DC Circuit Analysis PDF

Electrical Engineering Department Basic of Electrical Engineering (3110005) DC Circuit Analysis Presented By Prof. Naitik Trivedi GANDHINAGAR INSTITUTE OF TECHNOL...

Electrical Engineering Department Basic of Electrical Engineering (3110005) DC Circuit Analysis Presented By Prof. Naitik Trivedi GANDHINAGAR INSTITUTE OF TECHNOLGY Prof. Naitik Trivedi, EE 1 Department  Outline 1) DC circuit introduction 2) Different Sources and components like Resistor, Inductor and Capacitor 3) Ohm’s law and examples 4) KCL and KVL with node and mesh analysis 5) Current divider and Voltage divider rules 6) Superposition Theorem with numericals 7) Thevenin’s Theorem with numericals 8) Norton’s Theorem with numericals Prof. Naitik Trivedi, EE 2 Department  What is DC circuit? Direct current (DC) circuits basically consist of a loop of conducting wire (like copper) through which an electric current flows. An electric current consists of a flow of electric charges, analogous to the flow of water (water molecules) in a river. In addition to the copper wire in a circuit there usually are components such as resistors which restrict the flow of electric charge, similar to the way rocks and debris in a river restrict the flow of the river water. Prof. Naitik Trivedi, EE Department 3 Continue.. Fig 1 Common DC circuit diagram is shown in figure containing resistors and battery. Prof. Naitik Trivedi, EE Department 4 Voltage 𝑊𝑜𝑟𝑘 𝐷𝑜𝑛𝑒 𝑊 Voltage is measured in V= = volt ( V ). 𝐶ℎ𝑎𝑟𝑔𝑒 𝑄 Current 𝐶ℎ𝑎𝑟𝑔𝑒 𝑄 Current is measured in I= = amperes ( A ). 𝑡𝑖𝑚𝑒 𝑡 Power and Energy 𝑒𝑛𝑒𝑟𝑔𝑦 𝑊 Energy is measured in P= = joules ( J ) and power, in 𝑡𝑖𝑚𝑒 𝑡 watts ( W ). 𝑊𝑄 P= = 𝑉𝐼 Prof. Naitik Trivedi, EE Department 𝑄 𝑡 5  Voltage source A voltage source is a two terminal device which can maintain a fixed voltage. An ideal voltage source can maintain the fixed voltage independent of the load resistance or the output current. However, a real-world voltage source cannot supply unlimited current. A voltage source is the dual of a current source. Real-world sources of electrical energy, such as batteries, generators, and power systems, can be modeled for analysis purposes as a combination of an ideal voltage source and additional combinations of impedance elements. Prof. Naitik Trivedi, EE Department 6 Cont.. Fig 2 A schematic diagram of a real voltage source, V, driving a resistor, R, and creating a current I. Prof. Naitik Trivedi, EE Department 7  Fig 3.1 Cont.. Ideal Voltage Source Battery of cells Controlled Voltage Source Single cell Prof. Naitik Trivedi, EE Department 8  Current sources A current source is an electronic circuit that delivers or absorbs an electric current which is independent of the voltage across it. There are two types – an independent current source (or sink) delivers a constant current. A dependent current source delivers a current which is proportional to some other voltage or current in the circuit. Prof. Naitik Trivedi, EE 9 Department  Fig 3.2 Cont.. Controlled Current Source Ideal Current Source Prof. Naitik Trivedi, EE Department 10  Dependent and independent source Dependent sources:- In the theory of electrical networks, a dependent source is a voltage source or a current source whose value depends on a voltage or current somewhere else in the network. Dependent sources are useful, for example, in modeling the behavior of amplifiers. A bipolar junction transistor can be modeled as a dependent current source whose magnitude depends on the magnitude of the current fed into its controlling base terminal. Prof. Naitik Trivedi, EE Department 11  Classification Dependent sources can be classified as follows: a)Voltage-controlled voltage source: The source delivers the voltage as per the voltage of the dependent element. b)Voltage-controlled current source: The source delivers the current as per the voltage of the dependent element. c)Current-controlled current source: The source delivers the current as per the current of the dependent element. d)Current-controlled voltage source: The source delivers the voltage as per the current of the dependent element. Prof. Naitik Trivedi, EE 12 Department  Circuits Voltage-controlled voltage source Current controlled current source Voltage controlled current source Current controlled voltage source Prof. Naitik Trivedi, EE Fig 4 13 Department  Independent sources An independent voltage source maintains a voltage (fixed or varying with time) which is not affected by any other quantity. Similarly an independent current source maintains a current (fixed or time-varying) which is unaffected by any other quantity. The usual symbols are shown in figure Prof. Naitik Trivedi, EE 14 Department  Linear elements The resistance, inductance or capacitance offered by an element which changes linearly with the change in applied voltage or circuit current, the element is termed as linear element. Prof. Naitik Trivedi, EE V-I Characteristics of linear and non-linear elements15 Department  Non-linear elements A non-linear circuit element is one in which the current does not change linearly with the change in applied voltage. A semiconductor diode operating in the curved region of characteristics as shown in fig. is common example of non-linear element. Other example of non-linear elements are voltage depended resistance (VDR), voltage-dependent capacitor (varactor), temperature – dependent resistor (thermistor), light- dependent resistor (LDR), etc. linear elements obey ohm’s law where non- linear elements do not obey ohm’s law. Prof. Naitik Trivedi, EE 16 Department  Active elements An element which is a source of electrical signal or which is capable of increasing the level of signal energy is termed as active element. Battery, BJTs, FETs or Op-AMPs are treated as active elements because these can be used for the amplification or generation of signals. Prof. Naitik Trivedi, EE 17 Department  Passive elements Circuit elements, such as resistors, capacitors, inductors, VDR, LDR, thermistors, etc. The behavior of active elements can not be described by Ohm’s law. Prof. Naitik Trivedi, EE 18 Department  Resistance Resistance is the property of a material due to which it opposes the flow of electric current through it. Conductors – Metals, Acids and Salt solutions Insulators – Mica, Glass, Rubber, Bakelite Unit of resistance is ohm & represented by the symbol Ω. Prof. Naitik Trivedi, EE 19 Department  Resistance The resistance of a conductor depends on the following factors. – It is directly proportional to its length. – It is inversely proportional to the area of cross section of the conductor. – It depends on the nature of the material. – It also depends on the temperature of the conductor. Prof. Naitik Trivedi, EE 20 Department  Resistance Hence, 𝑙 𝑅∝ 𝐴 𝑙 𝑅= 𝜌 𝐴 Where, l is the length of the conductor. A is the cross-sectional area and 𝜌 is a constant known as specific resistance or resistivity of material Prof. Naitik Trivedi, EE 21 Department  Power dissipated in a Resistor We know that V=Ri Power absorbed by resistor is P=Vi Power dissipated in resistor is converted to heat which is given by 𝑡 𝑡 𝐸= 𝑣. 𝑖 𝑑𝑡 = 𝑅𝑖 𝑖 𝑑𝑡 = 𝑖 2 𝑅 𝑡 0 0 Prof. Naitik Trivedi, EE 22 Department Prof. Naitik Trivedi, EE 23 Department  SMR Prof. Naitik Trivedi, EE 24 Department  Inductor An inductor, also called a coil, choke, or reactor, is a passive two-terminal electrical component that stores energy in a magnetic field when electric current flows through it. The magnetic flux linkage generated by a given current depends on the geometric shape of the circuit. Prof. Naitik Trivedi, EE 25 Department  Current-Voltage relationships in inductor 𝑑𝑖 𝑡 𝑣=𝐿 1 𝑑𝑡 𝑖 𝑡 = 𝑣 𝑑𝑡 𝐿 1 0 𝑑𝑖 = 𝑣 𝑑𝑡 𝐿 Integrating both the sides, Energy stored in an inductor 𝑑𝑖 𝑖(𝑡) 𝑡 𝑣=𝐿 𝑑𝑡 1 𝑑𝑖 = 𝑣 𝑑𝑡 Energy supplied to the inductor 𝐿 𝑖(0) 0 during interval dt is given by 𝑡 1 𝑑𝑖 𝑖 𝑡 = 𝑣 𝑑𝑡 + 𝑖(0) 𝑑𝐸 = 𝑣 𝑖 𝑑𝑡 = 𝐿 𝑖 𝑑𝑡 = 𝐿 𝑖 𝑑𝑖 𝐿 𝑑𝑡 0 Prof. Naitik Trivedi, EE 26 Department  Current-Voltage relationships in inductor 𝐻𝑒𝑛𝑐𝑒 𝑡𝑜𝑡𝑎𝑙 𝑒𝑛𝑒𝑟𝑔𝑦 𝑠𝑢𝑝𝑝𝑙𝑖𝑒𝑑 𝑡𝑜 𝑡ℎ𝑒 𝑖𝑛𝑑𝑢𝑐𝑡𝑜𝑟 𝑤ℎ𝑒𝑛 𝑐𝑢𝑟𝑟𝑒𝑛𝑡 𝑖𝑠 𝑖𝑛𝑐𝑟𝑒𝑎𝑠𝑒𝑑 𝑓𝑟𝑜𝑚 0 𝑡𝑜 𝐼 𝑎𝑚𝑝𝑒𝑟𝑒𝑠 𝑖𝑠 𝐼 𝐼 1 2 𝐸= 𝑑𝐸 = 𝐿 𝑖 𝑑𝑖 = 𝐿 𝐼 2 0 0 Prof. Naitik Trivedi, EE 27 Department  Capacitor A capacitor is a device that stores electrical energy in an electric field. It is a passive electronic component with two terminals. Also defined as the ratio of the positive or negative charge Q on each conductor to the voltage V between them: Prof. Naitik Trivedi, EE 28 Department  Current-Voltage relationships in capacitor 𝑞 = 𝐶𝑣 𝑡 𝑑𝑞 𝑑 𝑑𝑣 1 𝑖= = 𝐶𝑣 = 𝐶 𝑣 𝑡 = 𝑖 𝑑𝑡 𝑑𝑡 𝑑𝑡 𝑑𝑡 𝐶 0 Expressing capacitor voltage Energy stored in an capacitor as a function of current, 𝑑𝑣 1 𝑖=𝐶 𝑑𝑣 = 𝑖 𝑑𝑡 𝑑𝑡 𝐶 Energy supplied to the capacitor Integrating both the sides, 𝑣(𝑡) 𝑡 during interval dt is given by 1 𝑑𝑣 = 𝑖 𝑑𝑡 𝐶 𝑑𝑣 𝑣(0) 0 𝑑𝐸 = 𝑣𝑖 𝑑𝑡 = 𝐶 𝑣 𝑑𝑡 𝑑𝑡 𝑡 1 v 𝑡 = 𝑖 𝑑𝑡 + 𝑣(0) 𝐶 Prof. Naitik Trivedi,0EE 29 Department  Current-Voltage relationships in capacitor 𝐻𝑒𝑛𝑐𝑒 𝑡𝑜𝑡𝑎𝑙 𝑒𝑛𝑒𝑟𝑔𝑦 𝑠𝑢𝑝𝑝𝑙𝑖𝑒𝑑 𝑡𝑜 𝑡ℎ𝑒 𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑜𝑟 𝑤ℎ𝑒𝑛 𝑝𝑜𝑡𝑒𝑛𝑡𝑖𝑎𝑙 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝑖𝑠 𝑖𝑛𝑐𝑟𝑒𝑎𝑠𝑒𝑑 𝑓𝑟𝑜𝑚 0 𝑡𝑜 𝑉 𝑣𝑜𝑙𝑡𝑠 𝑖𝑠 𝑉 𝑉 1 𝐸= 𝑑𝐸 = 𝐶𝑣 𝑑𝑣 = 𝐶 𝑣 2 2 0 0 Prof. Naitik Trivedi, EE 30 Department Prof. Naitik Trivedi, EE 31 Department  Ohm’s Law According to Ohm’s law, “The Current flowing between any two points on a conductor is directly proportional to the potential difference across them, provided the physical condition do not change.” (ex. material length, cross-sectional area and temperature of the conductor remain constant. ) 𝐼∝𝑉 𝑉 = constant 𝐼 V=IR Prof. Naitik Trivedi, EE 32 Department  Ohm’s Law Limitations – Ohm’s law does not apply to non-metallic conductors. – Ohm’s law also does not apply to non-linear devices such as zener diodes, etc. – Ohm’s law is true for metal conductors at constant temperature. If the temperature changes, the law is not applicable. Prof. Naitik Trivedi, EE 33 Department  Essential Terms – Node : A node is a point in the network where two or more circuit element are connected together. Referring to Fig., we find that A,B,C and D qualify as nodes in respect of the above definition. – Junction: A junction is that point in a network, where three or more circuit elements are joined. In Fig., we find that B and D are the junctions. – Branch: A branch is that part of a network which lies between two junction points. In Fig., BAD,BCD and BD qualify as branches. Prof. Naitik Trivedi, EE 34 Department  Essential Terms – Loop: A loop is any closed path of a network. Thus, in Fig., ABDA,BCDB and ABCDA are the loops. Prof. Naitik Trivedi, EE 35 Department  Kirchhoff’s Current Law States that the algebraic sum of currents meeting at any node is always zero. If the currents entering in to the node is (+ve) If the the currents leaving from the node is (-ve) which states that the sum of currents sum of currents entering a node = sum of currents leaving the entering node a node is equal to the sum of currents leaving the Prof. Naitik Trivedi, EE 36 node. Department  Kirchhoff’s Voltage Law States that the algebraic sum of voltages around any closed path in a circuit is zero. In general, the mathematical representation of Kirchhoff’s voltage law is Prof. Naitik Trivedi, EE 37 Department  Steps For Solving any Electrical Network using Kirchhoff’s Law Step1. Assume the current direction if not given. Step2. Mark +ve and –ve sign to the each end of resistance.  Mark the +ve sign when the current entering in to resistance.  Mark the –ve sign when the current leaving from the resistance. Step3. while traveling +ve to –ve there is a fall in potential  While traveling from negative (-ve) to positive (+ve) there is a rise in potential. Prof. Naitik Trivedi, EE 38 Department  Find the value of V1 and V2 using Kirchhoff’s law in below circuit. Prof. Naitik Trivedi, EE 39 Department Prof. Naitik Trivedi, EE 40 Department Prof. Naitik Trivedi, EE 41 Department Consider the circuit shown in figure. Find each branch current and voltage across each branch when R1 = 8Ω V2 = 10 volts i3 = 2A and R3 = 1Ω. Also find R2. Prof. Naitik Trivedi, EE 42 Department Consider the circuit shown in figure. Find each branch current and voltage across each branch when R1 = 8Ω V2 = 10 volts i3 = 2A and R3 = 1Ω. Also find R2. Prof. Naitik Trivedi, EE 43 Department Consider the circuit shown in figure. Find each branch current and voltage across each branch when R1 = 8Ω V2 = 10 volts i3 = 2A and R3 = 1Ω. Also find R2. Prof. Naitik Trivedi, EE 44 Department  (a)= 4A (b)= 0A (c)= Not Possible Prof. Naitik Trivedi, EE 45 Department  (a)= 12 V, (b)= -2.167 V Prof. Naitik Trivedi, EE 46 Department Prof. Naitik Trivedi, EE 47 Department Prof. Naitik Trivedi, EE 48 Department Prof. Naitik Trivedi, EE 49 Department Prof. Naitik Trivedi, EE 50 Department Prof. Naitik Trivedi, EE 51 Department Prof. Naitik Trivedi, EE 52 Department Prof. Naitik Trivedi, EE 53 Department Prof. Naitik Trivedi, EE 54 Department Prof. Naitik Trivedi, EE 55 Department Prof. Naitik Trivedi, EE 56 Department Prof. Naitik Trivedi, EE 57 Department Prof. Naitik Trivedi, EE 58 Department  Network Simplification Techniques Series Circuit Parallel Circuit Star – Delta conversion Node and Mesh analysis Prof. Naitik Trivedi, EE 59 Department  Series and Parallel Circuits Series Circuit Two or more circuit elements are said to be in series if the current from one element exclusively flows into the next elements. All series elements have the same current. Series Resistors Equivalent series resistance: 𝑵 𝑹𝑬𝑸 = 𝑹𝒏 = 𝑹𝟏 + 𝑹𝟐 + ⋯ + 𝑹𝑵 𝒏=𝟏 Prof. Naitik Trivedi, EE 60 Department  Series Circuits For the circuit shown, a) Find the equivalent resistance seen by the source. b) Find the current I. c) Calculate the voltage drop in each resistor. d) Calculate the power dissipated by each resistor. e) Find the power output of the source. Given: V = 24 V, R1 = 1 , R2 = 3 , and R3 = 4 . Prof. Naitik Trivedi, EE 61 Department  Series Circuits Solution: a) 𝑅𝐸𝑄 = 𝑅1 + 𝑅2 + 𝑅3 REQ =1+3+4 =8 𝑉 24 𝑉 b) 𝐼 = = =3𝐴 𝑅𝐸𝑄 8Ω c) 𝑉1 = 𝐼𝑅1 = 3 𝐴 1 Ω = 3 𝑉 𝑉2 = 𝐼𝑅2 = 3 𝐴 3 Ω = 9 𝑉 𝑉3 = 𝐼𝑅3 = 3 𝐴 4 Ω = 12 𝑉 Note: 𝑉1 + 𝑉2 + 𝑉3 = 3 + 9 + 12 = 24 𝑉 The total voltage drop is equal to the voltage output of the source. Prof. Naitik Trivedi, EE 62 Department  Series Circuits d) 𝑃1 = 𝐼2 𝑅1 𝑃2 = 𝐼2 𝑅2 𝑃3 = 𝐼2 𝑅3 2 2 = 3𝐴 1Ω = 3𝐴 2 3Ω = 3𝐴 4Ω =9𝑊 = 27 𝑊 = 36 𝑊 e) 𝑃 = 𝐼𝑉 = 3 𝐴 24 𝑉 = 72 𝑊 Note: 𝑃1 + 𝑃2 + 𝑃3 = 9 + 27 + 36 = 72 𝑊 The total power dissipated by the resistors is the same as the power output by the source. Prof. Naitik Trivedi, EE 63 Department  Parallel Circuits Parallel Circuit Two or more circuit elements are said to be in parallel if the elements share the same terminals. All parallel elements have the same voltage. Parallel Resistors Equivalent parallel resistance: 𝟏 𝟏 𝟏 𝟏 or = + + ⋯+ 𝑹𝑬𝑸 𝑹𝟏 𝑹𝟐 𝑹𝑵 𝟏 𝑹𝑬𝑸 = 𝟏 + 𝟏 + ⋯ + 𝟏 𝑹𝟏 𝑹𝟐 𝑹𝑵 where Prof. Naitik Trivedi, EE 𝑉1 = 𝑉2 = 𝑉3 = 𝑉4 64 Department  Parallel Circuits Various Parallel Resistors Networks 1 2 3 4 5 Prof. Naitik Trivedi, EE 65 Department  Parallel Circuits For the circuit shown, a) Find the equivalent resistance seen by the source. b) Find the total current I. c) Calculate the currents in each resistor. d) Calculate the power dissipated by each resistor. e) Find the power output of the source. Given: V = 24 V, R1 = 1 , R2 = 3 , and R3 = 4 . Prof. Naitik Trivedi, EE 66 Department  Parallel Circuits I REQ Solution: 1 1 1 1 1 1 1 a) = + + = + + = 1.583 Ω 𝑅𝐸𝑄 𝑅1 𝑅2 𝑅3 1 3 4 1 ∴ 𝑅𝐸𝑄 = = 0.632 Ω 1.583 𝑉 24 𝑉 b) 𝐼= = = 37.975 𝐴 ≈ 38 𝐴 𝑅𝐸𝑄 0.632 Ω Prof. Naitik Trivedi, EE 67 Department  Parallel Circuits d) 𝑃1 = 𝐼1 2 𝑅1 𝑃2 = 𝐼2 2 𝑅2 𝑃3 = 𝐼3 2 𝑅3 2 = 24 𝐴 1Ω = 8𝐴 2 3Ω = 6𝐴 2 4Ω = 576 𝑊 = 192 𝑊 = 144 𝑊 e) 𝑃 = 𝐼𝑉 = 38 𝐴 24 𝑉 = 912 𝑊 Note: 𝑃1 + 𝑃2 + 𝑃3 = 576 + 192 + 144 = 912 𝑉 The total power dissipated by the resistors is the same as the power output by the source. Prof. Naitik Trivedi, EE 68 Department Voltage Divider Rule (VDR) VDR is useful in determining the voltage drop across a resistance within a series circuit. + V1 - 𝑹𝑿 + 𝑽𝑿 = 𝑽𝑺 V2 𝑹𝑬𝑸 S - + V3 - where VX = the voltage drop across the measured resistor, RX = the resistance value of the measured resistor, REQ = the circuit total resistance, VS = the circuit applied voltage Prof. Naitik Trivedi, EE 69 Department  Determine the voltage across the R2 and the R3. Solution: 𝑅2 20 𝑉2 = 𝑉 = 60 = 20 V 𝑅𝐸𝑄 𝑆 10 + 20 + 30 𝑅3 30 𝑉3 = 𝑉𝑆 = 60 = 30 V 𝑅𝐸𝑄 10 + 20 + 30 Prof. Naitik Trivedi, EE 70 Department Current Divider Rule (CDR) CDR is useful in determining the current flow through one branch of a parallel circuit. 𝑹𝑬𝑸 𝑰𝑿 = 𝑰𝑺 𝑹𝑿 where IX = the current flow through any parallel branches, RX = the resistance of the branch through which the current is to be determined, REQ = the total resistance of the parallel branch, IS = the circuit applied current Prof. Naitik Trivedi, EE 71 Department  For the particular case of two parallel resistors, 𝑹𝟏 𝑹𝟐 I 𝑹𝑬𝑸 = 𝑹𝟏 𝑹𝟐 = 𝑹𝟏 + 𝑹𝟐 I1 I2 and, the current passing through R1 and R2 are 𝑅𝐸𝑄 𝑅1 𝑅2 𝑹𝟐 𝐼1 = ∙𝐼 = 𝑅1 +𝑅2 ∙𝐼  𝑰𝟏 = ∙𝑰 𝑅1 𝑅1 𝑹𝟏 + 𝑹𝟐 𝑅1 𝑅2 𝑅𝐸𝑄 𝑹𝟏  𝑅1 +𝑅2 𝐼2 = ∙𝐼 = ∙𝐼 𝑰𝟐 = ∙𝑰 𝑅2 𝑅2 𝑹𝟏 + 𝑹𝟐 Prof. Naitik Trivedi, EE Note: It only works for two parallel resistors. 72 Department  Find each of the branch currents in the figure shown below. Solution: 𝑅1 𝑅2 𝑅3 1 1 1 1 1 1 1 1 = + + = + + = 𝑅𝐸𝑄 𝑅1 𝑅2 𝑅3 3 𝑘Ω 8 𝑘Ω 24 𝑘Ω 2 𝑘Ω 𝑅𝐸𝑄 = 2 𝑘Ω Prof. Naitik Trivedi, EE Department 73  Thus, 𝑅𝐸𝑄 2 𝑘Ω 𝐼𝑅1 = ∙𝐼 = 12 𝑚𝐴 = 8 𝑚𝐴 𝑅1 3 𝑘Ω 𝑅𝐸𝑄 2 𝑘Ω 𝐼𝑅2 = ∙𝐼 = 12 𝑚𝐴 = 3 𝑚𝐴 𝑅2 8 𝑘Ω 𝑅𝐸𝑄 2 𝑘Ω 𝐼𝑅3 = ∙𝐼 = 12 𝑚𝐴 = 1 𝑚𝐴 𝑅3 24 𝑘Ω 𝐼𝑅1 + 𝐼𝑅2 + 𝐼𝑅3 = 8 + 3 + 1 = 12 𝑚𝐴 ( KCL  ) Prof. Naitik Trivedi, EE 74 Department  Inductor The function of the inductor is to store the magnetic field. The ratio of flux linkage to the current flowing through the coil is known as inductance. Unit is henry and represented by the symbol H. A coil is said to have an inductance of one henry if a current of one ampere when flowing through it produces flux linkages of one weber-turn in it. Prof. Naitik Trivedi, EE 75 Department  Inductance The inductance of an inductor depends on the following factors. – Directly proportional to the square of the number of turns. – Directly proportional to the area of cross section. – Inversely proportional to the length. – Depends on the absolute permeability of magnetic material. Hence, 𝑁2 𝐴 Where, 𝐿∝ 𝑙 𝜇 is the absolute permeability of magnetic material 𝑁2 𝐴 𝐿=𝜇 𝑙 Prof. Naitik Trivedi, EE 76 Department Prof. Naitik Trivedi, EE 77 Department  Capacitance The function of the capacitor is to store the charge. Capacitance is the property of a capacitor to store an electric charge when its plates are at different potentials. 𝑄 𝐶= 𝑉 Unit is farad and represented by symbol F. Capacitor is said to be capacitance of one farad if a charge of one coulomb is required to establish a potential difference of one volt between its plates. Prof. Naitik Trivedi, EE 78 Department  Capacitance The capacitance of capacitor depends on the following factors. – Directly proportional to the are of the plates. – Inversely proportional to the distance between two plates. – Depends on the absolute permittivity of the medium between the plates. Hence, 𝐴 𝐶∝ 𝑑 𝐴 𝐶=𝜀 𝑑 Where, d is distance between two plates, Prof. Naitik Trivedi, EE 79 Department  Star & Delta connection circuit The Y-Δ transform, also written wye-delta and also known by many other names, is a mathematical technique to simplify the analysis of an electrical network. The name derives from the shapes of the circuit diagrams, which look respectively like the letter Y and the Greek capital letter. This circuit transformation theory was published by Arthur Edwin Kennelly in 1899. It is widely used in analysis of three-phase electric power circuits. The Y-Δ transform can be considered a special case of the star- mesh transform for three resistors. Prof. Naitik Trivedi, EE 80 Department  Cont.. The transformation is used to establish equivalence for networks with three terminals. Where three elements terminate at a common node and none are sources, the node is eliminated by transforming the impedances. For equivalence, the impedance between any pair of terminals must be the same for both networks. The equations given here are valid for complex as well as real impedances. Equations for the transformation from Δ-load to Y-load 3- phase circuit The general idea is to compute the impedance at a terminal node of the Y circuit with impedances , to adjacent node in the Δ circuit by Prof. Naitik Trivedi, EE 81 Department  A A Ra R R ca ab Rc Rb C B B R bc C Star Connection Delta Connection Prof. Naitik Trivedi, EE 82 Department  Delta to Star Transformation: A A Rca Rab Ra Rc Rb C B B R bc C In Delta : Equivalent Resistance between A and B R ab (R bc  R ca )  R ab in parallel with (R bc  R ca )  R ab  R bc  R ca In Star : Equivalent Resistance between A and B  Ra  Rb R ab (R bc  R ca ) Therefore, R a  R b  R ab  R bc  R ca Prof. Naitik Trivedi, EE 83 Department  R ab (R bc  R ca ) Ra  Rb  R ab  R bc  R ca R bc (R ca  R ab ) Similarly, R b  R c  R ab  R bc  R ca R ca (R ab  R bc ) Rc  Ra  R ab  R bc  R ca Add any two equations and subtract the third equation: R ab R ca A Ra  R ab  R bc  R ca A Rca Rab R bc R ab Rb  Ra R ab  R bc  R ca C Rc Rb B R ca R bc R bc Rc  C B R ab  R bc  R ca Prof. Naitik Trivedi, EE 84 Department Star to Delta Transformation: Multiplying the above equations, 2 R ab R ca R bc RaRb  R ab  R bc  R ca 2 A A Ra 2 Rca Rab R bc R ab R ca Rc Rb R bR c  R ab  R bc  R ca  2 C B C B R bc 2 R ca R ab R bc R cR a  R ab  R bc  R ca 2 Adding equations, R ab R bc R ca R ab  R bc  R ca  R a R b  R bR c  R cR a  R ab  R bc  R ca 2 Prof. Naitik Trivedi, EE 85 Department  Star to Delta Transformation: Adding equations, R ab R bc R ca R ab  R bc  R ca  R a R b  R bR c  R cR a  R ab  R bc  R ca 2 R ab R bc R ca   R bc R a  R ca R b  R ab R c R ab  R bc  R ca  Ra.Rb A Rab  Ra  Rb  Rc A Ra Rb.Rc Rca Rab Rbc  Rb  Rc  Rc Rb Ra C B C B R.R R bc Rca  Rc  Ra  c a Rb Prof. Naitik Trivedi, EE 86 Department  Find the equivalent resistance between A and B in the network shown. Prof. Naitik Trivedi, EE 87 Department  Find the equivalent resistance between A and B in the network shown. Prof. Naitik Trivedi, EE 88 Department  Find the equivalent resistance between A and B in the network shown. Prof. Naitik Trivedi, EE 89 Department  Find the equivalent resistance between A and B in the network shown. Prof. Naitik Trivedi, EE 90 Department  Find the equivalent resistance between A and B in the network shown. Redrawing the network Prof. Naitik Trivedi, EE 91 Department  Find the equivalent resistance between A and B in the network shown. Prof. Naitik Trivedi, EE 92 Department  Steps of Nodal Analysis 1. Choose a reference (ground) node. 2. Assign node voltages to the other nodes. 3. Apply KCL to each node other than the reference node; express currents in terms of node voltages. 4. Solve the resulting system of linear equations for the nodal voltages. Prof. Naitik Trivedi, EE 93 Department  Calculate the current through 2 Ω resistor for the network shown in fig. Prof. Naitik Trivedi, EE 94 Department  Calculate the current through 2 Ω resistor for the network shown in fig. Assume that the current are moving away from nodes. Applying KCL at node A, 𝑉𝐴 − 20 𝑉𝐴 𝑉𝐴 − 𝑉𝐵 + + =0 1 1 0.5 1 1 1 1 20 + + 𝑉𝐴 − 𝑉𝐵 = 1 1 0.5 0.5 1 4𝑉𝐴 − 2𝑉𝐵 = 20 …(i) Applying KCL at node B, 𝑉𝐵 − 𝑉𝐴 𝑉𝐵 𝑉𝐵 − 20 + + =0 0.5 2 1 1 1 1 1 20 − 𝑉 + + + 𝑉 = 0.5 𝐴 0.5 2 1 𝐵 1 −2𝑉𝐴 + 3.5𝑉𝐵 = 20 …(ii) Prof. Naitik Trivedi, EE 95 Department  Calculate the current through 2 Ω resistor for the network shown in fig. Assume that the current are moving away from nodes. Applying KCL at node A, 𝑉𝐴 − 20 𝑉𝐴 𝑉𝐴 − 𝑉𝐵 + + =0 1 1 0.5 1 1 1 1 20 + + 𝑉𝐴 − 𝑉𝐵 = 1 1 0.5 0.5 1 4𝑉𝐴 − 2𝑉𝐵 = 20 …(i) Applying KCL at node B, 𝑉𝐵 − 𝑉𝐴 𝑉𝐵 𝑉𝐵 − 20 + + =0 0.5 2 1 1 1 1 1 20 − 𝑉 + + + 𝑉 = 0.5 𝐴 0.5 2 1 𝐵 1 −2𝑉𝐴 + 3.5𝑉𝐵 = 20 …(ii) Prof. Naitik Trivedi, EE 96 Department  Calculate the current through 2 Ω resistor for the network shown in fig. Solving Eqs (i) and (ii), 𝑉𝐴 = 11 𝑉 𝑉𝐵 = 12 𝑉 Current through the 2 Ω resistor 𝑉𝐵 12 = =6A 2 2 Prof. Naitik Trivedi, EE 97 Department  Find the voltage at nodes 1 and 2 for the network shown in fig. Prof. Naitik Trivedi, EE 98 Department  Find the voltage at nodes 1 and 2 for the network shown in fig. Assume that the current are moving away from nodes. Applying KCL at node 1, 𝑉1 − 𝑉2 𝑉1 + =1 2 2 1 1 1 + 𝑉 − 𝑉2 = 1 2 2 1 2 𝑉1 − 0.5𝑉2 = 1 …(i) Applying KCL at node 2, 𝑉2 − 𝑉1 𝑉2 + =2 2 1 1 1 − 𝑉1 + 1 + 𝑉 =2 2 2 2 −0.5𝑉1 + 1.5𝑉2 = 2 …(ii) Prof. Naitik Trivedi, EE 99 Department  Find the voltage at nodes 1 and 2 for the network shown in fig. Solving Eqs (i) and (ii), 𝑉1 = 2 𝑉 𝑉2 = 2 𝑉 Prof. Naitik Trivedi, EE 100 Department  Find the current in the 100 Ω resistor for the network shown in fig. Prof. Naitik Trivedi, EE 101 Department  Find the current in the 100 Ω resistor for the network shown in fig. Assume that the current are moving away from nodes. Applying KCL at node 1, 𝑉1 − 𝑉2 𝑉1 − 60 + =1 30 20 1 1 1 60 + 𝑉 − 𝑉 = +1 20 30 1 30 2 20 0.083 𝑉1 − 0.033 𝑉2 = 4 …(i) Applying KCL at node 2, 𝑉2 − 𝑉1 𝑉2 − 40 𝑉2 + + =0 30 50 100 1 1 1 1 40 − 𝑉1 + + + 𝑉2 = 30 30 50 100 50 −0.033 𝑉1 + 0.063 𝑉2 = 0.8 …(ii) Prof. Naitik Trivedi, EE 102 Department  Find the current in the 100 Ω resistor for the network shown in fig. Solving Eqs (i) and (ii), 𝑉1 = 67.25 𝑉 𝑉2 = 48 𝑉 Current through the 100 Ω resistor 𝑉1 48 = = 0.48 𝐴 100 100 Prof. Naitik Trivedi, EE 103 Department  Find the voltage drop across the 5 Ω resistor in the network shown in fig. Prof. Naitik Trivedi, EE 104 Department  Find the voltage drop across the 5 Ω resistor in the network shown in fig. Assume that the current are moving away from nodes. Applying KCL at node 1, 𝑉1 − 𝑉2 𝑉1 − 𝑉3 1+ + =0 5 4 1 1 1 1 + 𝑉1 − 𝑉2 − 𝑉3 = −1 5 4 5 4 0.45 𝑉1 − 0.2 𝑉2 − 0.25 𝑉3 = −1 …(i) Applying KCL at node 2, 𝑉2 − 𝑉1 𝑉2 + =2 5 1 1 1 − 𝑉1 + + 1 𝑉2 = 2 5 5 −0.2 𝑉1 + 1.2 𝑉2 = 2 …(ii) Prof. Naitik Trivedi, EE 105 Department  Find the voltage drop across the 5 Ω resistor in the network shown in fig. Applying KCL at node 3, 𝑉3 − 𝑉1 𝑉3 + +2=0 4 2 1 1 1 − 𝑉1 + + 𝑉 = −2 4 2 4 3 −0.25 𝑉1 + 0.75 𝑉3 = −2 …(iii) Solving Eqs (i), (ii) and (iii), 𝑉1 = −4 𝑉 𝑉2 = 1 𝑉 𝑉3 = −4 𝑉 𝑉5Ω = 𝑉2 − 𝑉1 = 1 − −4 = 5 𝑉 Prof. Naitik Trivedi, EE 106 Department  Use the node voltage method to find how much power the 2A source extracts from the circuit shown in Fig. Prof. Naitik Trivedi, EE 107 Department  Use the node voltage method to find how much power the 2A source extracts from the circuit shown in Fig. Prof. Naitik Trivedi, EE 108 Department  Refer the circuit shown in Fig. (a) Use the node voltage method to find the branch currents i1 to i6. (b) Test your solution for the branch currents by showing the total power dissipated equals the power developed. Prof. Naitik Trivedi, EE 109 Department  Refer the circuit shown in Fig. (a) Use the node voltage method to find the branch currents i1 to i6. (b) Test your solution for the branch currents by showing the total power dissipated equals the power developed. Prof. Naitik Trivedi, EE 110 Department  Refer the circuit shown in Fig. (a) Use the node voltage method to find the branch currents i1 to i6. (b) Test your solution for the branch currents by showing the total power dissipated equals the power developed. Prof. Naitik Trivedi, EE 111 Department  Refer the circuit shown in Fig. (a) Use the node voltage method to find the branch currents i1 to i6. (b) Test your solution for the branch currents by showing the total power dissipated equals the power developed. Prof. Naitik Trivedi, EE 112 Department  Refer the circuit shown in Fig. (a) Use the node voltage method to find the branch currents i1 to i6. (b) Test your solution for the branch currents by showing the total power dissipated equals the power developed. Prof. Naitik Trivedi, EE 113 Department  Mesh Analysis Mesh analysis: another procedure for analyzing circuits, applicable to planar circuit. A Mesh is a loop which does not contain any other loops within it. Prof. Naitik Trivedi, EE 114 Department  A circuit with two meshes. Prof. Naitik Trivedi, EE 115 Department  Apply KVL to each mesh. For mesh 1,  V1  R1i1  R3 (i1  i2 )  0 ( R1  R3 )i1  R3i2  V1 For mesh 2, R2i2  V2  R3 (i2  i1 )  0  R3i1  ( R2  R3 )i2  V2 Prof. Naitik Trivedi, EE 116 Department  Solve for the mesh currents.  R1  R3  R3   i1   V1     R3 R2  R3  i2   V2  Use i for a mesh current and I for a branch current. It’s evident from Fig. that I1  i1 , I 2  i2 , I 3  i1  i2 Prof. Naitik Trivedi, EE 117 Department  Find the branch current I1, I2, and I3 using mesh analysis. Prof. Naitik Trivedi, EE 118 Department  For mesh 1,  15  5i1  10(i1  i2 )  10  0 3i1  2i2  1 For mesh 2, 6i2  4i2  10(i2  i1 )  10  0 i1  2i2  1 Prof. Naitik Trivedi, EE 119 Department  Use mesh analysis to find the current I0 in the circuit. Prof. Naitik Trivedi, EE 120 Department  Apply KVL to each mesh. For mesh 1,  24  10(i1  i2 )  12(i1  i3 )  0 11i1  5i2  6i3  12 For mesh 2, 24i2  4(i2  i3 )  10(i2  i1 )  0  5i1  19i2  2i3  0 Prof. Naitik Trivedi, EE 121 Department  For mesh 3, 4 I 0  12(i3  i1 )  4(i3  i2 )  0 At node A, I 0  I1  i2 , 4(i1  i2 )  12(i3  i1 )  4(i3  i2 )  0  i1  i2  2i3  0 In matrix from become  11  5  6  i1  12  5 19  2 i2    0    1  1 2  i   0    3    we can calculate i1, i2 and i3 by Cramer’s rule, and find I0. Prof. Naitik Trivedi, EE 122 Department  Mesh Analysis with Current Sources A circuit with a current source. Prof. Naitik Trivedi, EE 123 Department  Case 1 – Current source exist only in one mesh i1  2A – One mesh variable is reduced Prof. Naitik Trivedi, EE 124 Department  Equivalent Circuits:- Source Transformation Rs + Vs Is Rs - Vs Vs  Rs I s Is  Rs PageEENo: Prof. Naitik Trivedi, 2.61 self making from Circuits and Networks (U.A.Patel) 125 Department  Up till now we seen following Methods of Analysis Introduction Nodal analysis Nodal analysis with voltage source Mesh analysis Mesh analysis with current source Prof. Naitik Trivedi, EE 126 Department  Superposition Theorem:- How to Apply Superposition? To find the contribution due to an individual independent source, zero out the other independent sources in the circuit. – Voltage source  short circuit. – Current source  open circuit. Solve the resulting circuit using your favorite techniques. – Nodal analysis – Loop analysis Prof. Naitik Trivedi, EE 127 Department  Superposition For the above case: Zero out Vs, we have : Zero out E2, we have : I I2’’ I2 ’ R1 R3 R2 R1 R3 R2 + R2 R3 E2 Vs_ E2 R2 / / R3  R2  R3 R1 R2  R2 R3  R1 R3 R1   R1 / / R3   R1 R3 R1 / / R3  R 21  R3 R2  R3 R1 R2  R2 R3  R1R3 Vs  R2  R3  R1   R1 / / R3   I R1  R3 R1 R2  R2 R3  R1 R3 E2  R1  R3  R3 Vs  R2  R3  I 2   I 2  I   R1 R2  R2 R3  R1R3 R2  R3 R1 R2  R2 R3  R1 R3 Prof. Naitik Trivedi, EE 128 Department  Superposition 12V 2k 4mA - + 2mA 1k 2k I0 selfEEmaking Prof. Naitik Trivedi, from Circuits and Networks (U.A.Patel) 129 Department  Superposition I 0    I 2  I1  I1  2mA 2k KVL for mesh 2:  I 2  I1  1k  I 2  2k  0 2mA I1 1k 2k 1 2 I2 I 2  I1   mA 3 3  2  I’o Mesh 2 I 0    I 2  I1       2   3  4   mA 3 selfEEmaking Prof. Naitik Trivedi, from Circuits and Networks (U.A.Patel) 130 Department  Superposition P2.7 I 0   I 2 I1 KVL for mesh 2: 2k 4mA I 2 1k   I 2  I1   0  I 2  2k  0 I2  0 1k I2 2k I o  0 Mesh 2 I’’0 selfEEmaking Prof. Naitik Trivedi, from Circuits and Networks (U.A.Patel) 131 Department  Superposition P2.7 12V I o   I 2 2k KVL for mesh 2: - + I 2 1k  12V  I 2  2k  0 1k 2k 12 I2 I2   4mA I’’’0 1k  2k Mesh 2 I o  4mA selfEEmaking Prof. Naitik Trivedi, from Circuits and Networks (U.A.Patel) 132 Department Superposition 12V 2k 4mA - + 2mA 1k 2k I0 I0 = I’0 +I’’0+ I’’’0 = -16/3 mA Prof. Naitik Trivedi, EE Department self making from Circuits and Networks (U.A.Patel) 133  Thevenin's theorem Any circuit with sources (dependent and/or independent) and resistors can be replaced by an equivalent circuit containing a single voltage source and a single resistor Thevenin’s theorem implies that we can replace arbitrarily complicated networks with simple networks for purposes of analysis Prof. Naitik Trivedi, EE 134 Department  Thevenin’s theorem Independent Sources RTh + Voc - Circuit with independent Thevenin equivalent sources circuit selfEEmaking Prof. Naitik Trivedi, from Circuits and Networks (U.A.Patel) 135 Department  Thevenin’s theorem No Independent Sources RTh Circuit without independent sources Thevenin equivalent circuit selfEEmaking Prof. Naitik Trivedi, from Circuits and Networks (U.A.Patel) 136 Department Prof. Naitik Trivedi, EE 137 Department Prof. Naitik Trivedi, EE 138 Department