EContent 11 2024 Electrical Engineering DC Circuits PDF

Reference Books Department of Electrical Engineering Unit No:- 1 Name:- Fundamental of DC Circuits Basics of Electrical & Electronics Engineering- 01EE1101  Disclaimer It is hereby declared that the production of the said content is meant for...

Reference Books Department of Electrical Engineering Unit No:- 1 Name:- Fundamental of DC Circuits Basics of Electrical & Electronics Engineering- 01EE1101  Disclaimer It is hereby declared that the production of the said content is meant for non-commercial, scholastic and research purposes only. We admit that some of the content or the images provided in this channel's videos may be obtained through the routine Google image searches and few of them may be under copyright protection. Such usage is completely inadvertent. It is quite possible that we overlooked to give full scholarly credit to the Copyright Owners. We believe that the non- commercial, only-for-educational use of the material may allow the video in question fall under fair use of such content. However we honour the copyright holder's rights and the video shall be deleted from our channel in case of any such claim received by us or reported to us.  Definition Current, Voltage, E.M.F., Power Energy, Resistance, Open circuit and Short circuit Outline Kirchoff’s Laws Nodal Analysis & Mesh Analysis of Electrical Networks  1 Charge (Q)  The unit of charge is the coulomb (C),  where one coulomb is one ampere second (1 coulomb = 6.24 ×1018 electrons).  The coulomb is defined as the quantity of electricity which flows past a Introduction given point in an electric circuit when a current of one ampere for one second.  Charge, in coulombs Q=It  Charge on ions is measured with excess and deficit of electron from atom,  So, Charge on ions Q=ne, n-number of electron, e- charge of one electron Introduction  Positively charged: electrons are removed making the object electron deficient.  Negatively charged: electrons are added giving the object an excess of electrons.  Current (I)  If an electric pressure or voltage is applied across any material there is a tendency for electrons to move in a particular direction.  This movement of free electrons, known as drift, constitutes an electric current flow. Thus current is the rate of movement of charge. Introduction  Unit: ampere (A)  If the drift of electrons in a conductor takes place at the rate of one coulomb per second the resulting current is said to be a current of one ampere.  Electric potential / Electromotive force (e.m.f)  The voltage developed by any source of electrical energy such as a battery or dynamo.  The unit of electric potential is the volt (V), where one volt is one joule per coulomb. Introduction  Ability of charged body to do work is called Electric Potential.  Electrical Potential = Work done/Charge, V=W/Q, V=dw/dq  A change in electric potential between two points in an electric circuit is called a potential difference.  Electrical power and energy  When a direct current of I amperes is flowing in an electric circuit and the voltage across the circuit is V volts, then, power, in watts P=VI  Electrical energy=Power × time =VIt joules  Power is rate of change of energy. If, certain amount of energy is used over a certain span of time, then Power=energy/time, P=W/t, p=dw/dt Introduction  the unit used for energy is the kilowatt hour (kWh) where,  1kWh = 1000watt hour = 1000 ×3600watt seconds or joules = 3600000J  Resistance (R)  The flow of electric current is subject to friction.  This friction, or opposition, is called resistance, R, and is the property of a conductor that limits current. Electrical circuit  The unit of resistance is the ohm (Ω); 1 ohm is defined as the resistance which will have a current of 1 ampere flowing through it when 1 volt is connected across it, elements  Resistance (R) = potential difference (V)/ Current (I)  R α l, R α 1/A, R α ρ, R= ρl/A  Specific Resistance / Resistivity (ρ)  Resistance offered by material having unite dimension is known as Resistivity of Specific Resistance of material. Electrical circuit elements  Unit : Ohm-m  Conductivity (σ)  Ohm’s law  Ohm’s law states that the current I flowing in a circuit is directly proportional to the applied voltage V and inversely proportional to the resistance R, provided the temperature remains constant. Electrical circuit elements  Ohm’s law Electrical circuit elements  Ohm’s law  Limitation of Ohm’s Law  This law cannot be applied to unilateral networks. A unilateral network has unilateral elements like diode, transistors, etc., which do not have same voltage current relation for both directions of current. Electrical  Ohm’s law is also not applicable for non – linear elements. circuit  Non-linear elements are those which do not give current through it, is elements not exactly proportional to the voltage applied, that means the resistance value of those elements changes for different values of voltage and current. Examples of non – linear elements are thyristor, electric arc, etc.  Factors Affecting Resistance  The resistance of any material mainly depend upon four factors:  Material  Length  Cross-sectional area  Temperature of the material Electrical  the higher the resistivity, the greater the resistance of a conductor circuit  the longer the conductor, the greater the resistance elements  the greater the area of a conductor, the less the resistance  Determine the relationship in series and parallel connected resistors. Resistors in Series: Resistors are said to be connected in “Series”, when they are daisy chained together in a single line. Then the amount of current that flows through a set of resistors in series will be the same at all points in a series resistor network. Series & Parallel The resistors R1, R2, R3, R4..., Rn are all connected together in series between points A and B with a common current. The total voltage in a series circuit which is the sum of all the individual voltages added together. VT = V1 + V2 + V3 + V4 + …….. + Vn  Determine the relationship in series and parallel connected resistors. Resistors in Series: As the resistors are connected together in series the same current passes through each resistor in the chain and the total resistance, RT of the circuit must be equal to the sum of all the individual resistors added together. Series & Parallel RT = R1 + R2 + R3 + R4 + …….. + Rn  Determine the relationship in series and parallel connected resistors. Resistors in Parallel: Resistors are said to be connected together in parallel when both of their terminals are respectively connected to each terminal of the other resistor or resistors. Unlike the previous series resistor circuit, in a parallel resistor network the circuit current can take more than one path as there are multiple paths for the current. Then parallel circuits are classed as current dividers. Series & Parallel The resistors R1, R2, R3, R4..., Rn are all connected together in parallel between points A and B with a common potential difference. The total current, IT entering a parallel resistive circuit is the sum of all the individual currents flowing in all the parallel branches. IT = I1 + I2 + I3 + I4 + …….. + In  Determine the relationship in series and parallel connected resistors. Resistors in Parallel: Equivalent resistor of parallel connected resistors is mathematically expressed as, Series & Parallel  Determine the relationship in series and parallel connected resistors. Example:- A battery with a terminal voltage of is connected to a circuit consisting of four and one resistors all in series. Assume the battery has negligible internal resistance. (a) Calculate the equivalent resistance of the circuit. (b) Calculate the current through each resistor. (c) Calculate the potential drop across each resistor. Series & Parallel (a) The equivalent resistance is the algebraic sum of the resistances: Req = R1 + R2 + R3 + R4 + R5 = 20 + 20 + 20 + 20 + 10 = 90 Ω  Determine the relationship in series and parallel connected resistors. Example:- A battery with a terminal voltage of is connected to a circuit consisting of four and one resistors all in series. Assume the battery has negligible internal resistance. (a) Calculate the equivalent resistance of the circuit. (b) Calculate the current through each resistor. (c) Calculate the potential drop across each resistor. Series & Parallel (b) The current through the circuit is the same for each resistor in a series circuit and is equal to the applied voltage divided by the equivalent resistance: V 9 I= = = 0.1A R eq 90  Determine the relationship in series and parallel connected resistors. Example:- A battery with a terminal voltage of is connected to a circuit consisting of four and one resistors all in series. Assume the battery has negligible internal resistance. (a) Calculate the equivalent resistance of the circuit. (b) Calculate the current through each resistor. (c) Calculate the potential drop across each resistor. Series & Parallel (c) The potential drop across each resistor can be found using Ohm’s law: V1 = V2 = V3 = V4 = (0.1) × 20 = 2 V V5 = (0.1) × 10 = 1 V V1 + V 2 + V3 + V4 + V5 = 9 V  Determine the relationship in series and parallel connected resistors. Example:- Calculate the total current ( IT) taken from the 12V supply. Series & Parallel The two resistors, R2 and R3 are actually both connected together in a “SERIES” combination so we can add them together. R2 + R3 = 8Ω + 4Ω = 12Ω  Determine the relationship in series and parallel connected resistors. Series & Parallel  Determine the relationship in series and parallel connected resistors. The resultant resistive circuit now looks something like this: Series & Parallel R(ab) = Rcomb + R1 = 6Ω + 6Ω = 12Ω  Determine the relationship in series and parallel connected resistors. Example:- Find the equivalent resistance, REQ for the following resistor combination circuit. Series & Parallel RA is in series with R7 therefore the total resistance will be RA + R7 = 4 + 8 = 12Ω.  Determine the relationship in series and parallel connected resistors. Series & Parallel This resistive value of 12Ω is now in parallel with R6 and can be calculated as RB. RB is in series with R5 therefore the total resistance will be RB + R5 = 4 + 4 = 8Ω as shown.  Determine the relationship in series and parallel connected resistors. Series & This resistive value of 8Ω is now in parallel with R4 and can be calculated as RC as shown. Parallel RC is in series with R3 therefore the total resistance will be RC + R3 = 8Ω as shown.  Determine the relationship in series and parallel connected resistors. RC is in series with R3 therefore the total resistance will be RC + R3 = 8Ω as shown. Series & Parallel This resistive value of 8Ω is now in parallel with R2 from which we can calculated RD as: RD is in series with R1 therefore the total resistance will be RD + R1 = 4 + 6 = 10Ω as shown.  Determine the relationship in series and parallel connected resistors. RD is in series with R1 therefore the total resistance will be RD + R1 = 4 + 6 = 10Ω as shown. Series & Parallel  Voltage Sources Voltage Sources: In general, there will be a current flowing through a voltage source. That current can be positive, negative, or zero, depending on how the source is connected into the circuit. Ideal Independent Voltage Source: The ideal independent voltage Voltage and source maintains a fixed voltage across its terminals regardless of the current through it. current sources Ideal Dependent Voltage Source: The ideal dependent (or controlled) voltage source maintains a voltage across its terminals that depends on either a voltage or current elsewhere in the circuit.  Current Sources Current Sources: In general, there will be a voltage across a current source. That voltage can be positive, negative, or 0 depending on how it is connected into the circuit. Ideal Independent Current Source: The ideal independent current Voltage and source maintains a fixed current through its terminals regardless of the voltage across it. current sources Ideal Dependent Current Source: The ideal dependent (or controlled) current source maintains a current through its terminals that depends on either a voltage or current elsewhere in the circuit.  Comparison between series and parallel circuits Comparision  Open Circuit and Short Circuit 2 There are five differences between open and short circuits: 1. Current passing through an open circuit is zero, while current through the short circuit is infinite. 2. An open circuit posses infinite resistance, while a short circuit posses zero resistance. 3. The voltage through the short circuit is zero, while voltage through the Analysis of short circuit in maximum. Circuits 4. An ohmmeter connected to short circuit displays ‘0’ ohms while an ohmmeter connected to open circuit displays ‘infinity’ or ‘0L’. 5. Practically, short circuit happens when a low resistance wire connects across the circuit and an open circuit occurs when a circuit breaks from some point.  Open Circuit and Short Circuit An open circuit is the one having a disconnection between components. The figure below displays an open: Analysis of Circuits A short circuit is the one where components are connected with a very small or zero resistance wire. The figure below displays an ideal short.  Open Circuit and Short Circuit Resistance An open circuit posses infinite resistance, while a short circuit posses zero resistance. Ω for open → Infinite Ω for short → Zero Analysis of An ohmmeter connected across a short displays ‘0’ ohms or very small Circuits values of ohms. An ohmmeter across open will display 1 or 0L. (Most multimeter manufacturers display 0L for open).  Open Circuit and Short Circuit Current Current always requires a path to flow. If it is open, no electrons will flow from one terminal to other and the resultant current will be zero. Similarly, resistance is the other current controlling factor. As per Ohm’s law, the higher resistance means lower current. In case of open, the infinite resistance means zero current, and zero resistance means infinite current. Analysis of Circuits From Ohm’s law, I = V/R. Current for open → I = V/R = V/Infinite = 0 Current for short → I = V/R = V/0 = Infinite Voltage The voltage across the short circuit is zero. However, the voltage across open terminals equals the supply voltage.  Open Circuit and Short Circuit Open Circuit: When there is nothing attached to the terminals; the circuit is open there. Open circuit means RL = ∞. The voltage across the terminals in this case is the open circuit voltage. Short Circuit: In this condition, there is a wire connected between the terminals; in other words, RL = 0. The current flowing through the wire is the short circuit current. Analysis of Circuits  Open Circuit and Short Circuit Analysis of Circuits Bottom line: if there is a wire across a resistor, the current in it is 0 and we can replace it with an open circuit (i.e., remove it). If the resistor is hanging with nothing connected to it, the voltage across it is 0 and we can replace it with a short circuit.  Verify Kirchhoff’s Current Law and Kirchoff’s Voltage Law in Electrical Networks 3 Kirchhoff’s Current Law: Kirchhoff’s first law or current law is based on the law of conservation of charge, which requires that the algebraic sum of charges within a system cannot change. Kirchoff’s “Kirchhoff’s Current Law States that the algebraic sum of currents entering a node is current and zero.” voltage laws Where N is the number of branches connected to the node and in is the nth current entering (or leaving) the node. By this law, currents entering a node may be regarded as positive, while currents leaving the node may be taken as negative or vice versa.  Verify Kirchhoff’s Current Law and Kirchoff’s Voltage Law in Electrical Networks Kirchhoff’s Current Law: Consider a node in Figure, Applying KCL gives, Kirchoff’s i1 + (- i2) + i3 + i4 + (-i5) = 0 current and voltage laws Since currents i1, i3 and i4 are entering the node, while current i2 and i5 are leaving it. By rearranging the terms, we get i1 + i3 + i4 = i2 + i5 The sum of the current entering a node is equal to the sum of the currents leaving the node.  Network: Inter connection of two or more elements is called an electric network. Circuit: If a network contains at least one closed path, it is called an electric circuit. Node: the point at which two or more elements are connected together is generally called as node. EX: A,B,C and E Junction: It is a point where three or more elements are connected together. EX: B and E Branch: A section or portion of a network or circuit which lies between two junction points is called as branch. EX: BE, BAE and BCE Loop: Any closed path in a network is called a loop. EX: ABEA, BCEB and ABCEA Mesh: It is the most elementary form of a loop and cannot be further divided into other loos. EX: ABEA and BCEB  Verify Kirchhoff’s Current Law and Kirchoff’s Voltage Law in Electrical Networks Kirchhoff’s Voltage Law: Kirchhoff’s second law is based on the principle of conservation of energy. “Kirchhoff’s Voltage Law (KVL) states that the algebraic sum of all voltages around a Kirchoff’s closed path (or loop) is zero.” current and voltage laws Where M is the number of voltages in the loop (or the number of branches in the loop) and vm is the mth voltage.  Verify Kirchhoff’s Current Law and Kirchoff’s Voltage Law in Electrical Networks Kirchhoff’s Voltage Law: To illustrate KVL, consider the circuit shown in Figure. The sign of each voltage is the polarity of the terminal encountered first as we travel around the loop. We can start with any branch and go around that loop either clockwise or counter-clockwise. Kirchoff’s current and voltage laws Suppose we start with the voltage source and go clockwise around the loop as shown, then voltages would be -v1,+v2,+v3,-v4 and +v5, in that order. For example, as we reach branch 3, the positive terminal is met first; hence we have +v3. For branch 4, we reach the negative terminal first, hence -v4.  Verify Kirchhoff’s Current Law and Kirchoff’s Voltage Law in Electrical Networks Kirchhoff’s Voltage Law: Kirchoff’s current and -v1 + v2 + v3 – v4 + v5 = 0 voltage laws v2 + v3 + v5 = v1 + v4 Sum of Voltage drops = Sum of Voltage Rises  Example based on KCL and KVL Circuit Analysis by Kirchhoff’s Laws Resistors of R1= 10Ω, R2 = 4Ω and R3 = 8Ω are connected up to two batteries (of negligible resistance) as shown. Find the current through each resistor. Examples-1  Circuit Analysis by Kirchhoff’s Laws Assume currents to flow in directions indicated by arrows. Apply KCL on Junctions C and A. Current in mesh ABC = i1 Current in Branch CA = i2 Examples Current in Mesh CDA = i1 – i2 Now, Apply KVL on Mesh ABC, 20V are acting in clockwise direction. Equating the sum of IR products, we get; 10 i1 + 4 i2 = 20 ……………….(1)  Circuit Analysis by Kirchhoff’s Laws In mesh ACD, 12 volts are acting in clockwise direction, then: 8(i1– i2) – 4i2 = 12 8i1 – 8i2 – 4i2= 12 Examples 8i1 – 12i2 = 12 ……………. (2) Multiplying equation (1) by 3; 30i1 + 12i2 = 60  Circuit Analysis by Kirchhoff’s Laws 30i1 + 12i2 = 60 8i1 – 12i2 = 12 Examples 38i1 = 72 i1 = 1.895 Amp Substituting this value in (1), we get: 10(1.895) + 4i2 = 20 4i2 = 20 – 18.95 Now, i1 – i2 = 1.895 – 0.263 = 1.632 Amp i2 = 0.263 Amp  Example based on KCL and KVL Find the current flowing in the 40Ω Resistor, R3. Examples-2 The circuit has 3 branches, 2 nodes (A and B) and 2 independent loops. Using Kirchhoffs Current Law, KCL the equations are given as: At node A : I1 + I2 = I3 At node B : I3 = I1 + I2  Example based on KCL and KVL Find the current flowing in the 40Ω Resistor, R3. Using KVL, the equations are given as: Examples Loop 1 is given as : 10 = R1I1 + R3I3 = 10I1 + 40I3 Loop 2 is given as : 20 = R2I2 + R3I3 = 20I2 + 40I3 Loop 3 is given as : 10 – 20 = 10I1 – 20I2  Example based on KCL and KVL Find the current flowing in the 40Ω Resistor, R3. Examples As I3 is the sum of I1 + I2 we can rewrite the equations as; Eq. No 1 : 10 = 10I1 + 40(I1 + I2) = 50I1 + 40I2 Eq. No 2 : 20 = 20I2 + 40(I1 + I2) = 40I1 + 60I2 We now have two “Simultaneous Equations” that can be reduced to give us the values of I1 and I2 Substitution of I1 in terms of I2 gives us the value of I1 as -0.143 Amps Substitution of I2 in terms of I1 gives us the value of I2 as +0.429 Amps  Example based on KCL and KVL Find the current flowing in the 40Ω Resistor, R3. Examples As : I3 = I1 + I2 The current flowing in resistor R3 is given as : -0.143 + 0.429 = 0.286 Amps and the voltage across the resistor R3 is given as : 0.286 x 40 = 11.44 volts  Example based on KCL and KVL Find the voltage V1. Examples By applying KVL to this closed loop, we can write as VED + VAE + VBA + VCB + VDC = 0  Example based on KCL and KVL Where Voltage of point E with respect to point D, VED = -50 V Voltage of point D with respect to point C, VDC = -50 V Voltage of point A with respect to point E. VAE = I * R VAE = 500m* 200 VAE = 100 V Examples Similarly Voltage at point C with respect to pint B, VCB = 350m*100 VCB = 35V Consider voltage at point A with respect to point B, VAB = V1 VBA= -V1  Example based on KCL and KVL -50 + 100 – V1 + 35 – 50 = 0 V1 = 35 Volts Examples  Example based on KCL and KVL Consider the below typical two loop circuit where we have to find the currents I1 and I2 by applying the Kirchhoff’s laws. Examples  Example based on KCL and KVL Examples  Example based on KCL and KVL Examples By applying KVL to these loops we get For first loop, 2 (I1 + I2) + 4I1 – 28 = 0 6I1 + 2I2 = 28 ——— (1) For second loop, -2(I1 + I2) – 1I2 + 7 = 0 -2I1 – 3I2 = -7 ——– (2) By solving the above 1 and 2 equations we get, I1 = 5A and I2 = -1 A  Nodal Analysis & Mesh Analysis of Electrical Networks 4 Mesh is a loop that doesn’t consists of any other loop inside it. Mesh analysis technique, uses mesh currents as variables, instead of currents in the elements to analyze the circuit. Therefore, this method absolutely reduces the number of equations to be Mesh Analysis solved. Mesh analysis applies the Kirchhoff’s Voltage Law (KVL) to determine the unknown currents in a given circuit. Mesh analysis is also called as mesh-current method or loop analysis. After finding the mesh currents using KVL, voltages anywhere in a given circuit can be determined by using Ohms law.  Nodal Analysis & Mesh Analysis of Electrical Networks Steps to Analyze the mesh analysis technique 1. Check whether there is a possibility to transform all current sources in the given circuit to voltage sources. 2. Assign the current directions to each mesh in a given circuit and follow Mesh Analysis the same direction for each mesh. 3. Apply KVL to each mesh and simplify the KVL equations. 4. Solve the simultaneous equations of various meshes to get the mesh currents and these equations are exactly equal to the number of meshes present in the network.  Example based simple circuit with DC excitation Mesh Current Analysis or Loop Analysis Examples-3 One simple method of reducing the amount of math’s involved is to analyze the circuit using Kirchhoff’s Current Law equations to determine the currents, I1 and I2 flowing in the two resistors. Then there is no need to calculate the current I3 as its just the sum of I1 and I2.  Example based simple circuit with DC excitation Examples Kirchhoff’s voltage law simply becomes: Equation No 1 : 10 = 50I1 + 40I2 Equation No 2 : 20 = 40I1 + 60I2  Example based simple circuit with DC excitation Examples Multiply Eqn no:1 by 4 and Eqn no: 2 by 5. Equation No 1 : 10 = 50I1 + 40I2 40 = 200I1 + 160I2 Equation No 2 : 20 = 40I1 + 60I2 100 = 200I1 + 300I2 -60 = - 140I2 I2 = 0.429 A  Example based simple circuit with DC excitation Examples Now, put the value I2 in equation no: 1 10 = 50I1 + 40×0.429 10 = 50I1 + 17.16 I1 = - 0.1432 A The current flowing through 40Ω resistor = I1 + I2 = - 0.1432 + 0.429 = 0.286 A  Nodal Analysis & Mesh Analysis of Electrical Networks Procedure of Nodal Analysis 1. Identify the principal nodes and choose one of them as reference node. We will treat that reference node as the Ground. 2. Label the node voltages with respect to Ground from all the principal Nodal nodes except the reference node. Analysis 3. Write nodal equations at all the principal nodes except the reference node. Nodal equation is obtained by applying KCL first and then Ohm’s law. 4. Solve the nodal equations obtained in Step 3 in order to get the node voltages.  Example based simple circuit with DC excitation Nodal Voltage Analysis Examples-4 At each node point write down Kirchhoff’s first law equation, that is: “the currents entering a node are exactly equal in value to the currents leaving the node” then express each current in terms of the voltage across the branch. For “n” nodes, one node will be used as the reference node and all the other voltages will be referenced or measured with respect to this common node.  Example based simple circuit with DC excitation Examples In the above circuit, node D is chosen as the reference node and the other three nodes are assumed to have voltages, Va, Vb and Vc with respect to node D. For example;  Example based simple circuit with DC excitation Examples As Va = 10V and Vc = 20V, Vb can be easily found by:  Example based simple circuit with DC excitation Mesh Current Analysis or Loop Analysis Find the current through each resistance. Step 1 − Identify the meshes and label the mesh currents in either clockwise or anti-clockwise direction. Step 2 − Observe the amount of current Examples-5 that flows through each element in terms of mesh currents. Step:1 Step 3 − Write mesh equations to all Identify loops 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.  Example based simple circuit with DC excitation Mesh Current Analysis or Loop Analysis Step:2 Label the Voltage Drop Polarities Examples Loop 1: Loop 2: 28 – 4I1 – 2(I1 + I2) = 0 2(I1 + I2) – 7 + 1I2 = 0 6I1 + 2I2 = 28 …………….. (1) 2I1 + 3I2 = 7 ……………… (2)  Example based simple circuit with DC excitation Mesh Current Analysis or Loop Analysis Multiply eqn no. 1 by 3 and equ no. by 2 18I1 + 6I2 = 84 4I1 + 6I2 = 14 14I1 = 70 Examples I1 = 5 A Put the value of I1 in eqn no. 1 6×5 + 2I2 = 28 30 + 2I2 = 28 2I2 = -2 I2 = -1 A  Example based simple circuit with DC excitation Mesh Current Analysis or Loop Analysis Step:3 Redraw the circuit Examples A current of 4 amps through R2 is resistance of 2 Ω gives us a voltage drop of 8 volts (E=IR),  Example based simple circuit with DC excitation Nodal Voltage Analysis Calculate Node Voltages in following circuit Examples-6 In the above circuit we have 3 nodes from which one is reference node and other two are non reference nodes – Node 1 and Node 2. Calculate Node Voltages in following circuit Calculate Node Voltages in following circuit Calculate Node Voltages in following circuit Calculate Node Voltages in following circuit Calculate Node Voltages in following circuit Calculate Node Voltages in following circuit Calculate Node Voltages in following circuit  Example based simple circuit with DC excitation Step I. Assign the nodes voltages as V1 and V2 and also mark the directions of branch currents with respect to the reference nodes. Step II. Apply KCL to Nodes 1 and 2 Step III. Apply Ohm’s Law to KCL equations Examples At Node 1 At Node 2  Example based simple circuit with DC excitation Step IV. Now solve the equations 3 and 4 to get the values of V1 and V2 as, And substituting value V2 = 20 Volts in equation (3) we get- Examples  Example based simple circuit with DC excitation Nodal Voltage Analysis Find the node voltages of the following circuit. Examples-7 Step-1 Identify all of the essential nodes and choose one of them as a reference node.  Example based simple circuit with DC excitation Step-2 Assign voltages variable to all nodes except the reference node. Examples Step-3 Apply the KCL at each node except the reference node. In this step for each node we assume the branch current is leaving from the node, and then we describe the branch current in term of node voltages.  Example based simple circuit with DC excitation Examples Apply KCL at Node # 1 Node#1 has three branches, so we apply the KCL to obtain an equation with three terms as follows:  Example based simple circuit with DC excitation Apply KCL at Node # 2 Examples Node#2 has three branches, so we apply the KCL to obtain an equation with three terms as follows: Step #4 Solving question 1 and 2 for the unknown node voltages (V1 and V2): V1 = - 6 V V2 = -15 V  Example based simple circuit with DC excitation Nodal Voltage Analysis Find i0 by using the nodal analysis: Examples-8 Step-1 Identify all of the essential nodes and choose one of them as a reference node.  Example based simple circuit with DC excitation Examples We do not need to apply the KCL at Node 1, because the node voltage at this node is known, V1 =12 V. Also, we do not need to apply the KCL at Node 3, because the node voltage at this node is known V3 = - 6 V. Notice that the V3 is negative because its polarity is opposite to the polarity of the voltage source. Notice that, for node voltages the negative sign of the polarity is at the reference node.  Example based simple circuit with DC excitation Examples Now, we have only one unknown node voltage (V2). Apply KCL at node #2:  Example based simple circuit with DC excitation Examples Substitute the values of V1 and V3: Find the current i0 using ohm’s law: i0 = V2 / 6 kΩ = 0.25 mA Chapter – 2 Electromechanical Energy Conversion 11/15/2024 1 Topics Principle, Singly excited magnetic system and doubly excited magnetic system, physical concept of torque production, Electromagnetic torque and reluctance torque. Principle and operation of DC machine, Induction motor and transformer. 11/15/2024 2 Introduction 11/15/2024 3 Introduction ▪ Electromechanical energy conversion – ▪ Electrical energy into mechanical energy ▪ Achieved by some motors. ▪ Motor can be AC or DC (input supply) ▪ Process is reversible ▪ Electrical generator converts the mechanical energy applied to its input to an electrical energy. ▪ Electromechanical energy conversion – ▪ Prime mover is required. ▪ Electrical system ▪ Can be AC or DC generator. ▪ Coupling field ▪ Mechanical system 11/15/2024 4 Singly excited magnetic system The electromechanical energy conversion is done through medium of magnetic field. As its name "Singly" suggests that there is only a coil is required to produce the magnetic field. There is one set of electrical input terminal and one set of mechanical output terminal in this excitation system. The electro – magnetic relay, solenoid coil, hysteresis motor etc are the examples of singly excited system. 11/15/2024 5  Figure shows a force producing device in which single coil acts as an input terminal and a movable plunger serves as an output terminal. The electrical input has two variable : e ( volts ) and I (current ) whereas the mechanical output has two variables : f ( force ) and x ( distance ). 11/15/2024 6 Double excited magnetic system As its name "Doubly" suggests that two coils are required to produce mechanical output force. There are two sets of electrical input terminals and one set of mechanical output terminal in this system. The synchronous motor, alternators, DC machines etc. are the examples of doubly excited system. 11/15/2024 7  There are two windings in the rotating systems. One winding is wound on the stator and the other on the rotor. The stator and rotor windings are connected to electrical sources V1 and V2 respectively. The motor can allow rotating between two poles. The magnetic field depends upon the current I1 and I2 and θ between stator and rotor. 11/15/2024 8  Electromagnetic torque: Two types of torque are produced in electro mechanical machines. First type of torque is produced due to the interaction of the field produced by the stator and rotor winding which move relative to each other. This toque is also known as electro mechanical torque or induced torque. Reluctance torque: When piece of ferromagnetic material is placed in a magnetic field, it tries to line up with the external magnetic field. This happens because the external magnetic field in the piece, So the torque is produced between the field in such a way that gives minimum reluctance for the magnetic flux. This torque is also called alignment torque or saliency torque. 11/15/2024 9 Types of DC machines Types of DC machines ▪ DC generator ▪ DC motor ▪ Windings in a DC machine ▪ Field winding – stationary winding ▪ Armature winding – rotatory winding (mounted on shaft) ▪ Construction of DC generator and motor are same so the same motor can be used as generator or motor. ▪ The DC generator thus takes the mechanical energy as an input energy from the prime mover and delivers an electrical energy to the load. 11/15/2024 10 Principle of operation ▪ Principle ▪ Work on the principle of Dynamically induced emf (Faraday’s Law) in a conductor. ▪ If the flux linkage with a conductor changes due to the relative motion between the magnetic field and conductor then emf is induced into the conductor. ▪ Relative motion is produced by either moving the magnetic flux or by moving the conductor. Practical DC generator – ▪ Parts of DC generator are – 1. An electromagnet to produce the magnetic flux. 2. A movable (rotatory) conductor placed in this field. 3. A mechanism to rotate the conductor. ▪ Stationary winding is called field winding – to produce flux. ▪ Rotatory winding is called armature winding – as movable conductor. ▪ Machine called prime mover – to rotate the armature winding (can be turbine, engine etc.) ▪ Direction of induced emf in armature winding can be given by Fleming's “right hand rule”. 11/15/2024 11 Principle of operation ▪ Fleming’s Right hand Rule – ▪ Notation: Left thumb – Motion of conductor ▪ First finger – lines of forces (N to S pole) ▪ Second finger – indicates the direction of induced emf (or current) 11/15/2024 12 Principle of operation Magnitude of Induced EMF – ▪ Induced emf depends upon the rate of change of magnetic flux linkage takes place with conductor. ▪ When the plane of conductor motion is parallel to plane of magnetic flux then induced emf is zero. ▪ When the plane of conductor motion is perpendicular to plane of magnetic flux then induced emf is maximum. Induced emf = 𝑩𝒍𝒗 𝐬𝐢𝐧 𝜽 volts where B = Magnetic flux density (Wb/m2) l = Length of the conductor v = Velocity of the moving conductor θ = Angle between the plane of conductor and the plane of the magnetic flux. 11/15/2024 13 Principle of operation Effect of angle on induced emf – ▪ When conductor movement takes place in parallel with the place of magnetic flux i.e. θ = 0°, then induced emf is zero. ▪ When the conductor movement takes place in perpendicular with the place of magnetic flux i.e. θ = 90°, then induced emf is maximum i.e. emax = Blv volts ▪ The plane of conductor making and angle θ with the place of magnetic flux hence the induced emf is in between 0 and maximum. Shape of induced emf – e = 𝑩𝒍𝒗 𝒔𝒊𝒏 𝜽 Hence if B, l and v are constant then e = 𝐤 𝒔𝒊𝒏 𝜽 Hence e is directly proportional to the sin θ. Therefore the shape of induced emf will be a sine wave. Induced voltage is maximum at θ = 90° & 270° and minimum at θ = 0°, 180° & 360°. One rotation of conductor corresponds to the one cycle of induced emf. Thus induced voltage in the armature winding of a DC generator is not DC but it is alternating. To convert it into a unidirectional (DC) signal, we use a rectifying device called Commutator. Commutator is a device used in DC generators to convert alternating induced voltage into DC voltage. 11/15/2024 14 Constructional Features Important parts of DC generator – ▪ Yoke ▪ Field Winding ▪ Poles, pole shoe and pole core ▪ Armature core ▪ Armature winding ▪ Commutator, brushes and gear ▪ Bearings 11/15/2024 15 Constructional Features Yoke – ▪ Also called Frame. ▪ Provide protection to rotating and other part of the machine from moisture, dust etc. ▪ Iron body which provides path for the flux & essential to complete the magnetic circuit. ▪ Mechanical support to poles. ▪ Basically made up of low reluctance materials such as cast iron, silicon steel, rolled steel, cast steel etc. ▪ Small machine – Cast iron ▪ Large machine – Cast steel & rolled steel. 11/15/2024 16 Constructional Features Poles, Pole shoe and Pole Core – ▪ Electromagnet & field winding wound over the poles. ▪ Produce magnetic flux when field winding is excited. ▪ Extended part of a pole. ▪ Due to its typical shape, it enlarges the area of pole. ▪ Due to this enlarged area, more flux can pass through the air gap to armature. ▪ Low reluctance magnetic material (i.e. cast steel or cast iron) used for pole & pole shoe. ▪ Construction of pole is done using the laminations of particular shape to reduce eddy currents losses. 11/15/2024 17 Constructional Features Field Winding – ▪ Coils wound around the poles cores. ▪ Field coils are connected in series to form field winding. ▪ Current is passed through the field winding in a specific direction, to magnetize the poles and pole shoes. The magnetic flux φ is thus produced in the air gap between the pole shoes and armature. ▪ It is also called exciting winding. ▪ Material used in copper for field winding. ▪ Due to current flowing through the field winding alternate N & S pole are produced. Which pole is produced at particular core is decided by the right hand thumb rule for a current carrying circular conductor. 11/15/2024 18 Constructional Features Armature core – ▪ It is cylindrical drum mounted on shaft. ▪ Provided with large number of slots all over the its periphery. ▪ All slots are parallel to the shaft axis. ▪ Armature conductor are placed in these slots. ▪ It provides low reluctance path to flux produced by the field winding. ▪ High permeability, low reluctance materials such as cast steel or cast iron are used for the armature core. ▪ Air holes are provided for the air circulation which helps in cooling the armature core. ▪ Laminated construction is used to produced the armature core to minimize the eddy current losses. 11/15/2024 19 Constructional Features Armature Winding – ▪ Armature conductors made of copper are placed in the armature slots present on the periphery of armature core. ▪ Armature conductors are interconnected to form the armature winding. ▪ when it is rotates using a prime mover, it cuts the magnetic flux lines and voltage gets induced in it. ▪ It is connected to the external circuit through the commutator and brushes. ▪ Armature winding is supposed to carry the entire load current so it should be made up of a conducting material such as copper. 11/15/2024 20 Constructional Features Commutator – ▪ Cylindrical drum mounted on shaft along with the armature core. ▪ Large number of wedge-shaped segments of hard drawn copper. ▪ Segments are insulated from each other by thin layer of mica. ▪ Armature winding are tapped at various points and these tapping's are successively connected at various segments of the commutator. ▪ Converts AC emf in DC (emf) voltage (mechanical rectifier). ▪ Collects currents from armature conductors and passes it into the external load via brushes. ▪ In motors, it helps to produce a unidirectional torque. ▪ Made up of copper and insulating material between the segments of mica. 11/15/2024 21 Constructional Features Brushes – ▪ Current is conducted from armature to the external circuit or load by the carbon brushes which are held against the surface of commutator by springs. ▪ Brushes wear with time. ▪ They should be inspected regularly and replaced occasionally. Function – ▪ To collect current from the commutator and apply to the external load. Material used – ▪ Made up of carbon & rectangular in shape. 11/15/2024 22  Induction Motor An Induction motor is the most widely used AC motor in industrial and domestic applications. Now you might think, why is it so? It is because of the low cost, simple and rugged construction of the motor. Moreover, it has good operating characteristics with an efficiency as high as 90%. An induction motor does not have any commutator, like the one we saw in a DC motor. Hence it provides a good speed regulation without any sparking. With that many advantages, it becomes essential to meet the mechanical power demand using an induction motor. 11/15/2024 23  Construction of an Induction Motor Frame: It is the outer body of the motor. It supports the stator core and protects the inner parts of the machine from the surroundings. 11/15/2024 24 Stator core: A stator consists of a stack of laminations of silicon steel in the form of a ring. It fits inside the stator frame and contains slots on its inner periphery. These slots carry the three-phase winding, separated by 120 degrees in space. Right here, the figure shows the distribution of three-phase winding in a stator. 11/15/2024 25 Rotor A rotor consists of a stack of laminations in the form of a cylinder. It has slots punched on its outer periphery, which contains rotor windings. Based on the winding type, there are two categories of rotors. Squirrel cage rotor This rotor uses copper or aluminum bars as the rotor conductors. Each slot of the rotor carries a conductor without any insulation from the core. All the conductors are shorted by annular rings, aka end rings. Wound rotor It uses wires or straps for the rotor windings. The distribution of the three-phase winding on the rotor is similar to that of the stator. The winding connects to an external resistance through slip rings and brushes. The wound rotor provides a higher starting torque as compared to the squirrel cage rotor. 11/15/2024 26 Induction motor working principle The induction motor follows these two laws to generate a unidirectional torque. Faraday’s law of electromagnetic induction It states that a conductor placed in a varying magnetic field induces an Electromagnetic force (EMF). The closed conductor leads to the flow of current, known as induced current. Lorentz force law It states that a current-carrying conductor placed in a magnetic field experiences a force. The force on the conductor is orthogonal to the direction of the current and magnetic field. 11/15/2024 27 How does the rotor rotate in an Induction motor? after the generation of the rotating magnetic field, the rotor conductors start interacting with the magnetic field. Assume a rotor conductor interacting with the magnetic field as shown in the figure. This interaction induces a current in the conductor (according to Faraday’s law of electromagnetic induction). Now, according to Lorentz law, a force starts acting on the conductor. This force tends to displace the rotor in a direction, as shown in the figure. 11/15/2024 28  Introduction 1. Need of transformer? 2. Definition – Transformer is a static device which is used to transfer electrical energy from one ac circuit to another ac circuit, with increase and decrease voltage/current and without changing frequency. 3. Step up & step down transformer 11/15/2024 29 Type of transformer 1. Step up and Step down transformer 2. Three phase and single phase transformer 3. Power transformer, distribution transformer & Instrument transformer 4. Two winding transformer and auto transformer 5. Outdoor and indoor transformer 6. Oil cooled and dry type transformer 7. Core type, Shell type and berry type transformer 11/15/2024 30 Principle of operation 1. As soon as the primary winding is connected to the single phase ac supply. An ac current starts flowing through it. 2. The ac primary current produces an alternating flux φ in the core. 3. Most of this changing flux gets linked with the secondary winding through the core. 4. The varying flux will induce voltage into the secondary winding according to the Faraday’s laws of electromagnetic induction. 11/15/2024 31 Construction Most Important part of transformer are windings & the core. Suitable tank Conservator Buchholz Relay Bushings Breather Explosion vent etc. 11/15/2024 32 Laminated Steel core 1. Material used for transformer core is silicon steel (High permeability & low magnetic reluctance). Due to this magnetic field produces in the core is very strong. 2. Stacks of laminated thin steel sheets which are electrically isolated from each other. Laminations thickness are typically 0.35mm to 0.5mm. 3. Core assembled in such a way that it should provides a continuous path to magnetic flux with minimum air gap. 11/15/2024 33 Transformer Limbs 1. The cross section of transformer limb of the core of small transformer is rectangular. 2. Windings wound around it are also rectangular. 3. Size of transformer increases by rectangular limb and rectangular winding. 4. Cross section of the limb of core of such transformer is either square or stepped. 5. As number of steps increase the cross section of the windings will be more & more close to the circular cross section & less copper is required. 6. Labor charge will increase due to stepped structure. 11/15/2024 34 Windings 1. Primary and secondary windings are mounted on same limb of the core. 2. Two type of windings 1. Concentric cylindrical winding 2. Sandwiched type winding 3. Concentric cylindrical winding ◦ Both windings as well as core are insulated from each other. ◦ Why LV is placed close to core (limb)? Sandwiched type windings – 1. HV & LV windings are divided into a number of small coils and then these small windings are interleaved. 2. Top and bottom windings are low voltage coils because they are close to the core. 11/15/2024 35 Windings 11/15/2024 36 Construction Transformer Tank – Whole assembly is placed in a sheet metal tank and transformer is immersed in oil (noninflammable insulating oil) which act as insulator as well as coolant. Functions - Heat generated at core and the windings should be removed efficiently. To avoid the insulation deterioration, the moisture should not be allowed to creep into the insulation. Increase the cooling surface exposed to ambient, tubes or fins are provided on the outside of the tank walls. 11/15/2024 37 Construction Conservator – Empty space is always provided above the oil level. This space is essential for letting the oil to expand or contract due to temperature changes. When the oil temperature increase, it expands and the air will be expelled out from the conservator. Whereas when the oil cools, it contracts and the outside air gets sucked inside the conservator. This process is called as the breathing of the transformer. The outside air which has being drawn in can have the moisture content. When such air comes into the contact with the oil, the oil will absorb the moisture content and losses its insulating properties, to some extent. This can be prevent by Conservator. 11/15/2024 38 Construction Breather – Apparatus through which breathing of the transformer takes place is called breather. Air goes in or out through the breather. To reduce the moisture content of this air, some drying agent (material absorb moisture) such as silica gel or calcium chloride is used in the breather. The dust particles present in the air are also removed by the breather. 11/15/2024 39 Construction Buchholz Relay – There is pipe connecting the tank and conservator. Buchholz relay is a protective device mounted on that pipe. Transformer is about to be faulty and draws large currents, the oil becomes very hot and decomposes. During this process different types of gases are liberated. The Buchholz relay get operated by these gases and gives an alarm to the operator. If the fault continues to persist, then the relay will trip off the main circuit breaker to protect the transformer. 11/15/2024 40 Construction Explosion Vent – An explosion vent or relief valve is the bent up pipe fitted on the main tank. It consists of a glass diaphragm or aluminum foil. When the transformer becomes faulty, the cooling oil will get decomposed and various type of gases are liberated. If the gas pressure reaches a certain level then the diaphragm in the explosion vent will brust to release the pressure. This will save the main tank from getting damaged. 11/15/2024 41 AC Circuits ❑Introduction to AC quantities ❑Phasor representation of alternating quantities ❑Analysis of series RL circuit, RC circuit, RLC circuit ❑Parallel and series-parallel AC circuits ❑phasor method, admittance method  Introduction The use of direct currents is limited to a few applications e.g. charging of batteries, electroplating, electric traction etc. For large scale power distribution there are, however, many advantages in using alternating current (a.c.). The a.c. system has offered so many advantages that at present electrical energy is universally generated, transmitted and used in the form of alternating current. Even when d.c. energy is necessary, it is a common practice to convert a.c. into d.c. by rectifiers. Three principal advantages are claimed for a.c. system over the d.c. system. First, alternating voltages can be stepped up or stepped down efficiently by means of a transformer. The transmission of electric power at high voltages to achieve economy and distribute the power at utilisation voltages. Generation of Alternating Voltages and Currents By rotating coil within By rotating magnetic field stationary magnetic field within stationary coil The first method is used for small a.c. generators while the second method is employed for large a.c. generators.  Generation of Alternating voltage Faraday’s Law of electromagnetic Induction A law stating that when the magnetic flux linking with conductor, an electromotive force is induced in the conductor which is proportional to the rate of change of the flux linkage. d e = − ( N ) dt  Equation of Alternating Voltage and Current Consider a rectangular coil of n turns rotating in anticlockwise direction with an angular velocity of w rad/sec in a uniform magnetic field. Instantaneous value. The value of an alternating quantity at any instant is called instantaneous value. The instantaneous values of alternating voltage and current are represented by v and i respectively. Cycle One complete set of positive and negative values of an alternating quantity is known as a cycle. Time period The time taken in seconds to complete one cycle of an alternating quantity is called its time period. It is generally represented by T. Frequency The number of cycles that occur in one second is called the frequency (f) of the alternating quantity. It is Hertz (Hz). Amplitude/Peak value/Crest value/Maximum value The maximum value (positive or negative) attained by an alternating quantity is called its amplitude or peak value. The amplitude of an alternating voltage or current is designated by Vm (or Em) or Im. Example: The maximum current in a sinusoidal a.c. circuit is 10A. What is the instantaneous current at 45º ? Example: An alternating current i is given by ; i = 141·4 sin 314 t Find (i) the maximum value (ii) frequency (iii) time period and (iv) the instantaneous value when t is 3 ms. Values of Alternating Voltage and Current In a d.c. system, the voltage and current are constant so that there is no problem of specifying their magnitudes. However, an alternating voltage or current varies from instant to instant. A natural question arises how to express the magnitude of an alternating voltage or current. There are four ways of expressing it, namely ; (i) Peak value (ii) Average value or mean value (iii) R.M.S. value or effective value (iv) Peak-to-peak value Although peak, average and peak-to-peak values may be important in some engineering applications, it is the r. m.s. or effective value which is used to express the magnitude of an alternating voltage or current. Peak Value It is the maximum value attained by an alternating quantity. The peak or maximum value of an alternating voltage or current is represented by Vm or Im. The knowledge of peak value is important in case of testing materials. However, peak value is not used to specify the magnitude of alternating voltage or current. Instead, we generally use r.m.s. values to specify alternating voltages and currents. Average Value The average value of a waveform is the average of all its values over a period of time. In performing such a computation, we regard the area above the time axis as positive area and area below the time axis as negative area. The algebraic signs of the areas must be taken into account when computing the total (net) area. The time interval over which the net area is computed is the period T of the waveform. (i) In case of *symmetrical waves (e.g. sinusoidal voltage or current), the average value over one cycle is zero. It is because positive half is exactly equal to the negative half so that net area is zero. However, the average value of positive or negative half is not zero. Hence in case of symmetrical waves, average value means the average value of half-cycle or one alternation. Average Value of Sinusoidal Current Form Factor and Peak Factor There exists a definite relation among the peak value, average value and r.m.s. value of an alternating quantity. The relationship is expressed by two factors, namely ; form factor and peak factor. (i) Form factor: The ratio of r.m.s. value to the average value of an alternating quantity is known as form factor i.e. (ii) Peak factor: The ratio of maximum value to the r.m.s. value of an alternating quantity is known as peak factor i.e. Example: Find the average value, r.m.s. value, form factor and peak factor for (i) halfwave rectified alternating current and (ii) full-wave rectified alternating current. (i) Half-wave rectified a.c. Example: An alternating voltage v = 200 sin 314t is applied to a device which offers an ohmic resistance of 20 Ω to the flow of current in one direction while entirely preventing the flow of current in the opposite direction. Calculate the r.m.s. value, average value and form factor.  Phasor representation of alternating quantities An alternating voltage or current may be represented in the form of (i) waves and (ii) equations. The waveform presents to the eye a very definite picture of what is happening at every instant. But it is difficult to draw the wave accurately. No doubt the current flowing at any instant can be determined from the equation form i = Im sin ωt but this equation presents no picture to the eye of what is happening in the circuit. The above difficulty has been overcome by representing sinusoidal alternating voltage or current by a line of definite length rotating in anticlockwise direction at a constant angular velocity (ω). Such a rotating line is called a phasor. The length of the phasor is taken equal to the maximum value (on a suitable scale) of the alternating quantity and angular velocity equal to the angular velocity of the alternating quantity. As we shall see presently, this phasor (i.e. rotating line) will generate a sine Phasor Representation of Sinusoidal Quantities Consider an alternating current represented by the equation i = Im sin ωt. Take a line OP to represent to scale the maximum value Im. Imagine the line OP (or phasor, as it is called) to be rotating in anticlockwise direction at an angular velocity ω rad/sec about the point O. Measuring the time from the instant when OP is horizontal, let OP rotate through an angle θ (= ωt) in the anticlockwise direction. The projection of OP on the Y-axis is OM. OM = OP sin θ = Im sin ωt = i, the value of current at that instant Addition of alternating quantities Following methods are used for addition of two or more sinusoidal quantities of the same kind. 1. Method of components 2. Analytical addition method 3. Parallelogram method 4. Waveform addition method  Method of Components. This method provides a very convenient means to add two or more phasors. Each phasor is resolved into horizontal and vertical components. The horizontals are summed up algebraically to give the resultant horizontal component X. The verticals are likewise summed up algebraically to give the resultant vertical component Y. Example: A circuit consists of four loads in series ; the voltage across these loads are given by the following relations measured in volts : v1 = 50 sin ω t ; v2 = 25 sin (ω t + 60º) v3 = 40 cos ω t ; v4 = 30 sin (ω t − 45º) Calculate the supply voltage giving the relation in similar form. R-L Series A.C. Circuit If the applied voltage is v = Vm sin ωt, then equation for the circuit current will be  Apparent, True and Reactive Powers ✓ Consider an inductive circuit in which circuit current I lags behind the applied voltage V by φ°. ✓ The phasor diagram of the circuit is shown in Fig. ✓ The current I can be resolved into two rectangular components: (i) I cos φ in phase with V. (ii) I sin φ ; 90° out of phase with V. 1. Apparent power ✓ The total power that appears to be transferred between the source and load is called apparent power.. Apparent power, S = V × I = VI ✓ It is measured in volt-ampers (VA). ✓ Apparent power has two components: true power and reactive power. 2. True power ✓ The power which is actually consumed in the circuit is called true power or active power. ✓ It is the useful component of apparent power. ✓ The product of voltage (V) and component of total current in phase with voltage (I cos φ) is equal to true power. ✓ It is measured in watts (W). The component I cos φ is called in-phase component or wattful component. 3. Reactive power ✓ The component of apparent power which is neither consumed nor does any useful work in the circuit is called reactive power. ✓ The power consumed (or true power) in L and C is zero because all the power received from the source in one quarter-cycle is returned to the source in the next quarter-cycle. This circulating power is called reactive power. ✓ The product of voltage (V) and component of total current 90° out of phase with voltage (I sin φ) is equal to reactive power ✓ It is measured in volt-amperes reactive (VAR). The component I sin φ is called the reactive component (or wattless component) R-C Series A.C. Circuit Power angle: R-L-C Series A.C. Circuit A coil having a resistance of 7  and an inductance of 31.8 mH is connected to 230 V, 50 Hz supply. Calculate : (i) the circuit current (ii) phase angle (iii) power factor (iv) power consumed and (v) voltage drop across resistor and inductor. A capacitor of capacitance 79.5 μ F is connected in series with a non-inductive resistance of 30  across 100 V, 50 Hz supply. Find: (i) impedance (ii) current (iii) phase angle and (iv) equation for the instantaneous value of current. A 230 V, 50 Hz a.c. supply is applied to a coil of 0.06 H inductance and 2.5 resistance connected in series with a 6.8 μF capacitor. Calculate: (i) impedance (ii) current (iii) phase angle between current and voltage (iv) power factor and (v) power consumed.  Resonance in A.C. Circuits ✓ An a.c. circuit containing reactive elements (L and C) is said to be in resonance when the circuit power factor is unity. ✓ Resonance means to be in step with. When applied voltage and circuit current in an a.c. circuit are in step with (i.e. phase angle is zero or p.f. is unity), the circuit is said to be in electrical resonance. ✓ If this condition exists in a series a.c. circuit, it is called series resonance. On the other hand, if this condition exists in a parallel a.c. circuit, it is called parallel resonance. ✓ The frequency at which resonance occurs is called resonant frequency (fr). Prof.Vatsal Patel Resonance in Series A.C. Circuit (Series Resonance) ✓ A series R-L-C a.c. circuit is said to be in resonance when circuit power factor is unity. ✓ The circuit impedance (Z) and circuit current (I) are given by ✓ The resonant frequency (fr) for R – L – C series a.c. circuit is defined as the frequency at which XL = XC. PROF.VATSAL PATEL PROF.VATSAL PATEL PROF.VATSAL PATEL  Graphical Explanation of series resonance 𝑋𝐿 ∝ 𝑓 1 𝑋𝐶 ∝ 𝑓 PROF.VATSAL PATEL Resonance Curve 𝑋𝐿 > 𝑋𝐶 𝑋𝐶 > 𝑋𝐿 PROF.VATSAL PATEL  Methods of Solving Parallel A.C. Circuits 1. By phasor diagram 2. By phasor algebra 3. Equivalent impedance method 4. Admittance method Prof.Vatsal Patel  Admittance Triangle For an inductive circuit ✓ Admittance angle is equal to the impedance angle but is negative. For this reason, BL will be along OY′-axis and hence negative. Prof.Vatsal Patel Prof.Vatsal Patel  For Capacitive circuit ✓ admittance angle is equal to the impedance angle but of opposite sign. For this reason, BC will lie along OY-axis and hence positive. Prof.Vatsal Patel Prof.Vatsal Patel Admittance Method for Parallel A.C. Circuit Solution ✓ Fig. shows two impedances Z1 = R1 – j XC1 and Z2 = R2 + jXL2 in parallel across an a.c. supply of V volts. We can convert the impedances into equivalent admittances as under : Prof.Vatsal Patel Admittance Method for Parallel A.C. Circuit Solution ✓ Fig. shows Y1 and Y2 resolved into conductances and suceptances. It may be noted that conductance and suceptance components of each admittance are paralleled elements. where G = G1 + G2 and B = B1 – B2 Prof.Vatsal Patel Basic of Electrical & Electronics Department of Electrical Engineering Engineering Polyphase System (01EE1101) Poly Phase System A polyphase alternator has two or more separate but identical windings (called phases) displaced from each other by **equal electrical angle and acted upon by the common uniform magnetic field. Phase Voltage It is defined as the voltage across either phase winding or load terminal. It is denoted by Vph. Phase voltage VRN, VYN and VBN are measured between R-N, Y-N, B-N for star connection and between R-Y, Y-B, B-R in delta connection. Line voltage It is defined as the voltage across any two-line terminal. It is denoted by VL. Line voltage VRY, VYB, VBR measure between R-Y, Y-B, B-R terminal for star and delta connection both. Phase current It is defined as the current flowing through each phase winding or load. It is denoted by Iph. Phase current IR(ph), IY(ph) and IB(Ph) measured in each phase of star and delta connection. respectively. Line current It is defined as the current flowing through each line conductor. It denoted by IL. Line current IR(line), IY(line), and IB((line) are measured in each line of star and delta connection. Phase sequence The order in which three coil emf or currents attain their peak values is called the phase sequence. It is customary to denoted the 3 phases by the three colours. i.e. red (R), yellow (Y), blue (B). Balance System A system is said to be balance if the voltages and currents in all phase are equal in magnitude and displaced from each other by equal angles. Unbalance System A system is said to be unbalance if the voltages and currents in all phase are unequal in magnitude and displaced from each other by unequal angles. Balance load In this type the load in all phase are equal in magnitude. It means that the load will have the same power factor equal currents in them. Unbalance load In this type the load in all phase have unequal power factor and currents. Relation between line and phase values for voltage and current in case of balanced delta connection. Relation between line and phase values for voltage and current in case of balanced star connection.  Recording Ch:4 Semiconductor Diodes Department of Energy Band Diagram of conductor, semiconductor and Electrical Engineering insulator; Crystal Structure of Semiconductor Materials, Intrinsic and Extrinsic Semiconductor Materials. Symbol and Unit no 4 Construction, Operating Characteristics in Forward and Reverse Unit title: Semiconductor Diodes Bias, Applications of Diode as Switch, Clipper, Clamper and Subject name and Rectifier; Special Purpose Diodes: Zener Diode; Optical Diodes code : BEEE (01EE0101) like LED, Photo Diode, Seven Segment Display Prof. Uvesh Sipai BE Prof. Nirav Tolia   The material can be classified according to their resistivity range as –  Conductors (1.6 X 10-8 to 10-6 ohm-m)  Semiconductors (10-4 to 106 ohm-m)  Insulators (107 to 1016 ohm-m)  Semiconductors are the materials whose resistivity (and Introduction hence the conductivity) lies between those of conductors and insulator  The semiconductor devices are highly compact, efficient, more reliable, low power consuming, free from mechanical noise and are cheap.  Most commonly used semiconductors are Si, Ge, GaAs, InP etc. BE Basic Electronics Prof. Uvesh Sipai Introduction BE Basic Electronics Prof. Uvesh Sipai  Properties of Semiconductors – 1. Usually high resistivity. 2. Semiconductors are unipolar. Electron 3. They have negative temperature coefficient. Energy States 4. They are metallic in nature. of an isolated 5. At 0K they behaves like insulators. Atom 6. Both electrons and holes can be charge carrier. BE Basic Electronics Prof. Uvesh Sipai  Types of Semiconductors 1. Elemental and compound semiconductors 2. Direct and indirect bandgap semiconductors Electron 3. Intrinsic and Extrinsic semiconductors Energy States of an isolated Atom BE Basic Electronics Prof. Uvesh Sipai  Intrinsic or pure semiconductors – Silicon and germanium have crystalline structure. Their atoms are arranged in an ordered array known as the crystal lattice. There material are tetravalent i.e. with four valence Intrinsic electrons in the outermost shell. semiconductor The neighboring atoms form covalent bonds by sharing four electrons with each other so as to achieve stable structure. BE Basic Electronics Prof. Uvesh Sipai  Crystal lattice structure and band gap diagram of pure germanium Energy gap – 0.72 eV for Ge. Valence Band remains full and conduction band is empty and material behaves as insulator. Intrinsic At room temp, valence band semiconductor electrons acquire thermal energy greater than Eg and hence they can jump to the higher energy conduction band. They are free now and can move under the influence of small applied field. BE Basic Electronics Prof. Uvesh Sipai  The absence of electrons in valence band is known as hole. Hence in semiconductor there are 2 types of charge carriers. The total current is the sum of electron and holes currents. At any given temperature, no of electrons = Intrinsic no of holes in valence band are same. semiconductor With the rise in temperature, more and more electrons hole pairs are formed and more charge carriers are available for conduction. Hence the conductivity of intrinsic semiconductors increases with rise in temperature. The intrinsic semiconductor have low conductivity. BE Basic Electronics Prof. Uvesh Sipai  Doped or Extrinsic Semiconductors – Doping is the process of adding a controlled quantity of impurity to an intrinsic semiconductor, so as to increase its conductivity. A semiconductor doped with impurity atoms is called an extrinsic semiconductor. The impurity is added by melting Ge or Si, then the Extrinsic crystal is grown in which the impurities are incorporated. semiconductor The impurity atoms occupy lattice positions which were occupied by Ge atoms in pure metal. Doping element is from III and V group elements. Two types of extrinsic semiconductors are produced depending upon the group of impurity atom. BE Basic Electronics Prof. Uvesh Sipai  N-type Semiconductors – The pentavalent impurities is added in Ge crystal lattice. It forms four covalent bonds with four neighboring Ge atoms. The 5th electron, not used in bonding, it loosely bound & with supply little energy, it can be made free leaving Extrinsic behind a positively charged immobile ion. semiconductor The impurity atoms donate free electrons to the crystal thereby increasing the conductivity of material. Hence they are called the donor impurities. The conductivity is due to negatively charged electrons. Hence the material is called N-type semiconductor. BE Basic Electronics Prof. Uvesh Sipai  Electron-hole pairs are generated in Ge due to thermal energy. For n-type the concentration of free electrons is far greater Extrinsic than concentration of holes. semiconductor Addition of donor impurities generates new energy levels in the band picture. The energy levels ED of neutral donor atoms lie very close to the lowed edge of EC of conduction band. BE Basic Electronics Prof. Uvesh Sipai  With the supply of little energy (0.01eV for Ge and Extrinsic 0.04eV for Si) the neutral donor atom loses fifth electron semiconductor for conduction and itself gets positively charged. BE Basic Electronics Prof. Uvesh Sipai  P-type Semiconductors – The trivalent impurities is added in Ge crystal lattice. Trivalent is one electron short of being able to complete the stable structure. The absence of electron in one of these bonds is a hole. Extrinsic With the small amount of energy, it can accept an electron semiconductor from the neighboring Ge atom and vacancy shifts there. The impurity atom becomes a negative charged ion on accepting the electron. Thus impurity atom supply holes which are ready to accept electrons. Hence it’s a acceptor impurity. BE Basic Electronics Prof. Uvesh Sipai  The holes concentration is much more than the electron concentration. The conductivity is due to the positively charged holes. Hence the semiconductor is called P-type semiconductor. The addition of impurity introduces additional energy Extrinsic levels EA, in the band picture, slightly above the top of the semiconductor valence band. With supply of little energy these vacancies can be occupied by electrons in VB and thus increasing the holes in VB. The extrinsic materials are electrically neutral at any given temperature. BE Basic Electronics Prof. Uvesh Sipai Extrinsic semiconductor BE Basic Electronics Prof. Uvesh Sipai   If a single piece of germanium doped with P-type material from one side and the other half is doped with N-type material, then the Ge is dividing into two zones forms a P- N junction.  the P-N junction is the basic element for semiconductor diodes. A Semiconductor diode facilitates the flow of electrons completely in one direction only – which is the P-N Junction main function of semiconductor diode.  It can also be used as a Rectifier. BE Basic Electronics Prof. Uvesh Sipai   There are two operating regions: P-type and N-type.  And based on the applied voltage, there are three possible “biasing” conditions for the P-N Junction Diode, which are as follows: Zero Bias – No external voltage is applied to the PN junction diode. P-N Junction Forward Bias– The voltage potential is connected positively to the P-type terminal and negatively to the N- type terminal of the Diode. Reverse Bias– The voltage potential is connected negatively to the P-type terminal and positively to the N- type terminal of the Diode. BE Basic Electronics Prof. Uvesh Sipai   There are two operating regions: P-type and N-type.  And based on the applied voltage, there are three possible “biasing” conditions for the P-N Junction Diode, which are as follows: P-N Junction – Zero Bias – No external voltage is applied to the PN junction diode. Zero Bias In this case, no external voltage is applied to the P-N junction diode; and therefore, the electrons diffuse to the P-side and simultaneously holes diffuse towards the N-side through the junction, and then combine with each other. Due to this an electric field is generated by these charge carriers. BE Basic Electronics Prof. Uvesh Sipai   When a p-n junction is formed, some of the free electrons in the n-region diffuse across the junction and combine with holes to form negative ions. In so doing they leave behind positive ions at the donor impurity sites. P-N Junction – Zero Bias BE Basic Electronics Prof. Uvesh Sipai P-N Junction BE Basic Electronics Prof. Uvesh Sipai   In the forward bias condition, the negative terminal of the P-N Junction – battery is connected to the N-type material and the positive terminal of the battery is connected to the P-Type Forward Bias material. This connection is also called as giving positive voltage.  Electrons from the N-region cross the junction and enters the P-region. Due to the attractive force that is generated in the P-region the electrons are attracted and move towards the positive terminal.  Simultaneously the holes are attracted to the negative terminal of the battery. By the movement of electrons and holes current flows.  In this condition, the width of the depletion region decreases due to the reduction in the number of positive and negative ions. BE Basic Electronics Prof. Uvesh Sipai   In the forward bias condition, the negative terminal of the battery is Forward Bias connected to the N-type material and the positive terminal of the battery is Condition connected to the P-Type material. This connection is also called as giving positive voltage.  Electrons from the N-region cross the junction and enters the P-region. Due to the attractive force that is generated in the P-region the electrons are attracted and move towards the positive terminal.  Simultaneously the holes are attracted to the negative terminal of the battery. By the movement of electrons and holes current flows.  In this condition, the width of the depletion region decreases due to the reduction in the number of positive and negative ions. BE Basic Electronics Prof. Uvesh Sipai 22   Forward Bias– The voltage potential is connected positively to the P-type terminal and negatively to the N-type terminal of the Diode. P-N Junction Diode BE Basic Electronics Prof. Uvesh Sipai 23   In the Reverse bias condition, the negative terminal of the P-N Junction – battery is connected to the P-type material and the positive terminal of the battery is connected to the N-type material. Reverse Bias  Hence, the electric field due to both the voltage and depletion layer is in the same direction. This makes the electric field stronger than before.  Due to this strong electric field, electrons and holes want more energy to cross the junction so they cannot diffuse to the opposite region.  Hence, there is no current flow due to the lack of movement of electrons and holes. BE Basic Electronics Prof. Uvesh Sipai   Reverse Bias– The voltage potential is connected positively to the N- type terminal and negatively to the P-type terminal of the Diode. P-N Junction Diode BE Basic Electronics Prof. Uvesh Sipai 25 V-I Characteristics BE Basic Electronics Prof. Uvesh Sipai 26 P-N Junction – Characteristics BE Basic Electronics Prof. Uvesh Sipai  Definition: The approximation technique that helps in analyzing the various initial criteria of the diode can be defined as Diode Approximations. Each approximates relates from assuming ideal conditions to reaching practical ones. Diode There are three (3) Diodes Approximations: Approximations  Ideal Diode (1st Approximation)  Second Approximation  Third Approximation BE Basic Electronics Prof. Uvesh Sipai 28  Ideal (1st-Approximation) - For the 1st-approx. assume the diode drop voltage is zero (Perfect closed switch) Second approximation - For the 2nd-approx. assume the diode drop voltage of 0.7 volts Diode Third Approximation - For the 3rd –approx. assume the diode drop Approximations voltage of 0.7 volt and consider the forward bulk resistance of the diode: Vd = 0.7 V + Id x Rb Ignore bulk resistance of the diode if Rb < 0.01 Rth BE Basic Electronics Prof. Uvesh Sipai 29   An ideal diode is a diode that acts like a perfect conductor when voltage is applied forward biased and like a perfect insulator when voltage is applied reverse biased.  So when positive voltage is applied across the anode to the cathode, the diode conducts forward current instantly. Ideal Diode  When voltage is applied in reverse, the diode conducts no current at all.  This diode operates like a switch.  Forward Bias – Closed switch,  Reverse Bias – Open Switch BE Basic Electronics Prof. Uvesh Sipai 30   Zero Resistance Characteristics of Ideal Diode - An ideal diode does not offer any resistance to the flow of Forward Biased current through it when it is in forward biased mode. This means that the ideal diode will be a perfect conductor when forward biased. From this property of the ideal diode, one can infer that the ideal diode does not have any barrier potential. BE Basic Electronics Prof. Uvesh Sipai 31   Infinite Amount of Current Characteristics of Ideal Diode - This property of the ideal diode can be directly implied from its Forward Biased previous property which states that the ideal diodes offer zero resistance when forward biased. The reason can be explained as follows. In electronic devices, the relationship between the current (I), voltage (V) and resistance (R) is expressed by Ohm’s law which is stated as I = V/R. Now, if R = 0, then I = ∞. This indicates that there is no higher limit for the current which can flow through the forward-biased ideal diode BE Basic Electronics Prof. Uvesh Sipai 32   Zero Threshold Voltage Characteristics of Ideal Diode - Even this characteristic of the ideal diode under the forward biased Forward Biased state can be referred from its first property of possessing zero resistance. This is because threshold voltage is the minimum voltage which is required to be provided to the diode to overcome its barrier potential and to start conducting. Now, if the ideal diode is void of depletion region itself, then the question of threshold voltage does not arise at all. This property of the ideal diode makes them conduct right at the instant of being biased, leading to the green-curve of diode characteristics BE Basic Electronics Prof. Uvesh Sipai 33   Infinite Resistance Characteristics of Ideal Diode An ideal diode is expected to fully inhibit the flow of current when Reverse through it under reverse biased condition. Biased In other words it is expected to mimic the behavior of a perfect insulator when reverse biased. BE Basic Electronics Prof. Uvesh Sipai 34   Zero Reverse Leakage Current Characteristics of Ideal Diode This property of the ideal diode can be directly implied from when Reverse its previous property which states that the ideal diodes possess Biased infinite resistance when operating in reverse biased mode. The reason can be understood by considering the Ohm’s law again which now takes the form Thus it means that there will be no current flowing through the ideal diode when it is reverse biased, no matter how high the reverse voltage applied be. BE Basic Electronics Prof. Uvesh Sipai 35   No Reverse Breakdown Voltage Characteristics Re