CPEN 206 Linear Circuits Lecture # 4 Network Analysis Dr. Godfrey A. Mills Email: gmills@ug.edu.gh Phone: 026-907-3163 February 22, 2016 Course TA David S. Tamakloe 1
What is Network Technique o Network analysis techniques are fundamental principles used to determine currents flowing through unknown branches and node voltages in an electric network. o With network analysis techniques, it is important to define all the relevant variables as clearly as possible and also in a systematic way. o With known and unknown variables identification, a set of equations relating these variables is constructed and are solved by suitable techniques. o Network analysis can be separated into 3 broad areas: n Resistive circuit analysis n Time domain analysis n Frequency domain analysis 2
What is Network Technique o Beyond the basic circuit laws of Ohms and Kirchhoff, the two most generally applied network analysis and solution techniques are: n Nodal analysis: o based on application of the KCL n Mesh analysis: o based on application of the KVL o Concept behind the nodal and mesh methods is to develop a set of simultaneous equations which can be solved to obtain the required values of current or voltage. 3
What is Network Technique o Other network methods used to find solutions are : n Superposition Principle: o based on separate operation of each independent source in a network n Source transformation: o based on transformation of independent voltage and current sources in the network with their equivalents n Thevenin Theorem: o based on concept of replacing network with an equivalent network with single voltage source. n Norton Theorem: o based on concept of replacing network with an equivalent network with single current source. 4
Why Network Techniques o Network analysis is one important area that helps Engineers to properly learn how to analyze a problem. n Example, in system design, analysis forms first step and then we do comparison between desired behavior (specifications) and predicted behavior (from analysis) that leads to the refinement in design o Network analysis provides process to learn how to: n be methodical n determine the goal of particular problem n work with the information given to develop a plan of attack n implement the plan to obtain solution n verify that the proposed solution is accurate 5
Nodal Analysis Method o This method provides a procedure for analyzing circuits using node voltages as the circuit variables. o In Node analysis, we are interested in finding the node voltages. The choice for node voltages rather than the element voltages in circuit gives much flexibility and reduces number of simultaneous equations to solve. o The method defines voltage at each node as an independent variable, with one node selected as a reference (usually ground), where each node voltages is referenced. o Once each node voltage is defined, the Ohms law is then applied between any two adjacent nodes to determine the current flowing in each branch. o With Node method, each branch current is expressed in terms of one or more node voltages, hence, currents do not explicitly get into the equation. 6
Nodal Analysis Method o For a given circuit with n nodes, we summarize the node procedure for circuit analysis as follows: n First, choose a reference node (usually ground). This node will be tied to most of the elements. o Assign virtual voltage to the node voltage V0. n Define the virtual node voltages for the (n-1) nodes in the network excluding reference node and source as V1, V2, V3, etc, to the n-1 nodes. n Apply KCL at each of the nodes in the network excluding that of the reference node. o Use Ohms law to express the branch currents in terms of the node voltages. n Solve the resulting sets of linear equations for the unknown node voltages V1, V2, V3, etc,. 7
Nodal Analysis Method o Node with Dependent Sources: n In circuits analysis, we sometimes meet cases of dependent voltage/current sources in the network. n To analysis such circuit networks with dependent sources, we employ the following strategies: o Treat all dependent source(s) whether current or voltage as independent sources o Apply the KCL to derive the equations o Determine the relationship between dependent source values and the controlling parameters. o Solve the equations for the unknowns. 8
Mesh Analysis Method o This provides a procedure for analyzing circuits using mesh currents as the circuit variables. o As already indicated, the Node analysis applies KCL to find unknown voltages in a given circuit but the mesh applies the KVL to find unknown currents. o Note that a mesh is a loop that does not contain any other loop within it and a loop is a closed path with no node passed more than once. o In mesh analysis, we are interested in finding the loop currents through a mesh in a circuit. o Choosing mesh current rather than element currents in the circuit gives much flexibility and also reduces the number of simultaneous equations to solve. 9
Mesh Analysis Method o To analyze a circuit with n loops using mesh analysis method, we employ the following procedure : n Select the n loops and assign mesh currents I 1, I 2, I 3, etc, to the n meshes of the circuit. n Apply the KVL to each of the meshes in the circuit. o Use ohms law to express the voltages in terms of the mesh currents. n Solve the resulting simultaneous linear equations to obtain the unknown mesh currents. 10
Mesh Analysis Method o Mesh with dependent source n We analyze circuits with dependent sources using the mesh analysis as follows: n Dependent voltage source: o Formulate and write the KVL mesh equations o Express dependency constraint in terms of mesh currents n Dependent current source: o Use supermesh, which is a mesh that avoids the branch containing the current source. o Apply KVL for this supermesh o Express dependency constraint in terms of mesh currents 11
Circuit Analysis Example 1 o Question 1: n A 30V voltage source provides supply to a 4Ω resistor that is connected to a 2Ω resistor which is parallelly connected to another 5Ω resistor. The 5Ω is also supplied by a 40V voltage supply. o Sketch the circuit diagram and find the currents in all the branches of the network using the following methods: n (a) nodal analysis n (b) mesh analysis 12
Nodal Approach Solution 1 o To use node analysis, we first establish our node points and assign virtual voltage to the nodes. o From the circuit, we have four node points with virtual voltages : V A, V B, V C, and V D. o We then take node D as our reference node. This means voltages from node A with reference to D is 30V and that of node C with reference to D is 40V. o Now, we assume that the current I 1 flows from the node A to node B, I 2 flows from node C to node B and finally, I 3 flows from node B to node D. o We next proceed to write down all the equations for the nodes. o For Node B we have: n I 3 = V B /2 ---------- (1) 13
Nodal Approach o If we apply the KCL at the node B, we have the following expressions: n I 1 + I 2 = I 3 ---------- (2) o I 1 = (30 V B )/4 o I 2 = (40 V B )/5 o I 3 = V B /2 n >> 150 5V B + 40 4V B = 5V B >> V B = 16.32V o Using this value we determine the currents as n I 1 = (30 16.3)/4 = 3.43A n I 2 = (40 16.3)/5 = 4.74A n I 3 = I 1 + I 2 = 8.17A 14
Mesh Approach 1a o We first identify the loops and assume the direction of current in loop 1 to flow from the 30V source and that in loop 2 to flow from the 40V source. o Write down the equations for Loop 1 and Loop 2. o Applying KVL to loop 1, we have n 4I 1 + 2(I 1 + I 2 ) = 30 ---------- (1) >> 6I 1 + 2I 2 = 30 o Applying KVL to loop 2, we have n 5I 2 + 2(I 2 + I 1 ) = 40 ---------- (2) >> 2I 1 + 7I 2 = 40 o Multiplying (2) by 3 and subtracting from (1) gives n >> I 2 = 4.7A o Substituting I 2 into the equation, we have n >> I 1 = 3.4A n Current through the 2Ω = (I 2 + I 1 ) = 8.17A 15
Mesh Approach - 1b o Now, lets assume that the direction of current in loop 1 flows from the 30V voltage source and that in loop 2 flows toward the 40V source. o Applying KVL to Loop 1, we have n 4I 1 + 2(I 1 I 2 ) = 30 ---------- (1) >> 6I 1-2I 2 = 30 o Applying KVL to loop 2, we have n 5I 2 + 2(I 2 I 1 ) + 40 = 0 ---------- (2) >> -2I 1 + 7I 2 = -40 o Multiplying (2) by 3 and adding to (1) gives n >> I 2 = -4.7A o Substituting I 2 into the equation, we have n >> I 1 = 3.4A n Since I 2 is negative, we conclude that the direction of current in Loop 2 is opposite to our assumption. 16
Circuit Analysis Example 2 o Find the current I that will flow through the circuit network below and determine the terminal voltage V ab across the circuit. 17
Circuit Analysis Solution 2 o Since we have a single loop, we can apply the KVL to the circuit. o From the loop, we have (note the direction of current flow and the polarity of the voltage) n 5*I -5 + 5*I + 10 10 = 0 n 10*I = 5 n I = 0.5A n Vab = 5*I + 10 = 12.5V 18
Circuit Analysis Example 3 o Find the current Io that is delivered by the network using the mesh analysis and calculate the power dissipated in the 400 ohms resistors. 19
Circuit Analysis Solution 3 o We first create our loops 1 and 2 and assign currents in the loops as Io and I1. o Applying KVL to loop 1, we have 400Io + 750Io 750I1 = 80 ---------- (1) >> 1150Io 750I1 = 80 -------------------2) o Applying KVL to loop 2, we have I1 = -0.6Io ---------- (3) o Putting (3) in (2), and solving, we have Io = 80/1600 = 50mA o To determine the power dissipated in the 400 ohms become n P2 = V*Io = I 2 R = (50mA) 2 *400 = 1W o How will you find the power in the 750 ohms resistors 20
Circuit Analysis Example 4 o Circuit below is a model for an inverting amplifier circuit derived from an op-amp. Find the current I delivered to the circuit and the dependent source voltage Vs. I 1kΩ 4kΩ 10kΩ 10V + - - V f + + - V s =100V f 21
Circuit Analysis Solution 4 o Since we have a loop, we use the Mesh analysis. o For the Step 1: we apply KVL around loop n -10V + 1kΩ I + 4kΩ I + 10KΩ I + 100 V f = 0 >> -10V + 15kΩ I + 100 V f = 0 o Step 2 : we find V f in terms of I: n V f + 10kΩ I + 100V f = 0 >> V f = 10kΩ/101 I 22
Circuit Analysis Solution 4 o Step 3 : Solve for I: n I = 1.961mA o Step 4 : Solve for V f : n V f = -0.194V o Step 5 : Solve for the source voltage: n V s = -19.4V 23
Circuit Analysis Example 5 o For the equivalent op-amp circuit below, find the output voltage V s, assuming a frequency of ω=5000 I 1kΩ 4kΩ 10nF 10V + - - V f + + - V s =100V f 24
Circuit Analysis Solution 5 o We first find the impedance of the circuits using the given angular frequency w=5000. o The next step is to use the same approach as in the previous example to determine the Current and voltage. I 1kΩ 4kΩ -j2kω 10V + - - V f + + - V s =100V f 25
Circuit Analysis Solution 5 o Step 1 : Apply KVL around loop: -10V + 1kΩ I + 4kΩ I - j2kω I + 100 V f = 0 o Step 2 : Get V f in terms of I: V f - j2kω I + 100 V f = 0 V f = j2kω/101 I 26
Circuit Analysis Solution 5 o Step 3 : Solve for I: I = 2mA -0.2 o Step 4 : Solve for V f : V f = 39.6mV 89.8 o Step 5 : Solve for source voltage: V s = 3.96V 89.8 27
Circuit Analysis Example 6 o The Circuit below is a small-signal linear equivalent circuit for a transistor amplifier. Find the Voltage Vx: 5mA 6kΩ 5 10-4 V 3kΩ + V X - 28
Circuit Analysis Solution 6 o Apply KCL at the top Node n 5mA = V X /6kΩ + 5 10-4 V X + V X /3kΩ n 5mA = 1.67 10-4 V X + 5 10-4 V X + 3.33 10-4 V X n V X =5mA(1.67 10-4 + 5 10-4 + 3.33 10-4 ) n V X =5V 29
Circuit Analysis Example 7 o Find the power delivered to the 2 ohms resistor in the circuit below using the node analysis method. 30
Circuit Analysis Solution 7 o We first create nodes 1 and 2 and assign voltages to them as V1 and V2. o Applying KCL at node 1, we have 2 V1/4 +(v2 V1)/2 = 0 ---------- (1) >> 3V1 2V2 = 8 -------------------2) o Applying KCL at node 2, we have (v1 V2)/2 +V1/3 - V2/8 = 0 ---------- (3) >> V1 = 3*V2/4 -------------------4) o Putting (4) in (2), and solving, we have V1 = 24V V2 = 32V o The power dissipated in the 2 ohms become P2 = V*I1 = V 2 /R = (V1-V2) 2 /R = 64/2 = 32W 31
Circuit Analysis Example 8 o Find the current in the network and determine the power in the 6 ohms resistor 10V 32
Circuit Analysis Solution 8 o Using the KVL for loop 1, we have n -10+ 4I1 2Ix + 6 (I1-I2) = 0 o Ix = I1 I2 >> 4I1-2I2 = 5 o Using KVL for loop 2 we have n 6(I2-I1) +2I2+12 = 0 o >> 3I1 4I2 = 6 o Solving the two simultaneous equations, we have n I1 = 0.8A n I2 = -0.9A o Since Ix = I1 I2 = (0.8+0.9) = 1.7A, we have the power dissipated in the 6 ohms as n P6 = V*Ix = I 2 R = (1.7) 2 *6 W 33
Circuit Analysis Example 9 o Find the voltage Vo of the circuit 34
Circuit Analysis Solution 9 o We first find the current I and use the value to compute the voltage V1 and finally solve for the required voltage Vo. o Thus, n I = V/Req = 240/(25k+75k) = 2.4mA n V1 = 75000*I = 75000*2.4mA = 180V o With V1 as a voltage source, we can use the KVL voltage divider method to compute V0 as n Vo = V1*120/(120+30) = 180*120/150 = 144V 35