23ECE103 Fund of Electrical Engg 1 23ECE103:FUNDAMENTALS OF ELECTRICAL ENGINEERING (3-0-0-3) Unit 1, Lecture 10 Topic: Nodal Analysis 30 October 2023
Outline Objective Steps in Nodal Analysis Inspection method To solve DC circuits using Nodal analysis. AC circuits using Nodal analysis. 23ECE103 Fund of Electrical Engg 2 30 October 2023
Objective 30 October 2023 23ECE103 Fund of Electrical Engg 3 To introduce Nodal analysis technique (Nodal Voltage method) To formulate Node equations To solve circuits using Nodal analysis To introduce nodal analysis by inspection method
Nodal Analysis 30 October 2023 23ECE103 Fund of Electrical Engg 4 An analysis technique to solve electrical circuit where the node voltages are used as the circuit variables It involves systematic steps with an objective to solve the node voltages Nodal Analysis method applies KCL to find the unknown voltages With node voltages known, current in every branch can be calculated 6 2 7 1A 4A Reference node v 1 , v 2 : node voltages If v 1 and v 2 are known, current in all branches can be calculated
Nodal Analysis Steps: Step 1. Determine the reference node: typically the one with the most branches (or ) bottom node Step 2. Assign the rest of the nodes with node voltages (referred to the reference node) Step 3. Write down equations using KCL for every non-reference node in terms of node voltages. Step 4. Obtain the node voltages by solving the simultaneous equations in step 3 30 October 2023 23ECE103 Fund of Electrical Engg 5
Nodal Analysis- Example 1 30 October 2023 23ECE103 Fund of Electrical Engg 6 6 2 7 1A 4A
Nodal Analysis- Example 1 Step 1. Determine the reference node: typically the one with the most branches (or ) bottom node 30 October 2023 23ECE103 Fund of Electrical Engg 7 6 2 7 1A 4A
Nodal Analysis- Example 1 Step 2. Assign the rest of the nodes with node voltages (referred to the reference node) 30 October 2023 23ECE103 Fund of Electrical Engg 8 6 2 7 1A 4A v 1 v 2
Nodal Analysis- Example 1 Step 3. Write down equations using KCL for every non-reference node in terms of node voltages if there are N nodes, there should be (N-1) equations 30 October 2023 23ECE103 Fund of Electrical Engg 9 6 2 7 1A 4A v 1 v 2
Nodal Analysis- Example 1 Step 3. Write down equations using KCL for every non-reference node in terms of node voltages if there are N nodes, there should be (N-1) equations 30 October 2023 23ECE103 Fund of Electrical Engg 10 6 2 7 1A 4A v 1 v 2 i1 i2 i3 i3 i4 i5
Nodal Analysis- Example 1 Step 3. Apply KCL to Node 1 i 1 = i 2 + i 3 ---- (1) 30 October 2023 23ECE103 Fund of Electrical Engg 11 6 2 7 1A 4A v 1 v 2 i1 i2 i3 i3 i4 i5
Nodal Analysis- Example 1 Step 3. Apply KCL to Node 2 i 3 + i 5 = i 4 ---- (2) 30 October 2023 23ECE103 Fund of Electrical Engg 12 6 2 7 1A 4A v 1 v 2 i1 i2 i3 i3 i4 i5
Nodal Analysis- Example 1 Step 4. Obtain the node voltages by solving the simultaneous equations in step 3 ---- (1) ---- (2) Rearrange and solve the equations, V1=-2v & V2 = -14 V 30 October 2023 23ECE103 Fund of Electrical Engg 13
Nodal Analysis- Example 2 Determine the current in every branch 30 October 2023 23ECE103 Fund of Electrical Engg 14 9k 9k 3k 3k 4k 6k + 12 V
Nodal Analysis- Example 2 Determine the current in every branch Step 1. Determine the reference node: typically the one with the most branches (or ) bottom node 30 October 2023 23ECE103 Fund of Electrical Engg 15 9k 9k 3k 3k 4k 6k + 12 V
Nodal Analysis- Example 2 Determine the current in every branch Step 2. Assign the rest of the nodes with node voltages (referred to the reference node) Note: If a voltage source is placed near a node without resistance, that node will have same voltage of voltage source. 30 October 2023 23ECE103 Fund of Electrical Engg 16 9k 9k 3k 3k 4k 6k + 12 V v 1 v 2
Nodal Analysis- Example 2 Determine the current in every branch Step 3. Write down equations using KCL for every non-reference node in terms of node voltages 30 October 2023 23ECE103 Fund of Electrical Engg 17 9k 9k 3k 3k 4k 6k + 12 V v 1 v 2 i1 i2 i3 i4 i5
Nodal Analysis- Example 2 Determine the current in every branch Step 3. KCL at node 1 ---- (1) 30 October 2023 23ECE103 Fund of Electrical Engg 18 9k 9k 3k 3k 4k 6k + 12 V v 1 v 2 i1 i2 i3 i4 i5
Nodal Analysis- Example 2 Determine the current in every branch Step 3. KCL at node 2 ---- (2) 30 October 2023 23ECE103 Fund of Electrical Engg 19 9k 9k 3k 3k 4k 6k + 12 V v 1 v 2 i1 i2 i3 i4 i5
Nodal Analysis- Example 2 Step 4: ---- (1) ---- (2) Solving (1) & (2); v1=3V& V2=1.5V , With these values find all branch currents. 30 October 2023 23ECE103 Fund of Electrical Engg 20
Nodal Analysis- Example 2 Determine the current in every branch v1=3V& V2=1.5V , I1 = (12-V1)/9k = 9/9 =1 mA I2= V1/6k = ½ mA I3= (v1-V2) / 3k =1.5/3k =1/2 mA 30 October 2023 23ECE103 Fund of Electrical Engg 21 9k 9k 3k 3k 4k 6k + 12 V v 1 v 2 i1 i2 i3 i4 i5
Nodal Analysis- Example 2 Determine the current in every branch v1=3V& V2=1.5V , I3= (v1-V2) / 3k =1.5/3k =1/2 mA I4 = V2/4k = 1.5/4k = 4.5/12 mA I5 =V2/12k = 1.5/12 mA 30 October 2023 23ECE103 Fund of Electrical Engg 22 9k 9k 3k 3k 4k 6k + 12 V v 1 v 2 i1 i2 i3 i4 i5
Nodal Analysis- Example 3 Determine the unknown voltages 30 October 2023 23ECE103 Fund of Electrical Engg 23 3k Ω
Nodal Analysis- Example 3 Determine the unknown voltages Step 1. Determine the reference node: typically the one with the most branches (or ) bottom node 30 October 2023 23ECE103 Fund of Electrical Engg 24 3k Ω
Nodal Analysis- Example 3 Determine the unknown voltages Step 2. Assign the rest of the nodes with node voltages (referred to the reference node) Note: If a voltage source is placed near a node without resistance, that node will have same voltage of voltage source. 30 October 2023 23ECE103 Fund of Electrical Engg 25 3k Ω
Nodal Analysis- Example 3 Determine the unknown voltages Step 3. Write down equations using KCL for every non-reference node in terms of node voltages 30 October 2023 23ECE103 Fund of Electrical Engg 26 i1 i2 i4 3k Ω i3 i5 i6
Nodal Analysis- Example 3 Determine the unknown voltages Step 3. KCL at node 1 ---- (1) 30 October 2023 23ECE103 Fund of Electrical Engg 27 i1 i2 i4 3k Ω i3 i5 i6
Nodal Analysis- Example 3 Determine the unknown voltages Step 3. KCL at node 2 ---- (2) 30 October 2023 23ECE103 Fund of Electrical Engg 28 i1 i2 i4 3k Ω i3 i5 i6
Nodal Analysis- Example 3 Determine the unknown voltages Step 3. KCL at node 3 --- (3) 30 October 2023 23ECE103 Fund of Electrical Engg 29 i1 i2 i4 3k Ω i3 i5 i6
Nodal Analysis- Example 3 Step 4: ---- (1) ---- (2) --- (3) Solving (1), (2) & (3); v1=? ,V2=? & V3=? 30 October 2023 23ECE103 Fund of Electrical Engg 30
Nodal Analysis- Example 4 Use Nodal analysis to find node voltages in the given circuit. Also find the power drop across 2 ohm resistor. Ans: V 1 =20V, V 2 =13.33V; P 2ohms =? 30 October 2023 23ECE103 Fund of Electrical Engg 31
Nodal Analysis- Example 5 Use Nodal analysis to find node voltages in the given circuit. Also find V . 30 October 2023 23ECE103 Fund of Electrical Engg 32
Nodal Analysis- Example 6 Use Nodal analysis to find node voltages in the given circuit. Also find V . 30 October 2023 23ECE103 Fund of Electrical Engg 33 Ans: V 1 =30V, V 2 =20V; V =?
Nodal Analysis- Example 7 Use Nodal analysis to find node voltages in the given circuit. 30 October 2023 23ECE103 Fund of Electrical Engg 34 How many node voltages need to be found to solve the circuit? a) 5 b) 2 c) 3 d) 4 2. What will be the voltage at the reference node? 0V 12V voltage across 20 ohms voltage across 10 ohms 3. What is the equation of current through the 4 ohm resistor?
Nodal Analysis – Inspection method Steps: Step 1: First, convert all the voltage sources to equivalent current sources. Step 2: The conductances of all branches connected to node ‘ i ’ are added and denoted by G ii . G ii is called the self conductance of node ‘ i ’ . Step 3: All the conductances connected to nodes ‘ i ’ and ‘j’ are added and denoted by G ij . G ij is called mutual conductance of nodes ‘ i ’ and ‘j’ . This G ij is written with negative sign. If no conductance is connected between nodes ‘ i ’ and ‘j’ then G ij = 0, G ij = G ji . Step 4: I i denotes the value of the current source to node ‘ i ’ and is written on the right hand side of the equation. The sign I i is positive if it is flowing towards node ‘ i ’ , otherwise it is negative. If no current source is connected to node ‘ i ’ , then I i = 0. 30 October 2023 23ECE103 Fund of Electrical Engg 35
Nodal Analysis – Inspection method 30 October 2023 23ECE103 Fund of Electrical Engg 36
Nodal Analysis – Inspection method 30 October 2023 23ECE103 Fund of Electrical Engg 37
Nodal Analysis- Example 1 30 October 2023 23ECE103 Fund of Electrical Engg 38 6 2 7 1A 4A v 1 v 2
Nodal Analysis- Example 1 30 October 2023 23ECE103 Fund of Electrical Engg 39 6 2 7 1A 4A v 1 v 2 solve the equations, V1=-2v & V2 = -14 V
Nodal Analysis- Example 4 Use Nodal analysis to find node voltages in the given circuit. Also find the power drop across 2 ohm resistor. 30 October 2023 23ECE103 Fund of Electrical Engg 40
Nodal Analysis- Example 4 30 October 2023 23ECE103 Fund of Electrical Engg 41
Nodal Analysis- Example 4 Use Nodal analysis to find node voltages in the given circuit. Also find the power drop across 2 ohm resistor. Ans: V 1 =20V, V 2 =13.33V; P 2ohms = V 2 2 /R = ? 30 October 2023 23ECE103 Fund of Electrical Engg 42
30 October 2023 23ECE103 Fund of Electrical Engg 43
Nodal Analysis for AC circuits - Example 1 If Z L = – j 2Ω, find VL in the circuit using nodal analysis. 30 October 2023 23ECE103 Fund of Electrical Engg 44
Nodal Analysis for AC circuits - Example 1 Step 1: Let us mark nodes along with the node voltages V 1 and V 2 in the given figure. Step 2: At node (1): At node (2): i.e. 30 October 2023 23ECE103 Fund of Electrical Engg 45
Nodal Analysis for AC circuits - Example 1 Comparing: 30 October 2023 23ECE103 Fund of Electrical Engg 46
Nodal Analysis for AC circuits - Example 1 To calculate V 2 Load voltage 30 October 2023 23ECE103 Fund of Electrical Engg 47
Nodal Analysis for AC circuits - Example 2 30 October 2023 23ECE103 Fund of Electrical Engg 48 Use Nodal Analysis to calculate V 2
Nodal Analysis for AC circuits - Example 2 30 October 2023 23ECE103 Fund of Electrical Engg 49 Calculating the parallel branches of j3 and –j5 ohms, resulting j7.5 Ohms.
Nodal Analysis for AC circuits - Example 2 30 October 2023 23ECE103 Fund of Electrical Engg 50
Nodal Analysis for AC circuits - Example 2 30 October 2023 23ECE103 Fund of Electrical Engg 51
30 October 2023 23ECE103 Fund of Electrical Engg 52 THANK YOU