Norton s equivalent: By current division, i x = 30 7 [ = 7 +3] 3 7 [ 7 16 ] = 15 8 A 3. Problems with only Dependent Sources Such circuits can t function on their own so Vth and Isc doesn t exists but still they exhibit resistance, that resistance can be indirectly determined by V/I method by placing an active source across the terminals. 1. Place a voltage source of 1V across the terminal and find the current (IT) flowing through it. Then, 7 P a g e
8 R th = R N = 1V I T (or) 2. Place a current source of 1A across the terminals and find the voltage (Vt) across the current source. Then, R th = R N = V T 1A Example 3: Find the Thevenin s resistance across terminals xy Solution: 1. by placing 1V voltage source across terminals xy By nodal analysis, V 1 2V 3 o + V 1 1 = 0 5 And, 8 P a g e
9 V o = V 1 ( 2 3 ) V 1 3 4V V = 0 4V 1 5 I T = 1 V 1 5 = 1 5 V 1 = 1 4 = 1 ( 1 4 ) = 1 A 5 4 R x y = R Th = R N = 1V I T = = 4Ω 2. by placing 1A current source across terminals xy By nodal analysis V 1 3 2V o 1 = 0 And, V o = 2V 1 3 V 1 3 4V = 0 V 1 = 1 By KVL, V V T = V T = 0 V T = 4V R x y = R th = R N = V T 1A = 4 1 = 4Ω 9 P a g e
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