Category: network Theory

Network Theory – Equivalent Circuits ”; Previous Next If a circuit consists of two or more similar passive elements and are connected in exclusively of series type or parallel type, then we can replace them with a single equivalent passive element. Hence, this circuit is called as an equivalent circuit. In this chapter, let us discuss about the following two equivalent circuits. Series Equivalent Circuit Parallel Equivalent Circuit Series Equivalent Circuit If similar passive elements are connected in series, then the same current will flow through all these elements. But, the voltage gets divided across each element. Consider the following circuit diagram. It has a single voltage source (VS) and three resistors having resistances of R1, R2 and R3. All these elements are connected in series. The current IS flows through all these elements. The above circuit has only one mesh. The KVL equation around this mesh is $$V_S = V_1 + V_2 + V_3$$ Substitute $V_1 = I_S R_1, : V_2 = I_S R_2$ and $V_3 = I_S R_3$ in the above equation. $$V_S = I_S R_1 + I_S R_2 + I_S R_3$$ $$Rightarrow V_S = I_S(R_1 + R_2 + R_3)$$ The above equation is in the form of $V_S = I_S R_{Eq}$ where, $$R_{Eq} = R_1 + R_2 + R_3$$ The equivalent circuit diagram of the given circuit is shown in the following figure. That means, if multiple resistors are connected in series, then we can replace them with an equivalent resistor. The resistance of this equivalent resistor is equal to sum of the resistances of all those multiple resistors. Note 1 − If ‘N’ inductors having inductances of L1, L2, …, LN are connected in series, then the equivalent inductance will be $$L_{Eq} = L_1 + L_2 + … + L_N$$ Note 2 − If ‘N’ capacitors having capacitances of C1, C2, …, CN are connected in series, then the equivalent capacitance will be $$frac{1}{C_{Eq}} = frac{1}{C_1} + frac{1}{C_2} + … + frac{1}{C_N}$$ Parallel Equivalent Circuit If similar passive elements are connected in parallel, then the same voltage will be maintained across each element. But, the current flowing through each element gets divided. Consider the following circuit diagram. It has a single current source (IS) and three resistors having resistances of R1, R2, and R3. All these elements are connected in parallel. The voltage (VS) is available across all these elements. The above circuit has only one principal node (P) except the Ground node. The KCL equation at this principal node (P) is $$I_S = I_1 + I_2 + I_3$$ Substitute $I_1 = frac{V_S}{R_1}, : I_2 = frac{V_S}{R_2}$ and $I_3 = frac{V_S}{R_3}$ in the above equation. $$I_S = frac{V_S}{R_1} + frac{V_S}{R_2} + frac{V_S}{R_3}$$ $$Rightarrow I_S = V_S lgroup frac{1}{R_1} + frac{1}{R_2} + frac{1}{R_3} rgroup$$ $$Rightarrow V_S = I_Sleft [ frac{1}{lgroup frac{1}{R_1} + frac{1}{R_2} + frac{1}{R_3} rgroup} right ]$$ The above equation is in the form of VS = ISREq where, $$R_{Eq} = frac{1}{lgroup frac{1}{R_1} + frac{1}{R_2} + frac{1}{R_3} rgroup}$$ $$frac{1}{R_{Eq}} = frac{1}{R_1} + frac{1}{R_2} + frac{1}{R_3}$$ The equivalent circuit diagram of the given circuit is shown in the following figure. That means, if multiple resistors are connected in parallel, then we can replace them with an equivalent resistor. The resistance of this equivalent resistor is equal to the reciprocal of sum of reciprocal of each resistance of all those multiple resistors. Note 1 − If ‘N’ inductors having inductances of L1, L2, …, LN are connected in parallel, then the equivalent inductance will be $$frac{1}{L_{Eq}} = frac{1}{L_1} + frac{1}{L_2} + … + frac{1}{L_N}$$ Note 2 − If ‘N’ capacitors having capacitances of C1, C2, …, CN are connected in parallel, then the equivalent capacitance will be $$C_{Eq} = C_1 + C_2 + … + C_N$$ Print Page Previous Next Advertisements ”;

Network Theory – Thevenin’s Theorem ”; Previous Next Thevenin’s theorem states that any two terminal linear network or circuit can be represented with an equivalent network or circuit, which consists of a voltage source in series with a resistor. It is known as Thevenin’s equivalent circuit. A linear circuit may contain independent sources, dependent sources, and resistors. If the circuit contains multiple independent sources, dependent sources, and resistors, then the response in an element can be easily found by replacing the entire network to the left of that element with a Thevenin’s equivalent circuit. The response in an element can be the voltage across that element, current flowing through that element, or power dissipated across that element. This concept is illustrated in following figures. Thevenin’s equivalent circuit resembles a practical voltage source. Hence, it has a voltage source in series with a resistor. The voltage source present in the Thevenin’s equivalent circuit is called as Thevenin’s equivalent voltage or simply Thevenin’s voltage, VTh. The resistor present in the Thevenin’s equivalent circuit is called as Thevenin’s equivalent resistor or simply Thevenin’s resistor, RTh. Methods of Finding Thevenin’s Equivalent Circuit There are three methods for finding a Thevenin’s equivalent circuit. Based on the type of sources that are present in the network, we can choose one of these three methods. Now, let us discuss two methods one by one. We will discuss the third method in the next chapter. Method 1 Follow these steps in order to find the Thevenin’s equivalent circuit, when only the sources of independent type are present. Step 1 − Consider the circuit diagram by opening the terminals with respect to which the Thevenin’s equivalent circuit is to be found. Step 2 − Find Thevenin’s voltage VTh across the open terminals of the above circuit. Step 3 − Find Thevenin’s resistance RTh across the open terminals of the above circuit by eliminating the independent sources present in it. Step 4 − Draw the Thevenin’s equivalent circuit by connecting a Thevenin’s voltage VTh in series with a Thevenin’s resistance RTh. Now, we can find the response in an element that lies to the right side of Thevenin’s equivalent circuit. Example Find the current flowing through 20 Ω resistor by first finding a Thevenin’s equivalent circuit to the left of terminals A and B. Step 1 − In order to find the Thevenin’s equivalent circuit to the left side of terminals A & B, we should remove the 20 Ω resistor from the network by opening the terminals A & B. The modified circuit diagram is shown in the following figure. Step 2 − Calculation of Thevenin’s voltage VTh. There is only one principal node except Ground in the above circuit. So, we can use nodal analysis method. The node voltage V1 and Thevenin’s voltage VTh are labelled in the above figure. Here, V1 is the voltage from node 1 with respect to Ground and VTh is the voltage across 4 A current source. The nodal equation at node 1 is $$frac{V_1 – 20}{5} + frac{V_1}{10} – 4 = 0$$ $$Rightarrow frac{2V_1 – 40 + V_1 – 40}{10} = 0$$ $$Rightarrow 3V_1 – 80 = 0$$ $$Rightarrow V_1 = frac{80}{3}V$$ The voltage across series branch 10 Ω resistor is $$V_{10 Omega} = (-4)(10) = -40V$$ There are two meshes in the above circuit. The KVL equation around second mesh is $$V_1 – V_{10 Omega} – V_{Th} = 0$$ Substitute the values of $V_1$ and $V_{10 Omega}$ in the above equation. $$frac{80}{3} – (-40) – V_{Th} = 0$$ $$V_{Th} = frac{80 + 120}{3} = frac{200}{3}V$$ Therefore, the Thevenin’s voltage is $V_{Th} = frac{200}{3}V$ Step 3 − Calculation of Thevenin’s resistance RTh. Short circuit the voltage source and open circuit the current source of the above circuit in order to calculate the Thevenin’s resistance RTh across the terminals A & B. The modified circuit diagram is shown in the following figure. The Thevenin’s resistance across terminals A & B will be $$R_{Th} = lgroup frac{5 times 10}{5 + 10} rgroup + 10 = frac{10}{3} + 10 = frac{40}{3} Omega$$ Therefore, the Thevenin’s resistance is $mathbf {R_{Th} = frac{40}{3} Omega}$. Step 4 − The Thevenin’s equivalent circuit is placed to the left of terminals A & B in the given circuit. This circuit diagram is shown in the following figure. The current flowing through the 20 Ω resistor can be found by substituting the values of VTh, RTh and R in the following equation. $$l = frac{V_{Th}}{R_{Th} + R}$$ $$l = frac{frac{200}{3}}{frac{40}{3} + 20} = frac{200}{100} = 2A$$ Therefore, the current flowing through the 20 Ω resistor is 2 A. Method 2 Follow these steps in order to find the Thevenin’s equivalent circuit, when the sources of both independent type and dependent type are present. Step 1 − Consider the circuit diagram by opening the terminals with respect to which, the Thevenin’s equivalent circuit is to be found. Step 2 − Find Thevenin’s voltage VTh across the open terminals of the above circuit. Step 3 − Find the short circuit current ISC by shorting the two opened terminals of the above circuit. Step 4 − Find Thevenin’s resistance RTh by using the following formula. $$R_{Th} = frac{V_{Th}}{I_{SC}}$$ Step 5 − Draw the Thevenin’s equivalent circuit by connecting a Thevenin’s voltage VTh in series with a Thevenin’s resistance RTh. Now, we can find the response in an element that lies to the right side of the Thevenin’s equivalent circuit. Print Page Previous Next Advertisements ”;

Network Theory – Example Problems ”; Previous Next We discussed the types of network elements in the previous chapter. Now, let us identify the nature of network elements from the V-I characteristics given in the following examples. Example 1 The V-I characteristics of a network element is shown below. Step 1 − Verifying the network element as linear or non-linear. From the above figure, the V-I characteristics of a network element is a straight line passing through the origin. Hence, it is a Linear element. Step 2 − Verifying the network element as active or passive. The given V-I characteristics of a network element lies in the first and third quadrants. In the first quadrant, the values of both voltage (V) and current (I) are positive. So, the ratios of voltage (V) and current (I) gives positive impedance values. Similarly, in the third quadrant, the values of both voltage (V) and current (I) have negative values. So, the ratios of voltage (V) and current (I) produce positive impedance values. Since, the given V-I characteristics offer positive impedance values, the network element is a Passive element. Step 3 − Verifying the network element as bilateral or unilateral. For every point (I, V) on the characteristics, there exists a corresponding point (-I, -V) on the given characteristics. Hence, the network element is a Bilateral element. Therefore, the given V-I characteristics show that the network element is a Linear, Passive, and Bilateral element. Example 2 The V-I characteristics of a network element is shown below. Step 1 − Verifying the network element as linear or non-linear. From the above figure, the V-I characteristics of a network element is a straight line only between the points (-3A, -3V) and (5A, 5V). Beyond these points, the V-I characteristics are not following the linear relation. Hence, it is a Non-linear element. Step 2 − Verifying the network element as active or passive. The given V-I characteristics of a network element lies in the first and third quadrants. In these two quadrants, the ratios of voltage (V) and current (I) produce positive impedance values. Hence, the network element is a Passive element. Step 3 − Verifying the network element as bilateral or unilateral. Consider the point (5A, 5V) on the characteristics. The corresponding point (-5A, -3V) exists on the given characteristics instead of (-5A, -5V). Hence, the network element is a Unilateral element. Therefore, the given V-I characteristics show that the network element is a Non-linear, Passive, and Unilateral element. Print Page Previous Next Advertisements ”;

Network Theory – Passive Elements ”; Previous Next In this chapter, we will discuss in detail about the passive elements such as Resistor, Inductor, and Capacitor. Let us start with Resistors. Resistor The main functionality of Resistor is either opposes or restricts the flow of electric current. Hence, the resistors are used in order to limit the amount of current flow and / or dividing (sharing) voltage. Let the current flowing through the resistor is I amperes and the voltage across it is V volts. The symbol of resistor along with current, I and voltage, V are shown in the following figure. According to Ohm’s law, the voltage across resistor is the product of current flowing through it and the resistance of that resistor. Mathematically, it can be represented as $V = IR$ Equation 1 $Rightarrow I = frac{V}{R}$Equation 2 Where, R is the resistance of a resistor. From Equation 2, we can conclude that the current flowing through the resistor is directly proportional to the applied voltage across resistor and inversely proportional to the resistance of resistor. Power in an electric circuit element can be represented as $P = VI$Equation 3 Substitute, Equation 1 in Equation 3. $P = (IR)I$ $Rightarrow P = I^2 R$ Equation 4 Substitute, Equation 2 in Equation 3. $P = V lgroup frac{V}{R} rgroup$ $Rightarrow P = frac{V^2}{R}$ Equation 5 So, we can calculate the amount of power dissipated in the resistor by using one of the formulae mentioned in Equations 3 to 5. Inductor In general, inductors will have number of turns. Hence, they produce magnetic flux when current flows through it. So, the amount of total magnetic flux produced by an inductor depends on the current, I flowing through it and they have linear relationship. Mathematically, it can be written as $$Psi : alpha : I$$ $$Rightarrow Psi = LI$$ Where, Ψ is the total magnetic flux L is the inductance of an inductor Let the current flowing through the inductor is I amperes and the voltage across it is V volts. The symbol of inductor along with current I and voltage V are shown in the following figure. According to Faraday’s law, the voltage across the inductor can be written as $$V = frac{dPsi}{dt}$$ Substitute Ψ = LI in the above equation. $$V = frac{d(LI)}{dt}$$ $$Rightarrow V = L frac{dI}{dt}$$ $$Rightarrow I = frac{1}{L} int V dt$$ From the above equations, we can conclude that there exists a linear relationship between voltage across inductor and current flowing through it. We know that power in an electric circuit element can be represented as $$P = VI$$ Substitute $V = L frac{dI}{dt}$ in the above equation. $$P = lgroup L frac{dI}{dt}rgroup I$$ $$Rightarrow P = LI frac{dI}{dt}$$ By integrating the above equation, we will get the energy stored in an inductor as $$W = frac{1}{2} LI^2$$ So, the inductor stores the energy in the form of magnetic field. Capacitor In general, a capacitor has two conducting plates, separated by a dielectric medium. If positive voltage is applied across the capacitor, then it stores positive charge. Similarly, if negative voltage is applied across the capacitor, then it stores negative charge. So, the amount of charge stored in the capacitor depends on the applied voltage V across it and they have linear relationship. Mathematically, it can be written as $$Q : alpha : V$$ $$Rightarrow Q = CV$$ Where, Q is the charge stored in the capacitor. C is the capacitance of a capacitor. Let the current flowing through the capacitor is I amperes and the voltage across it is V volts. The symbol of capacitor along with current I and voltage V are shown in the following figure. We know that the current is nothing but the time rate of flow of charge. Mathematically, it can be represented as $$I = frac{dQ}{dt}$$ Substitute $Q = CV$ in the above equation. $$I = frac{d(CV)}{dt}$$ $$Rightarrow I = C frac{dV}{dt}$$ $$Rightarrow V = frac{1}{C} int I dt$$ From the above equations, we can conclude that there exists a linear relationship between voltage across capacitor and current flowing through it. We know that power in an electric circuit element can be represented as $$P = VI$$ Substitute $I = C frac{dV}{dt}$ in the above equation. $$P = V lgroup C frac{dV}{dt} rgroup$$ $$Rightarrow P = CV frac{dV}{dt}$$ By integrating the above equation, we will get the energy stored in the capacitor as $$W = frac{1}{2}CV^2$$ So, the capacitor stores the energy in the form of electric field. Print Page Previous Next Advertisements ”;

Network Theory – Two-Port Networks ”; Previous Next In general, it is easy to analyze any electrical network, if it is represented with an equivalent model, which gives the relation between input and output variables. For this, we can use two port network representations. As the name suggests, two port networks contain two ports. Among which, one port is used as an input port and the other port is used as an output port. The first and second ports are called as port1 and port2 respectively. One port network is a two terminal electrical network in which, current enters through one terminal and leaves through another terminal. Resistors, inductors and capacitors are the examples of one port network because each one has two terminals. One port network representation is shown in the following figure. Here, the pair of terminals, 1 & 1’ represents a port. In this case, we are having only one port since it is a one port network. Similarly, two port network is a pair of two terminal electrical network in which, current enters through one terminal and leaves through another terminal of each port. Two port network representation is shown in the following figure. Here, one pair of terminals, 1 & 1’ represents one port, which is called as port1 and the other pair of terminals, 2 & 2’ represents another port, which is called as port2. There are four variables V1, V2, I1 and I2 in a two port network as shown in the figure. Out of which, we can choose two variables as independent and another two variables as dependent. So, we will get six possible pairs of equations. These equations represent the dependent variables in terms of independent variables. The coefficients of independent variables are called as parameters. So, each pair of equations will give a set of four parameters. Two Port Network Parameters The parameters of a two port network are called as two port network parameters or simply, two port parameters. Following are the types of two port network parameters. Z parameters Y parameters T parameters T’ parameters h-parameters g-parameters Now, let us discuss about these two port network parameters one by one. Z parameters We will get the following set of two equations by considering the variables V1 & V2 as dependent and I1 & I2 as independent. The coefficients of independent variables, I1 and I2 are called as Z parameters. $$V_1 = Z_{11} I_1 + Z_{12} I_2$$ $$V_2 = Z_{21} I_1 + Z_{22} I_2$$ The Z parameters are $$Z_{11} = frac{V_1}{I_1}, : when : I_2 = 0$$ $$Z_{12} = frac{V_1}{I_2}, : when : I_1 = 0$$ $$Z_{21} = frac{V_2}{I_1}, : when : I_2 = 0$$ $$Z_{22} = frac{V_2}{I_2}, : when : I_1 = 0$$ Z parameters are called as impedance parameters because these are simply the ratios of voltages and currents. Units of Z parameters are Ohm (Ω). We can calculate two Z parameters, Z11 and Z21, by doing open circuit of port2. Similarly, we can calculate the other two Z parameters, Z12 and Z22 by doing open circuit of port1. Hence, the Z parameters are also called as open-circuit impedance parameters. Y parameters We will get the following set of two equations by considering the variables I1 & I2 as dependent and V1 & V2 as independent. The coefficients of independent variables, V1 and V2 are called as Y parameters. $$I_1 = Y_{11} V_1 + Y_{12} V_2$$ $$I_2 = Y_{21} V_1 + Y_{22} V_2$$ The Y parameters are $$Y_{11} = frac{I_1}{V_1}, : when : V_2 = 0$$ $$Y_{12} = frac{I_1}{V_2}, : when : V_1 = 0$$ $$Y_{21} = frac{I_2}{V_1}, : when : V_2 = 0$$ $$Y_{22} = frac{I_2}{V_2}, : when : V_1 = 0$$ Y parameters are called as admittance parameters because these are simply, the ratios of currents and voltages. Units of Y parameters are mho. We can calculate two Y parameters, Y11 and Y21 by doing short circuit of port2. Similarly, we can calculate the other two Y parameters, Y12 and Y22 by doing short circuit of port1. Hence, the Y parameters are also called as short-circuit admittance parameters. T parameters We will get the following set of two equations by considering the variables V1 & I1 as dependent and V2 & I2 as independent. The coefficients of V2 and -I2 are called as T parameters. $$V_1 = A V_2 – B I_2$$ $$I_1 = C V_2 – D I_2$$ The T parameters are $$A = frac{V_1}{V_2}, : when : I_2 = 0$$ $$B = -frac{V_1}{I_2}, : when : V_2 = 0$$ $$C = frac{I_1}{V_2}, : when : I_2 = 0$$ $$D = -frac{I_1}{I_2}, : when : V_2 = 0$$ T parameters are called as transmission parameters or ABCD parameters. The parameters, A and D do not have any units, since those are dimension less. The units of parameters, B and C are ohm and mho respectively. We can calculate two parameters, A and C by doing open circuit of port2. Similarly, we can calculate the other two parameters, B and D by doing short circuit of port2. T ’ parameters We will get the following set of two equations by considering the variables V2 & I2 as dependent and V1 & I1 as independent. The coefficients of V1 and -I1 are called as T’ parameters. $$V_2 = A” V_1 – B” I_1$$ $$I_2 = C” V_1 – D” I_1$$ The T’ parameters are $$A” = frac{V_2}{V_1}, : when: I_1 = 0$$ $$B” = -frac{V_2}{I_1}, : when: V_1 = 0$$ $$C” = frac{I_2}{V_1}, : when: I_1 = 0$$ $$D” = -frac{I_2}{I_1}, : when : V_1 = 0$$ T’ parameters are called as inverse transmission parameters or A’B’C’D’ parameters. The parameters A’ and D’ do not have any units, since those are dimension less. The units of parameters, B’ and C’, are Ohm and Mho respectively. We can calculate two parameters, A’ and C’, by doing an open circuit of port1. Similarly, we can calculate the other two parameters, B’ and D’, by

Discuss Network Theory ”; Previous Next This tutorial is meant to provide the readers the know-how to analyze and solve any electric circuit or network. After completing this tutorial, you will understand the laws and methods that can be applied to specific electric circuits and networks. Print Page Previous Next Advertisements ”;

Network Theory – Useful Resources ”; Previous Next The following resources contain additional information on Network Theory. Please use them to get more in-depth knowledge on this. Useful Video Courses Electromagnetic Theory Course for GATE 101 Lectures 9.5 hours Tutorialspoint More Detail Theory of Machines Online Course 169 Lectures 13.5 hours Tutorialspoint More Detail Graph Theory Online Course 98 Lectures 7 hours Tutorialspoint More Detail Kinetic Theory of Gases 9 Lectures 1 hours Tutorialspoint More Detail Graph Theory Algorithms 33 Lectures 6.5 hours William Fiset More Detail Basic Theory of Chinese Medicine: Chapter 1-10 57 Lectures 9.5 hours TzuChen Kao More Detail Print Page Previous Next Advertisements ”;

Network Theory – Quick Guide ”; Previous Next Network Theory – Overview 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. Network Theory – Example Problems We discussed the types of network elements in the previous chapter. Now, let us identify the nature of network elements from the V-I characteristics given in the following examples. Example 1 The V-I characteristics of a network element is shown below. Step 1 − Verifying the network element as linear or non-linear. From the above figure, the V-I characteristics of a network element is a straight line passing through the origin. Hence, it is a Linear element. Step 2 − Verifying the network element as active or passive. The given V-I characteristics of a network element lies in the first and third quadrants. In the first quadrant, the values of both voltage (V) and current (I) are positive. So, the ratios of voltage (V) and current (I) gives positive

Network Theory – Response of DC Circuits ”; Previous Next If the output of an electric circuit for an input varies with respect to time, then it is called as time response. The time response consists of following two parts. Transient Response Steady state Response In this chapter, first let us discuss about these two responses and then observe these two responses in a series RL circuit, when it is excited by a DC voltage source. Transient Response After applying an input to an electric circuit, the output takes certain time to reach steady state. So, the output will be in transient state till it goes to a steady state. Therefore, the response of the electric circuit during the transient state is known as transient response. The transient response will be zero for large values of ‘t’. Ideally, this value of ‘t’ should be infinity. But, practically five time constants are sufficient. Presence or Absence of Transients Transients occur in the response due to sudden change in the sources that are applied to the electric circuit and / or due to switching action. There are two possible switching actions. Those are opening switch and closing switch. The transient part will not present in the response of an electrical circuit or network, if it contains only resistances. Because resistor is having the ability to adjust any amount of voltage and current. The transient part occurs in the response of an electrical circuit or network due to the presence of energy storing elements such as inductor and capacitor. Because they can’t change the energy stored in those elements instantly. Inductor Behavior Assume the switching action takes place at t = 0. Inductor current does not change instantaneously, when the switching action takes place. That means, the value of inductor current just after the switching action will be same as that of just before the switching action. Mathematically, it can be represented as $$i_L (0^+) = i_L (0^-)$$ Capacitor Behavior The capacitor voltage does not change instantaneously similar to the inductor current, when the switching action takes place. That means, the value of capacitor voltage just after the switching action will be same as that of just before the switching action. Mathematically, it can be represented as $$v_c (0^+) = v_c (0^-)$$ Steady state Response The part of the time response that remains even after the transient response has become zero value for large values of ‘t’ is known as steady state response. This means, there won’t be any transient part in the response during steady state. Inductor Behavior If the independent source is connected to the electric circuit or network having one or more inductors and resistors (optional) for a long time, then that electric circuit or network is said to be in steady state. Therefore, the energy stored in the inductor(s) of that electric circuit is of maximum and constant. Mathematically, it can be represented as $W_L = frac{L {i_L}^2}{2} = $ Maximum & constant $Rightarrow i_L = $ Maximum & constant Therefore, inductor acts as a constant current source in steady state. The voltage across inductor will be $$V_L = L frac{di_{L}}{dt} = 0V$$ So, the inductor acts as a short circuit in steady state. Capacitor Behavior If the independent source is connected to the electric circuit or network having one or more capacitors and resistors (optional) for a long time, then that electric circuit or network is said to be in steady state. Therefore, the energy stored in the capacitor(s) of that electric circuit is of maximum and constant. Mathematically, it can be represented as $W_c = frac{C{v_c}^2}{2} = $ Maximum & constant $Rightarrow v_c = $Maximum & constant Therefore, capacitor acts as a constant voltage source in steady state. The current flowing through the capacitor will be $$i_c = Cfrac{dv_c}{dt} = 0A$$ So, the capacitor acts as an open circuit in steady state. Finding the Response of Series RL Circuit Consider the following series RL circuit diagram. In the above circuit, the switch was kept open up to t = 0 and it was closed at t = 0. So, the DC voltage source having V volts is not connected to the series RL circuit up to this instant. Therefore, there is no initial current flows through inductor. The circuit diagram, when the switch is in closed position is shown in the following figure. Now, the current i flows in the entire circuit, since the DC voltage source having V volts is connected to the series RL circuit. Now, apply KVL around the loop. $$V = Ri + L frac{di}{dt}$$ $frac{di}{dt} + lgroup frac{R}{L} rgroup i = frac{V}{L}$Equation 1 The above equation is a first order differential equation and it is in the form of $frac{dy}{dt} + Py = Q$Equation 2 By comparing Equation 1 and Equation 2, we will get the following relations. $$x = t$$ $$y = i$$ $$P = frac{R}{L}$$ $$Q = frac{V}{L}$$ The solution of Equation 2 will be $ye^{int p dx} = int Q e^{int p dx} dx + k$Equation 3 Where, k is the constant. Substitute, the values of x, y, P & Q in Equation 3. $ie^{int {lgroup frac{R}{L} rgroup}dt} = int (frac{V}{L}) lgroup e^{int {lgroup frac{R}{L} rgroup}dt} rgroup dt + k$ $Rightarrow ie^{lgroup frac{R}{L} rgroup t} = frac{V}{L} int e^{lgroup frac{R}{L} rgroup t} dt + k$ $Rightarrow ie^{lgroup frac{R}{L} rgroup t} = frac{V}{L} lbrace frac{e^{lgroup frac{R}{L} rgroup}t}{frac{R}{L}} rbrace + k$ $Rightarrow i = frac{V}{R} + k e^{-lgroup frac{R}{L} rgroup}t$Equation 4 We know that there is no initial current in the circuit. Hence, substitute, t = 0 and 𝑖 = 0 in Equation 4 in order to find the value of the constant k. $$0 = frac{V}{R} + ke^{-lgroup frac{R}{L} rgroup(0)}$$ $$0 = frac{V}{R} + k(1)$$ $$k = – frac{V}{R}$$ Substitute, the value of k in Equation 4. $$i = frac{V}{R} + lgroup – frac{V}{R} rgroup e^{-lgroup frac{R}{L} rgroup t}$$ $$i = frac{V}{R} – frac{V}{R}e^{-lgroup frac{R}{L} rgroup t}$$ Therefore, the current flowing through the circuit is $i

Network Theory – Coupled Circuits ”; Previous Next An electric circuit is said to be a coupled circuit, when there exists a mutual inductance between the coils (or inductors) present in that circuit. Coil is nothing but the series combination of resistor and inductor. In the absence of resistor, coil becomes inductor. Sometimes, the terms coil and inductor are interchangeably used. In this chapter, first let us discuss about the dot convention and then will discuss about classification of coupling. Dot Convention Dot convention is a technique, which gives the details about voltage polarity at the dotted terminal. This information is useful, while writing KVL equations. If the current enters at the dotted terminal of one coil (or inductor), then it induces a voltage at another coil (or inductor), which is having positive polarity at the dotted terminal. If the current leaves from the dotted terminal of one coil (or inductor), then it induces a voltage at another coil (or inductor), which is having negative polarity at the dotted terminal. Classification of Coupling We can classify coupling into the following two categories. Electrical Coupling Magnetic Coupling Now, let us discuss about each type of coupling one by one. Electrical Coupling Electrical coupling occurs, when there exists a physical connection between two coils (or inductors). This coupling can be of either aiding type or opposing type. It is based on whether the current enters at the dotted terminal or leaves from the dotted terminal. Coupling of Aiding type Consider the following electric circuit, which is having two inductors that are connected in series. Since the two inductors are connected in series, the same current I flow through both inductors having self-inductances L1 and L2. In this case, the current, I enter at the dotted terminal of each inductor. Hence, the induced voltage in each inductor will be having positive polarity at the dotted terminal due to the current flowing in another coil. Apply KVL around the loop of the above electric circuit or network. $$V – L_1 frac{dI}{dt} – M frac{dI}{dt} – L_2 frac{dI}{dt} – M frac{dI}{dt} = 0$$ $$V = L_1 frac{dI}{dt} + L_2 frac{dI}{dt} + 2M frac{dI}{dt}$$ $$V = (L_1 + L_2 + 2M)frac{dI}{dt}$$ The above equation is in the form of $mathbf{mathit{V = L_{Eq} frac{dI}{dt}}}$ Therefore, the equivalent inductance of series combination of inductors shown in the above figure is $$L_{Eq} = L_1 + L_2 + 2M$$ In this case, the equivalent inductance has been increased by 2M. Hence, the above electrical circuit is an example of electrical coupling which is of aiding type. Coupling of Opposing type Consider the following electric circuit, which is having two inductors that are connected in series. In the above circuit, the current I enters at the dotted terminal of the inductor having an inductance of L1. Hence, it induces a voltage in the other inductor having an inductance of L2. So, positive polarity of the induced voltage is present at the dotted terminal of this inductor. In the above circuit, the current I leaves from the dotted terminal of the inductor having an inductance of L2. Hence, it induces a voltage in the other inductor having an inductance of L1. So, negative polarity of the induced voltage is present at the dotted terminal of this inductor. Apply KVL around the loop of the above electric circuit or network. $$V – L_1 frac{dI}{dt} + M frac{dI}{dt} – L_2 frac{dI}{dt} + M frac{dI}{dt} = 0$$ $$Rightarrow V = L_1 frac{dI}{dt} + L_2 frac{dI}{dt} – 2M frac{dI}{dt}$$ $$Rightarrow V = (L_1 + L_2 – 2M)frac{dI}{dt}$$ The above equation is in the form of $mathbf{mathit{V = L_{Eq} frac{dI}{dt}}}$ Therefore, the equivalent inductance of series combination of inductors shown in the above figure is $$L_{Eq} = L_1 + L_2 – 2M$$ In this case, the equivalent inductance has been decreased by 2M. Hence, the above electrical circuit is an example of electrical coupling which is of opposing type. Magnetic Coupling Magnetic coupling occurs, when there is no physical connection between two coils (or inductors). This coupling can be of either aiding type or opposing type. It is based on whether the current enters at the dotted terminal or leaves from the dotted terminal. Coupling of Aiding type Consider the following electrical equivalent circuit of transformer. It is having two coils and these are called as primary and secondary coils. The currents flowing through primary and secondary coils are i1 and i2 respectively. In this case, these currents enter at the dotted terminal of respective coil. Hence, the induced voltage in each coil will be having positive polarity at the dotted terminal due to the current flowing in another coil. Apply KVL around primary coil. $$v_1 – L_1 frac{d i_1}{dt} – M frac{d i_2}{dt} = 0$$ $Rightarrow v_1 = L_1 frac{d i_1}{dt} + M frac{d i_2}{dt}$Equation 1 Apply KVL around secondary coil. $$v_2 – L_2 frac{d i_2}{dt} – M frac{d i_1}{dt} = 0$$ $Rightarrow v_2 = L_2 frac{d i_2}{dt} + M frac{d i_1}{dt}$Equation 2 In Equation 1 and Equation 2, the self-induced voltage and mutually induced voltage have the same polarity. Hence, the above transformer circuit is an example of magnetic coupling, which is of aiding type. Coupling of Opposing Type Consider the following electrical equivalent circuit of transformer. The currents flowing through primary and secondary coils are i1 and i2 respectively. In this case, the current, i1 enters at the dotted terminal of primary coil. Hence, it induces a voltage in secondary coil. So, positive polarity of the induced voltage is present at the dotted terminal of this secondary coil. In the above circuit, the current, i2 leaves from the dotted terminal of secondary coil. Hence, it induces a voltage in primary coil. So, negative polarity of the induced voltage is present at the dotted terminal of this primary coil. Apply KVL around primary coil. $$v_1 – L_1 frac{d i_1}{dt} + M frac{d i_2}{dt} = 0$$ $Rightarrow v_1 = L_1 frac{d i_1}{dt} – M frac{d i_2}{dt}$Equation 3 Apply KVL around secondary coil.