Category: network Theory

Network Theory – Series Resonance ”; Previous Next Resonance occurs in electric circuits due to the presence of energy storing elements like inductor and capacitor. It is the fundamental concept based on which, the radio and TV receivers are designed in such a way that they should be able to select only the desired station frequency. There are two types of resonances, namely series resonance and parallel resonance. These are classified based on the network elements that are connected in series or parallel. In this chapter, let us discuss about series resonance. Series Resonance Circuit Diagram If the resonance occurs in series RLC circuit, then it is called as Series Resonance. Consider the following series RLC circuit, which is represented in phasor domain. Here, the passive elements such as resistor, inductor and capacitor are connected in series. This entire combination is in series with the input sinusoidal voltage source. Apply KVL around the loop. $$V – V_R – V_L – V_C = 0$$ $$Rightarrow V – IR – I(j X_L) – I(-j X_C) = 0$$ $$Rightarrow V = IR + I(j X_L) + I(-j X_C)$$ $Rightarrow V = I[R + j(X_L – X_C)]$Equation 1 The above equation is in the form of V = IZ. Therefore, the impedance Z of series RLC circuit will be $$Z = R + j(X_L – X_C)$$ Parameters & Electrical Quantities at Resonance Now, let us derive the values of parameters and electrical quantities at resonance of series RLC circuit one by one. Resonant Frequency The frequency at which resonance occurs is called as resonant frequency fr. In series RLC circuit resonance occurs, when the imaginary term of impedance Z is zero, i.e., the value of $X_L – X_C$ should be equal to zero. $$Rightarrow X_L = X_C$$ Substitute $X_L = 2 pi f L$ and $X_C = frac{1}{2 pi f C}$ in the above equation. $$2 pi f L = frac{1}{2 pi f C}$$ $$Rightarrow f^2 = frac{1}{(2 pi)^2 L C}$$ $$Rightarrow f = frac{1}{(2 pi) sqrt{LC}}$$ Therefore, the resonant frequency fr of series RLC circuit is $$f_r = frac{1}{(2 pi) sqrt{LC}}$$ Where, L is the inductance of an inductor and C is the capacitance of a capacitor. The resonant frequency fr of series RLC circuit depends only on the inductance L and capacitance C. But, it is independent of resistance R. Impedance We got the impedance Z of series RLC circuit as $$Z = R + j(X_L – X_C)$$ Substitute $X_L = X_C$ in the above equation. $$Z = R + j(X_C – X_C)$$ $$Rightarrow Z = R + j(0)$$ $$Rightarrow Z = R$$ At resonance, the impedance Z of series RLC circuit is equal to the value of resistance R, i.e., Z = R. Current flowing through the Circuit Substitute $X_L – X_C = 0$ in Equation 1. $$V = I[R + j(0)]$$ $$Rightarrow V = IR$$ $$Rightarrow I = frac{V}{R}$$ Therefore, current flowing through series RLC circuit at resonance is $mathbf{mathit{I = frac{V}{R}}}$. At resonance, the impedance of series RLC circuit reaches to minimum value. Hence, the maximum current flows through this circuit at resonance. Voltage across Resistor The voltage across resistor is $$V_R = IR$$ Substitute the value of I in the above equation. $$V_R = lgroup frac{V}{R} rgroup R$$ $$Rightarrow V_R = V$$ Therefore, the voltage across resistor at resonance is VR = V. Voltage across Inductor The voltage across inductor is $$V_L = I(jX_L)$$ Substitute the value of I in the above equation. $$V_L = lgroup frac{V}{R} rgroup (jX_L)$$ $$Rightarrow V_L = j lgroup frac{X_L}{R} rgroup V$$ $$Rightarrow V_L = j QV$$ Therefore, the voltage across inductor at resonance is $V_L = j QV$. So, the magnitude of voltage across inductor at resonance will be $$|V_L| = QV$$ Where Q is the Quality factor and its value is equal to $frac{X_L}{R}$ Voltage across Capacitor The voltage across capacitor is $$V_C = I(-j X_C)$$ Substitute the value of I in the above equation. $$V_C = lgroup frac{V}{R} rgroup (-j X_C)$$ $$Rightarrow V_C = -j lgroup frac{X_C}{R} rgroup V$$ $$Rightarrow V_C = -jQV$$ Therefore, the voltage across capacitor at resonance is $mathbf{mathit{V_C = -jQV}}$. So, the magnitude of voltage across capacitor at resonance will be $$|V_C| = QV$$ Where Q is the Quality factor and its value is equal to $frac{X_{C}}{R}$ Note − Series resonance RLC circuit is called as voltage magnification circuit, because the magnitude of voltage across the inductor and the capacitor is equal to Q times the input sinusoidal voltage V. Print Page Previous Next Advertisements ”;

Network Theory – Superposition Theorem ”; Previous Next Superposition theorem is based on the concept of linearity between the response and excitation of an electrical circuit. It states that the response in a particular branch of a linear circuit when multiple independent sources are acting at the same time is equivalent to the sum of the responses due to each independent source acting at a time. In this method, we will consider only one independent source at a time. So, we have to eliminate the remaining independent sources from the circuit. We can eliminate the voltage sources by shorting their two terminals and similarly, the current sources by opening their two terminals. Therefore, we need to find the response in a particular branch ‘n’ times if there are ‘n’ independent sources. The response in a particular branch could be either current flowing through that branch or voltage across that branch. Procedure of Superposition Theorem Follow these steps in order to find the response in a particular branch using superposition theorem. Step 1 − Find the response in a particular branch by considering one independent source and eliminating the remaining independent sources present in the network. Step 2 − Repeat Step 1 for all independent sources present in the network. Step 3 − Add all the responses in order to get the overall response in a particular branch when all independent sources are present in the network. Example Find the current flowing through 20 Ω resistor of the following circuit using superposition theorem. Step 1 − Let us find the current flowing through 20 Ω resistor by considering only 20 V voltage source. In this case, we can eliminate the 4 A current source by making open circuit of it. The modified circuit diagram is shown in the following figure. There is only one principal node except Ground in the above circuit. So, we can use nodal analysis method. The node voltage V1 is labelled in the following figure. Here, V1 is the voltage from node 1 with respect to ground. The nodal equation at node 1 is $$frac{V_1 – 20}{5} + frac{V_1}{10} + frac{V_1}{10 + 20} = 0$$ $$Rightarrow frac{6V_1 – 120 + 3V_1 + V_1}{30} = 0$$ $$Rightarrow 10V_1 = 120$$ $$Rightarrow V_1 = 12V$$ The current flowing through 20 Ω resistor can be found by doing the following simplification. $$I_1 = frac{V_1}{10 + 20}$$ Substitute the value of V1 in the above equation. $$I_1 = frac{12}{10 + 20} = frac{12}{30} = 0.4 A$$ Therefore, the current flowing through 20 Ω resistor is 0.4 A, when only 20 V voltage source is considered. Step 2 − Let us find the current flowing through 20 Ω resistor by considering only 4 A current source. In this case, we can eliminate the 20 V voltage source by making short-circuit of it. The modified circuit diagram is shown in the following figure. In the above circuit, there are three resistors to the left of terminals A & B. We can replace these resistors with a single equivalent resistor. Here, 5 Ω & 10 Ω resistors are connected in parallel and the entire combination is in series with 10 Ω resistor. The equivalent resistance to the left of terminals A & B will be $$R_{AB} = lgroup frac{5 times 10}{5 + 10} rgroup + 10 = frac{10}{3} + 10 = frac{40}{3} Omega$$ The simplified circuit diagram is shown in the following figure. We can find the current flowing through 20 Ω resistor, by using current division principle. $$I_2 = I_S lgroup frac{R_1}{R_1 + R_2} rgroup$$ Substitute $I_S = 4A,: R_1 = frac{40}{3} Omega$ and $R_2 = 20 Omega$ in the above equation. $$I_2 = 4 lgroup frac{frac{40}{3}}{frac{40}{3} + 20} rgroup = 4 lgroup frac{40}{100} rgroup = 1.6 A$$ Therefore, the current flowing through 20 Ω resistor is 1.6 A, when only 4 A current source is considered. Step 3 − We will get the current flowing through 20 Ω resistor of the given circuit by doing the addition of two currents that we got in step 1 and step 2. Mathematically, it can be written as $$I = I_1 + I_2$$ Substitute, the values of I1 and I2 in the above equation. $$I = 0.4 + 1.6 = 2 A$$ Therefore, the current flowing through 20 Ω resistor of given circuit is 2 A. Note − We can’t apply superposition theorem directly in order to find the amount of power delivered to any resistor that is present in a linear circuit, just by doing the addition of powers delivered to that resistor due to each independent source. Rather, we can calculate either total current flowing through or voltage across that resistor by using superposition theorem and from that, we can calculate the amount of power delivered to that resistor using $I^2 R$ or $frac{V^2}{R}$. Print Page Previous Next Advertisements ”;

Network Theory – Network Topology ”; Previous Next Network topology is a graphical representation of electric circuits. It is useful for analyzing complex electric circuits by converting them into network graphs. Network topology is also called as Graph theory. Basic Terminology of Network Topology Now, let us discuss about the basic terminology involved in this network topology. Graph Network graph is simply called as graph. It consists of a set of nodes connected by branches. In graphs, a node is a common point of two or more branches. Sometimes, only a single branch may connect to the node. A branch is a line segment that connects two nodes. Any electric circuit or network can be converted into its equivalent graph by replacing the passive elements and voltage sources with short circuits and the current sources with open circuits. That means, the line segments in the graph represent the branches corresponding to either passive elements or voltage sources of electric circuit. Example Let us consider the following electric circuit. In the above circuit, there are four principal nodes and those are labelled with 1, 2, 3, and 4. There are seven branches in the above circuit, among which one branch contains a 20 V voltage source, another branch contains a 4 A current source and the remaining five branches contain resistors having resistances of 30 Ω, 5 Ω, 10 Ω, 10 Ω and 20 Ω respectively. An equivalent graph corresponding to the above electric circuit is shown in the following figure. In the above graph, there are four nodes and those are labelled with 1, 2, 3 & 4 respectively. These are same as that of principal nodes in the electric circuit. There are six branches in the above graph and those are labelled with a, b, c, d, e & f respectively. In this case, we got one branch less in the graph because the 4 A current source is made as open circuit, while converting the electric circuit into its equivalent graph. From this Example, we can conclude the following points − The number of nodes present in a graph will be equal to the number of principal nodes present in an electric circuit. The number of branches present in a graph will be less than or equal to the number of branches present in an electric circuit. Types of Graphs Following are the types of graphs − Connected Graph Unconnected Graph Directed Graph Undirected Graph Now, let us discuss these graphs one by one. Connected Graph If there exists at least one branch between any of the two nodes of a graph, then it is called as a connected graph. That means, each node in the connected graph will be having one or more branches that are connected to it. So, no node will present as isolated or separated. The graph shown in the previous Example is a connected graph. Here, all the nodes are connected by three branches. Unconnected Graph If there exists at least one node in the graph that remains unconnected by even single branch, then it is called as an unconnected graph. So, there will be one or more isolated nodes in an unconnected graph. Consider the graph shown in the following figure. In this graph, the nodes 2, 3, and 4 are connected by two branches each. But, not even a single branch has been connected to the node 1. So, the node 1 becomes an isolated node. Hence, the above graph is an unconnected graph. Directed Graph If all the branches of a graph are represented with arrows, then that graph is called as a directed graph. These arrows indicate the direction of current flow in each branch. Hence, this graph is also called as oriented graph. Consider the graph shown in the following figure. In the above graph, the direction of current flow is represented with an arrow in each branch. Hence, it is a directed graph. Undirected Graph If the branches of a graph are not represented with arrows, then that graph is called as an undirected graph. Since, there are no directions of current flow, this graph is also called as an unoriented graph. The graph that was shown in the first Example of this chapter is an unoriented graph, because there are no arrows on the branches of that graph. Subgraph and its Types A part of the graph is called as a subgraph. We get subgraphs by removing some nodes and/or branches of a given graph. So, the number of branches and/or nodes of a subgraph will be less than that of the original graph. Hence, we can conclude that a subgraph is a subset of a graph. Following are the two types of subgraphs. Tree Co-Tree Tree Tree is a connected subgraph of a given graph, which contains all the nodes of a graph. But, there should not be any loop in that subgraph. The branches of a tree are called as twigs. Consider the following connected subgraph of the graph, which is shown in the Example of the beginning of this chapter. This connected subgraph contains all the four nodes of the given graph and there is no loop. Hence, it is a Tree. This Tree has only three branches out of six branches of given graph. Because, if we consider even single branch of the remaining branches of the graph, then there will be a loop in the above connected subgraph. Then, the resultant connected subgraph will not be a Tree. From the above Tree, we can conclude that the number of branches that are present in a Tree should be equal to n – 1 where ‘n’ is the number of nodes of the given graph. Co-Tree Co-Tree is a subgraph, which is formed with the branches that are removed while forming a Tree. Hence, it is called as Complement of a Tree. For every Tree, there will be a corresponding Co-Tree and its branches are called as links

Equivalent Circuits Example Problem ”; Previous Next In the previous chapter, we discussed about the equivalent circuits of series combination and parallel combination individually. In this chapter, let us solve an example problem by considering both series and parallel combinations of similar passive elements. Example Let us find the equivalent resistance across the terminals A & B of the following electrical network. We will get the equivalent resistance across terminals A & B by minimizing the above network into a single resistor between those two terminals. For this, we have to identify the combination of resistors that are connected in series form and parallel form and then find the equivalent resistance of the respective form in every step. The given electrical network is modified into the following form as shown in the following figure. In the above figure, the letters, C to G, are used for labelling various terminals. Step 1 − In the above network, two 6 Ω resistors are connected in parallel. So, the equivalent resistance between D & E will be 3 Ω. This can be obtained by doing the following simplification. $$R_{DE} = frac{6 times 6}{6 + 6} = frac{36}{12} = 3 Omega$$ In the above network, the resistors 4 Ω and 8 Ω are connected in series. So, the equivalent resistance between F & G will be 12 Ω. This can be obtained by doing the following simplification. $$R_{FG} = 4 + 8 = 12 Omega$$ Step 2 − The simplified electrical network after Step 1 is shown in the following figure. In the above network, two 3 Ω resistors are connected in series. So, the equivalent resistance between C & E will be 6 Ω. This can be obtained by doing the following simplification. $$R_{CE} = 3 + 3 = 6 Omega$$ Step 3 − The simplified electrical network after Step 2 is shown in the following figure. In the above network, the resistors 6 Ω and 12 Ω are connected in parallel. So, the equivalent resistance between C & B will be 4 Ω. This can be obtained by doing the following simplification. $$R_{CB} = frac{6 times 12}{6 + 12} = frac{72}{18} = 4 Omega$$ Step 4 − The simplified electrical network after Step 3 is shown in the following figure. In the above network, the resistors 2 Ω and 4 Ω are connected in series between the terminals A & B. So, the equivalent resistance between A & B will be 6 Ω. This can be obtained by doing the following simplification. $$R_{AB} = 2 + 4 = 6 Omega$$ Therefore, the equivalent resistance between terminals A & B of the given electrical network is 6 Ω. Print Page Previous Next Advertisements ”;

Network Theory – Mesh Analysis ”; Previous Next In Mesh analysis, we will consider the currents flowing through each mesh. Hence, Mesh analysis is also called as Mesh-current method. A branch is a path that joins two nodes and it contains a circuit element. If a branch belongs to only one mesh, then the branch current will be equal to mesh current. If a branch is common to two meshes, then the branch current will be equal to the sum (or difference) of two mesh currents, when they are in same (or opposite) direction. Procedure of Mesh Analysis Follow these steps while solving any electrical network or circuit using Mesh analysis. Step 1 − Identify the meshes and label the mesh currents in either clockwise or anti-clockwise direction. Step 2 − Observe the amount of current that flows through each element in terms of mesh currents. Step 3 − Write mesh equations to all meshes. Mesh equation is obtained by applying KVL first and then Ohm’s law. Step 4 − Solve the mesh equations obtained in Step 3 in order to get the mesh currents. Now, we can find the current flowing through any element and the voltage across any element that is present in the given network by using mesh currents. Example Find the voltage across 30 Ω resistor using Mesh analysis. Step 1 − There are two meshes in the above circuit. The mesh currents I1 and I2 are considered in clockwise direction. These mesh currents are shown in the following figure. Step 2 − The mesh current I1 flows through 20 V voltage source and 5 Ω resistor. Similarly, the mesh current I2 flows through 30 Ω resistor and -80 V voltage source. But, the difference of two mesh currents, I1 and I2, flows through 10 Ω resistor, since it is the common branch of two meshes. Step 3 − In this case, we will get two mesh equations since there are two meshes in the given circuit. When we write the mesh equations, assume the mesh current of that particular mesh as greater than all other mesh currents of the circuit. The mesh equation of first mesh is $$20 – 5I_1 -10(I_1 – I_2) = 0$$ $$Rightarrow 20 – 15I_1 + 10I_2 = 0$$ $$Rightarrow 10I_2 = 15I_1 – 20$$ Divide the above equation with 5. $$2I_2 = 3I_1 – 4$$ Multiply the above equation with 2. $4I_2 = 6I_1 – 8$ Equation 1 The mesh equation of second mesh is $$-10(I_2 – I_1) – 30I_2 + 80 = 0$$ Divide the above equation with 10. $$-(I_2 – I_1) – 3I_2 + 8 = 0$$ $$Rightarrow -4I_2 + I_1 + 8 = 0$$ $4I_2 = I_1 + 8$ Equation 2 Step 4 − Finding mesh currents I1 and I2 by solving Equation 1 and Equation 2. The left-hand side terms of Equation 1 and Equation 2 are the same. Hence, equate the right-hand side terms of Equation 1 and Equation 2 in order find the value of I1. $$6I_1 – 8 = I_1 + 8$$ $$Rightarrow 5I_1 = 16$$ $$Rightarrow I_1 = frac{16}{5} A$$ Substitute I1 value in Equation 2. $$4I_2 = frac{16}{5} + 8$$ $$Rightarrow 4I_2 = frac{56}{5}$$ $$Rightarrow I_2 = frac{14}{5} A$$ So, we got the mesh currents I1 and I2 as $mathbf{frac{16}{5}}$ A and $mathbf{frac{14}{5}}$ A respectively. Step 5 − The current flowing through 30 Ω resistor is nothing but the mesh current I2 and it is equal to $frac{14}{5}$ A. Now, we can find the voltage across 30 Ω resistor by using Ohm’s law. $$V_{30 Omega} = I_2 R$$ Substitute the values of I2 and R in the above equation. $$V_{30 Omega} = lgroup frac{14}{5} rgroup 30$$ $$Rightarrow V_{30 Omega} = 84V$$ Therefore, the voltage across 30 Ω resistor of the given circuit is 84 V. Note 1 − From the above example, we can conclude that we have to solve ‘m’ mesh equations, if the electric circuit is having ‘m’ meshes. That’s why we can choose Mesh analysis when the number of meshes is less than the number of principal nodes (except the reference node) of any electrical circuit. Note 2 − We can choose either Nodal analysis or Mesh analysis, when the number of meshes is equal to the number of principal nodes (except the reference node) in any electric circuit. Print Page Previous Next Advertisements ”;

Network Theory – Overview ”; Previous Next Network theory is the study of solving the problems of electric circuits or electric networks. In this introductory chapter, let us first discuss the basic terminology of electric circuits and the types of network elements. Basic Terminology In Network Theory, we will frequently come across the following terms − Electric Circuit Electric Network Current Voltage Power So, it is imperative that we gather some basic knowledge on these terms before proceeding further. Let’s start with Electric Circuit. Electric Circuit An electric circuit contains a closed path for providing a flow of electrons from a voltage source or current source. The elements present in an electric circuit will be in series connection, parallel connection, or in any combination of series and parallel connections. Electric Network An electric network need not contain a closed path for providing a flow of electrons from a voltage source or current source. Hence, we can conclude that “all electric circuits are electric networks” but the converse need not be true. Current The current “I” flowing through a conductor is nothing but the time rate of flow of charge. Mathematically, it can be written as $$I = frac{dQ}{dt}$$ Where, Q is the charge and its unit is Coloumb. t is the time and its unit is second. As an analogy, electric current can be thought of as the flow of water through a pipe. Current is measured in terms of Ampere. In general, Electron current flows from negative terminal of source to positive terminal, whereas, Conventional current flows from positive terminal of source to negative terminal. Electron current is obtained due to the movement of free electrons, whereas, Conventional current is obtained due to the movement of free positive charges. Both of these are called as electric current. Voltage The voltage “V” is nothing but an electromotive force that causes the charge (electrons) to flow. Mathematically, it can be written as $$V = frac{dW}{dQ}$$ Where, W is the potential energy and its unit is Joule. Q is the charge and its unit is Coloumb. As an analogy, Voltage can be thought of as the pressure of water that causes the water to flow through a pipe. It is measured in terms of Volt. Power The power “P” is nothing but the time rate of flow of electrical energy. Mathematically, it can be written as $$P = frac{dW}{dt}$$ Where, W is the electrical energy and it is measured in terms of Joule. t is the time and it is measured in seconds. We can re-write the above equation a $$P = frac{dW}{dt} = frac{dW}{dQ} times frac{dQ}{dt} = VI$$ Therefore, power is nothing but the product of voltage V and current I. Its unit is Watt. Types of Network Elements We can classify the Network elements into various types based on some parameters. Following are the types of Network elements − Active Elements and Passive Elements Linear Elements and Non-linear Elements Bilateral Elements and Unilateral Elements Active Elements and Passive Elements We can classify the Network elements into either active or passive based on the ability of delivering power. Active Elements deliver power to other elements, which are present in an electric circuit. Sometimes, they may absorb the power like passive elements. That means active elements have the capability of both delivering and absorbing power. Examples: Voltage sources and current sources. Passive Elements can’t deliver power (energy) to other elements, however they can absorb power. That means these elements either dissipate power in the form of heat or store energy in the form of either magnetic field or electric field. Examples: Resistors, Inductors, and capacitors. Linear Elements and Non-Linear Elements We can classify the network elements as linear or non-linear based on their characteristic to obey the property of linearity. Linear Elements are the elements that show a linear relationship between voltage and current. Examples: Resistors, Inductors, and capacitors. Non-Linear Elements are those that do not show a linear relation between voltage and current. Examples: Voltage sources and current sources. Bilateral Elements and Unilateral Elements Network elements can also be classified as either bilateral or unilateral based on the direction of current flows through the network elements. Bilateral Elements are the elements that allow the current in both directions and offer the same impedance in either direction of current flow. Examples: Resistors, Inductors and capacitors. The concept of Bilateral elements is illustrated in the following figures. In the above figure, the current (I) is flowing from terminals A to B through a passive element having impedance of Z Ω. It is the ratio of voltage (V) across that element between terminals A & B and current (I). In the above figure, the current (I) is flowing from terminals B to A through a passive element having impedance of Z Ω. That means the current (–I) is flowing from terminals A to B. In this case too, we will get the same impedance value, since both the current and voltage having negative signs with respect to terminals A & B. Unilateral Elements are those that allow the current in only one direction. Hence, they offer different impedances in both directions. Print Page Previous Next Advertisements ”;