Select the branches d, e & f of this directed graph as twigs. Consider the point (5A, 5V) on the characteristics. $$V_1 = h_{11} I_1 + h_{21} \lbrace \lgroup \frac{-h_{21}}{h_{22}} \rgroup I_1 + \lgroup \frac{1}{h_{22}} \rgroup I_2 \rbrace$$, $$\Rightarrow V_1 = \lgroup \frac{h_{11}h_{22} - h_{12}h_{21}}{h_{22}} \rgroup I_1 + \lgroup \frac{h_{12}}{h_{22}} \rgroup I_2$$. Find the current flowing through 20 Ω resistor by first finding a Thevenin’s equivalent circuit to the left of terminals A and 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. So, the capacitor acts as an open circuit in steady state. Therefore, we need to find the response in a particular branch ‘n’ times if there are ‘n’ independent sources. Follow these steps in order to find the response in a particular branch using superposition theorem. The coefficients of V2 and -I2 are called as T parameters. The given electrical network is modified into the following form as shown in the following figure. According to the network approach (Johanson and Mattson 1988) internationalization is seen as a process in which relationships are continuously established, developed, maintained and dissolved with the aim of achieving the objectives of the company. Step 3 − Find the short circuit current ISC by shorting the two opened terminals of the above circuit. In this chapter, let us discuss about the passive filters. Because, every Tree will be having one Fundamental loop matrix. This coupling can be of either aiding type or opposing type. $$P_S = 2\lgroup \frac{V_{Th}}{2 R_{Th}} \rgroup ^2 R_{Th}$$, $$\Rightarrow P_S = 2\lgroup \frac{{V_{Th}}^2}{4 {R_{Th}}^2} \rgroup R_{Th}$$, $$\Rightarrow P_S = \frac{{V_{Th}}^2}{2 R_{Th}}$$. Repeat the above step for all the nodes of the given directed graph. Substitute the values of VTh and RTh in the following formula of Norton’s current. We can consider the arbitrary direction of current flow in each branch. Mathematically, it can be represented as. $W_c = \frac{C{v_c}^2}{2} = $ Maximum & constant. Note − KCL is independent of the nature of network elements that are connected to a node. W is the electrical energy and it is measured in terms of Joule. In the above figure, V1 is the voltage from node 1 with respect to ground and V2 is the voltage from node 2 with respect to ground. The currents flowing through primary and secondary coils are i1 and i2 respectively. We know that there are two practical sources, namely, voltage source and current source. Therefore, the response of the electric circuit during the transient state is known as transient response. So, just by doing the inverse of Z parameters matrix, we will get Y parameters matrix. Here, $$Z_{11} = \frac{h_{11}h_{22} - h_{12}h_{21}}{h_{22}}$$, Step 5 − Therefore, the Z parameters matrix is, $$\begin{bmatrix}Z_{11} & Z_{12} \\Z_{21} & Z_{22} \end{bmatrix} = \begin{bmatrix}\frac{h_{11}h_{22} - h_{12}h_{21}}{h_{22}} & \frac{h_{12}}{h_{22}} \\\frac{-h_{21}}{h_{22}} & \frac{1}{h_{22}} \end{bmatrix}$$. The first and second terms represent the transient response and steady state response of the current respectively. 1 an interconnected group or system. $\frac{di}{dt} + \lgroup \frac{R}{L} \rgroup i = \frac{V}{L}$Equation 1, The above equation is a first order differential equation and it is in the form of. 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. $$P_{L, Max} = \frac{{V_{Th}}^2}{4 R_{Th}}$$. Here, we have to represent Z parameters in terms of Y parameters. We know that the current is nothing but the time rate of flow of charge. Mathematically, it can be written as. So, the number of branches and/or nodes of a subgraph will be less than that of the original graph. Im is the mth branch current leaving the node. This means, there won’t be any transient part in the response during steady state. This connected subgraph contains all the four nodes of the given graph and there is no loop. Sometimes, they may absorb the power like passive elements. This is the Method 3 for finding a Thevenin’s equivalent circuit. The above circuit has only one mesh. In these two quadrants, the ratios of voltage (V) and current (I) produce positive impedance values. In the absence of resistor, coil becomes inductor. Step 1 − Find the response in a particular branch by considering one independent source and eliminating the remaining independent sources present in the network. The modified circuit diagram is shown in the following figure. Similarly, in the third quadrant, the values of both voltage (V) and current (I) have negative values. That means, a branch current leaves from one node and enters at another single node only. There are four variables V1, V2, I1 and I2 in a two port network as shown in the figure. $$\Rightarrow \frac{I_2}{V_2} = \frac{C}{D}$$. Among which, one port is used as an input port and the other port is used as an output port. We will get the equivalent resistance across terminals A & B by minimizing the above network into a single resistor between those two terminals. The modified circuit diagram is shown in the following figure. Step 1 − We know that, the following set of two equations of two port network regarding Z parameters. Hence, the independent ideal current sources do not exist practically, because there will be some internal resistance. It consists of a set of nodes connected by branches. So, negative polarity of the induced voltage is present at the dotted terminal of this primary coil. Mathematically, it can be represented as. 4 (Radio, television) a group of broadcasting stations that all transmit the same programme simultaneously. This can be obtained by doing the following simplification. The V-I characteristics of an independent ideal current source is a constant line, which is always equal to the source current (IS) irrespective of the voltage value (V). Passive Elements can’t deliver power (energy) to other elements, however they can absorb power. This concept is illustrated in following figures. Here. $$\Rightarrow I_1 + I_2 + I_3 = I_4 + I_5$$. Similarly, we can calculate the other two Y parameters, Y12 and Y22 by doing short circuit of port1. The circuit diagram, when the switch is in closed position is shown in the following figure. Now, we can find the response in an element that lies to the right side of Norton’s equivalent circuit. If only two circuit elements are connected to a node, then it is said to be simple node. 0000002961 00000 n Hence, the induced voltage in each coil will be having positive polarity at the dotted terminal due to the current flowing in another coil. $$h_{11} = \frac{V_1}{I_1},\: when\: V_2 = 0$$, $$h_{12} = \frac{V_1}{V_2},\: when\: I_1 = 0$$, $$h_{21} = \frac{I_2}{I_1},\: when\: V_2 = 0$$, $$h_{22} = \frac{I_2}{V_2},\: when\: I_1 = 0$$. Step 1 − Find a Thevenin’s equivalent circuit between the desired two terminals. So, each pair of equations will give a set of four parameters. We know that there is no initial current in the circuit. The amount of power dissipated across the load resistor is. Now, let us identify the nature of network elements from the V-I characteristics given in the following examples. In this case, the current, i1 enters at the dotted terminal of primary coil. 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. Consider the following connected subgraph of the graph, which is shown in the Example of the beginning of this chapter. We will get the following set of two equations by considering the variables V1 & I2 as dependent and I1 & V2 as independent. Hence, the above electrical circuit is an example of electrical coupling which is of aiding type. Linear Elements are the elements that show a linear relationship between voltage and current. The following equations represent the equivalent resistance between two terminals of star network, when the third terminal is kept open. The transfer function of the above network is, $$H(s) = \frac{V_o(s)}{V_i(s)} = \frac{\frac{1}{sC}}{R + \frac{1}{sC}}$$. Hence, the order of incidence matrix will be n × b. Similarly, Maximum power transfer theorem states that the AC voltage source will deliver maximum power to the variable complex load only when the load impedance is equal to the complex conjugate of source impedance. In the above circuit, the switch was kept open up to t = 0 and it was closed at t = 0. The symbol of resistor along with current, I and voltage, V are shown in the following figure. Fundamental cut set matrix is represented with letter C. This matrix gives the relation between branch voltages and twig voltages. 0000002803 00000 n network, can have a significant impact on the outcomes of similar legislative outcomes in other states. !n network contexts the complexity related to ~trz_aegic pt,.sitioning increases because a position is me sum of relationships willa their technical, social and cognitive dimensions. Step 3 − In this case, we will get two nodal equations, since there are two principal nodes, 1 and 2, other than Ground. Let us find the equivalent resistance across the terminals A & B of the following electrical network. Take a look at the following Tree of directed graph, which is considered for incidence matrix. Now, let us derive the values of parameters and electrical quantities at resonance of series RLC circuit one by one. Therefore, the magnitude of transfer function of Band stop filter will vary from 1 to 0 & 0 to 1 as ω varies from 0 to ∞. In graphs, a node is a common point of two or more branches. The Thevenin’s resistance across terminals A & B will be, $$R_{Th} = \lgroup \frac{5 \times 10}{5 + 10} \rgroup + 10 = \frac{10}{3} + 10 = \frac{40}{3} \Omega$$. It is useful for analyzing complex electric circuits by converting them into network graphs. Hence, they offer different impedances in both directions. The parameters, A and D do not have any units, since those are dimension less. Here, 5 Ω & 10 Ω resistors are connected in parallel and the entire combination is in series with 10 Ω resistor. Therefore, there is no initial current flows through the inductor. $$I_S = \frac{V_S}{R_1} + \frac{V_S}{R_2} + \frac{V_S}{R_3}$$, $$\Rightarrow I_S = V_S \lgroup \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} \rgroup$$, $$\Rightarrow V_S = I_S\left [ \frac{1}{\lgroup \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} \rgroup} \right ]$$. $$V_1 = A V_2 - B \lgroup \frac{C}{D} \rgroup V_2$$, $$\Rightarrow V_1 = \lgroup \frac{AD - BC}{D} \rgroup V_2$$, $$\Rightarrow \frac{V_1}{V_2} = \frac{AD - BC}{D}$$, $$\Rightarrow h_{12} = \frac{AD - BC}{D}$$, Step 6 − Therefore, the h-parameters matrix is, $$\begin{bmatrix}h_{11} & h_{12} \\h_{21} & h_{22} \end{bmatrix} = \begin{bmatrix}\frac{B}{D} & \frac{AD - BC}{D} \\-\frac{1}{D} & \frac{C}{D} \end{bmatrix}$$. VN is the voltage across Nth passive element. The above equation is in the form of VS = ISREq where, $$R_{Eq} = \frac{1}{\lgroup \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} \rgroup}$$, $$\frac{1}{R_{Eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}$$. There are two basic methods that are used for solving any electrical network: Nodal analysis and Mesh analysis. That’s why the left-hand side contains only one term. $$H(s) = \frac{V_o(s)}{V_i(s)} = \frac{R}{R + \frac{1}{sC} + sL}$$, $$\Rightarrow H(s) = \frac{s CR}{s^2 LC + sCR + 1}$$, $$H(j \omega) = \frac{j \omega CR}{1 - \omega^2 LC + j \omega CR}$$, $$|H(j \omega)| = \frac{\omega CR}{\sqrt{(1 - \omega^2 LC)^2 + (\omega CR)^2}}$$. $$V - L_1 \frac{dI}{dt} + M \frac{dI}{dt} - L_2 \frac{dI}{dt} + M \frac{dI}{dt} = 0$$, $$\Rightarrow V = L_1 \frac{dI}{dt} + L_2 \frac{dI}{dt} - 2M \frac{dI}{dt}$$, $$\Rightarrow V = (L_1 + L_2 - 2M)\frac{dI}{dt}$$. If positive voltage is applied across the capacitor, then it stores positive charge. In general, this frequency band lies in between low frequency range and high frequency range. $$h_{11} = \frac{V_1}{I_1}, \: when \: V_2 = 0$$, $$h_{12} = \frac{V_1}{V_2}, \: when \: I_1 = 0$$, $$h_{21} = \frac{I_2}{I_1}, \: when \: V_2 = 0$$, $$h_{22} = \frac{I_2}{V_2}, \: when \: I_1 = 0$$. Step 3 − Find the Norton’s resistance RN across the open terminals of the circuit considered in Step1 by eliminating the independent sources present in it. Following are the three matrices that are used in Graph theory. $$\Rightarrow I_L = -j \lgroup \frac{R}{X_L} \rgroup I$$. 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. Now, let us discuss two methods one by one. This is due to the voltage drop across the internal resistance (RS) of an independent practical voltage source. So, the fundamental cut set matrix of the above considered Tree will be, $$C = \begin{bmatrix}1 & -1 & 0 & 1 & 0 & 0\\0 & -1 & 1 & 0 & 1 & 0\\1 & 0 & -1 & 0 & 0 & 1 \end{bmatrix}$$. At resonance, the impedance Z of series RLC circuit is equal to the value of resistance R, i.e., Z = R. Substitute $X_L - X_C = 0$ in Equation 1. Consider the following electrical equivalent circuit of transformer. H��W]o�8}0����"fD��u�t&�i'h�C��Lۚ8�W����~�%%$�g�3(B���8��s//n��?_|��y���/���.��V��L���0g�����^�M�����V�n��`Uyv�+~���.gg�Y��-�A�fs��͞�Hx ����Ω�T Kㄋ8J���l�I7϶z`�ͦ�y�b�ewM�7z �s�/��6 �*e®�mB��^lV��g�� �"��.�L���lt���ei��VW�F���zvv��mr�CF��'�rsif۴�s�d[��l�$KÔ�_��3��. The coefficients of independent variables, V1 and V2 are called as Y parameters. There are three methods for finding a Norton’s equivalent 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. Here,$. Follow these steps in order to find the Thevenin’s equivalent circuit, when the sources of both independent type and dependent type are present. So, we have to eliminate the remaining independent sources from the circuit. 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. The resistor present in the Thevenin’s equivalent circuit is called as Thevenin’s equivalent resistor or simply Thevenin’s resistor, RTh. Next January, I will be finishing college. But, the difference of two mesh currents, I1 and I2, flows through 10 Ω resistor, since it is the common branch of two meshes. In this method, we will consider only one independent source at a time. Step 1 − We know that, the following matrix equation of two port network regarding Z parameters as, $\begin{bmatrix}V_1 \\V_2 \end{bmatrix} = \begin{bmatrix}Z_{11} & Z_{12} \\Z_{21} & Z_{22} \end{bmatrix} \begin{bmatrix}I_1 \\I_2 \end{bmatrix}$Equation 3, Step 2 − We know that, the following matrix equation of two port network regarding Y parameters as, $$\begin{bmatrix}I_1 \\I_2 \end{bmatrix} = \begin{bmatrix}Y_{11} & Y_{12} \\Y_{21} & Y_{22} \end{bmatrix} \begin{bmatrix}V_1 \\V_2 \end{bmatrix}$$, $\begin{bmatrix}V_1 \\V_2 \end{bmatrix} = \begin{bmatrix}Y_{11} & Y_{12} \\Y_{21} & Y_{22} \end{bmatrix}^{-1} \begin{bmatrix}I_1 \\I_2 \end{bmatrix}$Equation 4, Step 4 − By equating Equation 3 and Equation 4, we will get, $$\begin{bmatrix}Z_{11} & Z_{12} \\Z_{21} & Z_{22} \end{bmatrix} = \begin{bmatrix}Y_{11} & Y_{12} \\Y_{21} & Y_{22} \end{bmatrix}^{-1}$$, $$\Rightarrow \begin{bmatrix}Z_{11} & Z_{12} \\Z_{21} & Z_{22} \end{bmatrix} = \frac{\begin{bmatrix}Y_{22} & - Y_{12} \\- Y_{21} & Y_{11} \end{bmatrix}}{\Delta Y}$$, $$\Delta Y = Y_{11} Y_{22} - Y_{12} Y_{21}$$. At ω = ∞, the magnitude of transfer function is equal to 0. Hence, the independent ideal voltage sources do not exist practically, because there will be some internal resistance. The number of Fundamental loop matrices of a directed graph will be equal to the number of Trees of that directed graph. VS is the input voltage, which is present across the entire combination of series passive elements. Now, we can find the response in an element that lies to the right side of Thevenin’s equivalent circuit. So, positive polarity of the induced voltage is present at the dotted terminal of this inductor. Move that external device to the receiver PC 3. Examples: Voltage sources and current sources. In Equation 3 and Equation 4, self-induced voltage and mutually induced voltage are having opposite polarity. Follow these steps in order to find the fundamental cut set matrix of given directed graph. So, the ratios of voltage (V) and current (I) gives positive impedance values. Mathematically, it can be represented as. According to this approach , internationalization of the firm can be achieved through creating relationships in foreign country network s that are new to it ; the development of relationships and increasing resource commitments in those network s in which the company already has a position (penetration) or connecting the existing network s in different countries. The elements of fundamental loop matrix will be having one of these three values, +1, -1 and 0. The concept of Bilateral elements is illustrated in the following figures. Following are the two possible source transformations −, The transformation of practical voltage source into a practical current source is shown in the following figure. In the above circuit, the switch was kept open up to t = 0 and it was closed at t = 0. Step 1 − Verifying the network element as linear or non-linear. Band pass filter as the name suggests, it allows (passes) only one band of frequencies. Substitute $V_1 = I_S R_1, \: V_2 = I_S R_2$ and $V_3 = I_S R_3$ in the above equation. This example is an illustration of the “collective impact” approach, which recognizes that when many nonprofits work together, sharing resources and learning … The mesh currents I1 and I2 are considered in clockwise direction. H&M has outsourced the production and processing of their goods to different countries majorly Asian and South East Asian countries.The figure above explains the fundamentals of a network organizational structure. By observing the above incidence matrix, we can conclude that the summation of column elements of incidence matrix is equal to zero. We can calculate the steady state response of an electric circuit, when it is excited by a sinusoidal voltage source using Laplace Transform approach. So, in this case T parameters are the desired parameters and Z parameters are the given parameters. There are six branches in the above graph and those are labelled with a, b, c, d, e & f respectively. In the above figure, the branch currents I1, I2 and I3 are entering at node P. So, consider negative signs for these three currents. In this chapter, let us discuss about series resonance. We will get the following set of two equations by considering the variables V2 & I2 as dependent and V1 & I1 as independent. In the previous chapter, we discussed about six types of two-port network parameters. So, the number of f-loops will be equal to the number of links. At ω = 0, the magnitude of transfer function is equal to 1. Z parameters are called as impedance parameters because these are simply the ratios of voltages and currents. Step 1 − We know that, the following set of two equations, which represents a two port network in terms of T parameters. The transient response will be zero for large values of ‘t’. Step 4 − The Thevenin’s equivalent circuit is placed to the left of terminals A & B in the given circuit. Many intentional networks are hybrids of these two. The infor­mation, usually represented by a network, includes the sequences, interdependencies, interre­lationships, and criticality of various activities of the project. Step 1 − We know that, the following h-parameters of a two port network. Here, $$C = \frac {Y_{12} Y_{21} - Y_{11} Y_{22}}{Y_{21}}$$, $$\begin{bmatrix}A & B \\C & D \end{bmatrix} = \begin{bmatrix}\frac{-Y_{22}}{Y_{21}} & \frac{-1}{Y_{21}} \\\frac{Y_{12}Y_{21} - Y_{11}Y_{22}}{Y_{21}} & \frac{-Y_{11}}{Y_{21}} \end{bmatrix}$$. Here, we have to represent T parameters in terms of Y parameters. The value of elements will be 0 for the remaining links and twigs, which are not part of the selected f-loop. $$V_{30 \Omega} = \lgroup \frac{14}{5} \rgroup 30$$. 0000001266 00000 n Here, the passive elements such as resistor, inductor and capacitor are connected in parallel. $$V_2 = \lgroup \frac {V_S}{R_1 + R_2} \rgroup R_2$$, $$\Rightarrow V_2 = V_S \lgroup \frac {R_2}{R_1 + R_2} \rgroup$$. We will get the following set of two equations by considering the variables I1 & V2 as dependent and V1 & I2 as independent. Electron current is obtained due to the movement of free electrons, whereas, Conventional current is obtained due to the movement of free positive charges. It consists of a current source (IS) in parallel with a resistor (RS). But, the voltage gets divided across each element. Step 2 − We know that the following set of two equations of two port network regarding Y parameters. Independent ideal voltage source and its V-I characteristics are shown in the following figure. At ω = ∞, the magnitude of transfer function is equal to 1. In the above circuit, the current I leaves from the dotted terminal of the inductor having an inductance of L2. This concept is illustrated in the following figure. Here. These three f-loops are shown in the following figure. In this chapter, let us discuss about the following two division principles of electrical quantities. Dependent sources can be further divided into the following two categories −, A dependent voltage source produces a voltage across its two terminals. If three or more circuit elements are connected to a node, then it is said to be Principal Node. 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. $$R_{AB} = \frac{(R_1 + R_3)R_2}{R_1 + R_2 + R_3}$$, $$R_{BC} = \frac{(R_1 + R_2)R_3}{R_1 + R_2 + R_3}$$, $$R_{CA} = \frac{(R_2 + R_3)R_1}{R_1 + R_2 + R_3}$$. Following are the two types of subgraphs. In this introductory chapter, let us first discuss the basic terminology of electric circuits and the types of network elements. The modified circuit diagram is shown in the following figure. Substitute $I_1 = \frac{V_S}{R_1}$ and $I_2 = \frac{V_S}{R_2}$ in the above equation. We will get the Norton’s equivalent circuit. In general, a capacitor has two conducting plates, separated by a dielectric medium. $$\Rightarrow V - IR - I(j X_L) - I(-j X_C) = 0$$, $$\Rightarrow V = IR + I(j X_L) + I(-j X_C)$$, $\Rightarrow V = I[R + j(X_L - X_C)]$Equation 1. Its unit is Watt. M is the number of branches that are connected to a node. Input voltage is applied across this entire combination and the output is considered as the voltage across resistor. The resistor present in the Norton’s equivalent circuit is called as Norton’s equivalent resistor or simply Norton’s resistor RN. Mathematically, it can be written as. Psychiatric symptoms have been argued as reciprocal rather than common cause e ects. What is Network Approach 1. The resultant circuit diagram is shown in the following figure. 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. The Co-Tree corresponding to the above Tree is shown in the following figure. Because, every Tree will be having one Fundamental cut set matrix. $$\Rightarrow I_2 - h_{21} I_1 = h_{22} V_2$$, $$\Rightarrow V_2 = \frac{I_2 - h_{21} I_1}{h_{22}}$$, $$\Rightarrow V_2 = \lgroup \frac{-h_{21}}{h_{22}} \rgroup I_1 + \lgroup \frac{1}{h_{22}} \rgroup I_2$$, The above equation is in the form of $V_2 = Z_{21} I_1 + Z_{22} I_2. In the above equation, the left-hand side represents the sum of entering currents, whereas the right-hand side represents the sum of leaving currents. Substitute, $X_L = X_C$ in the above equation. The simplified circuit diagram is shown in the following figure. Following are the types of filters. Examples: Resistors, Inductors and capacitors. The given V-I characteristics of a network element lies in the first and third quadrants. The above statement of KCL can also be expressed as "the algebraic sum of currents entering a node is equal to the algebraic sum of currents leaving a node". Substitute, the values of x, y, P & Q in Equation 3. So, the capacitor stores the energy in the form of electric field. That means, this filter allows (passes) both low and high frequency components. Analysis: The requirements of the applications and the organizatio… Norton’s theorem is similar to Thevenin’s theorem. Hence, substitute, t = 0 and 𝑖 = 0 in Equation 4 in order to find the value of the constant k. $$0 = \frac{V}{R} + ke^{-\lgroup \frac{R}{L} \rgroup(0)}$$. 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. So, the magnitude of current flowing through inductor at resonance will be, Where, Q is the Quality factor and its value is equal to $\frac{R}{X_L}$, $$\Rightarrow I_C = j \lgroup \frac{R}{X_C} \rgroup I$$, Therefore, the current flowing through capacitor at resonance is $I_C = jQI$, So, the magnitude of current flowing through capacitor at resonance will be, Where, Q is the Quality factor and its value is equal to $\frac{R}{X_C}$. So, the connecting of branches to a node is called as incidence. 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. In the previous chapter, we discussed the importance of series resonance. 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