Unit 1: DC Circuits

UNIT I 1. DC CIRCUITS 1.1 BASIC DEFINITIONS 1.1.1 Charge (Q or q): Charge is the characteristic property of the elementary particles of the matter. The elementary particles are electron, proton and neutron. There are two types of charges in nature. They one positive and negative charges. The symbol of charge is ‘Q’ or ‘q’. The unit of charge is Coulomb. 1 coulomb = charge on 6.28 X 1018 electrons. 1.1.2 Electric Current (I): The flow of free electrons in a metal is called Electric Current. The unit of Current is “Ampere” A. Or The flow of free electrons in any conductor is called Electric Current. It is also defined as the rate of change of charge with respect to time. (ie) Current (I) = Q . t or I= dq . dt Q is measured in Coulomb. t is measured in Seconds. 1.1.3 Voltage (V): The difference in potential energy between the charges is called potential difference. In electrical terminology the potential difference is called voltage. or The potential difference between two points in an electric circuit is called voltage. 1 The unit of voltage is volt. It is represented by “V” or “v”. 1.1.4 Electro-motive force (emf): It is the force which causes to flow the electrons in any closed circuit. The unit of electro motive force is Volt. 1.1.5 Power: Power is the rate of doing work. In other words, power is the work done per unit time. Power = work done time The unit of power is joules/sec or watts. Power can also be expressed as (kw) kilo watts. Some times power is measured in Horse power (HP) = 735 watts. (Mechanical Horse power) Power P = VI watts (or) P = E.I (or) P = (IR)I = I2R watts. V = IR Also, I =  power, V R  V  V2   watts. P V   R   R Note:1) 1 KW = 1000 watts. 2) 1 HP = 746 watts. (Electrical Horse power) 1.1.6 Energy: Energy is the capacity to do work.  Energy = Power  time = p  t Unit of electric energy is watt sec. or joules. Since the watt-sec. is too small for practical use, the Kilo Watt Hour (kwh) is generally used. 1.1.7 Torque (T): Torque is the twisting moment or turning moment of a circulating body. The Unit of torque is Newton-meter. [1kg-m = 9.81 N-m]. 1.1.8 Resistance (R): The opposition offered by a substance to the flow of electric current is called Resistance. 2 1) Resistance may also be defined as the physical property of a substance due to which it opposes (or) restricts the flow of electricity through it. The Unit of resistance is Ohm (). 2) A device has a resistance of 1 Ohm if a current 1 A flows when a potential difference of 1 V exists across it. Potential Difference V = I  R = 11 = 1V 1.1.9 Conductance (G): Conductance is the inducement to the flow of current, while resistance is the opposition to the flow of current. Hence, it is the reciprocal of resistance. i.e., (G  1/ R) The Unit of conductance is mho . 1.2 OHM’S LAW: This law states that “the current flowing through a conductor is directly proportional to voltage across the conductor, provided the temperature remains constant.” “At constant temperature, the potential difference between the two ends of a conductor is directly proportional to the current flowing through the conductor.” ie., I  V V  IR Where, R is the resistance of the conductor. The relationship between potential difference (V), Current (I) and resistance (R) in a d.c circuit is first discovered by the scientist “George Simon Ohm” is called ohm’s law. 1.2.1 Limitation of Ohm’s law: 1) This law is not applicable for all non-metallic conductors or semi conductors. 2) Not applicable for non-linear device such as zener diodes, arc lamps etc. 3) If the temperature changes, ohm’s law is not applicable for metallic conductors also. 3 1.2.2 Special note on Ohm’s Law: Ohm’s law can be expressed in three forms: V i.e.,  R; I V  IR; V I  . R 1.3 Classification of Circuit Elements: 1. Active elements and 2. Passive elements. 1) Active elements: These elements supply Voltage or Current to the circuit to operate it. They are Voltage sources and Current sources. Examples: Generators, Transistor, Vaccum tubes etc., 2) Passive elements: These elements either dissipate energy in the form of heat or store energy from the external sources. Examples: Resistor (R), Inductance (L) and Capacitor (C). Resistor (R) Inductor (L) Capacitor (C) - Dissipate energy in the form of heat. Stores energy in its magnetic field. Stores energy in its electrostatic field. Problems: 1.1 When a current flowing to a conductor is 2 Amps and the resistance is 5. Find the voltage across the conductor? Given: R  5 I2A To find V =? Formula, V = IR [As per ohm’s Law] V  2 5 V  10V 4 1.2 Find the resistance value when the current flows at 3 amps and 18 volt? Given: V  18 volt I  3 amps. To find R =? As per ohm’s law, V 18 R   6 . I 3  R  6 . 1.3 Find the current flowing through a conductor. Resistance is 20  and Voltage is 40 V. Given: R = 20, V = 40 V, I = ? V 40 I   2 A. R 20 1.4 A resistor carrying a current of 5A has a voltage of 6 V across it. What is its resistance and what power does it dissipate? Given: V=6V I=5A Solution: Resistance, R  V I  6  1.2. 5 P  VI Power = 6  5 = 30w P  30w Also, P  I 2R = 52  1.2 = 25  1.2 = 30 watts. (or) V 2 62  R 1.2 P  30 watts. 1.4 Electrical Network Circuit: P 5 Any arrangement of electrical energy source (i.e. batteries, generator, etc.,) resistance and other circuit elements (e.g., inductance, capacitance) is called as electrical network. Symbols of various Circuit Elements: Active element: The elements which supply electrical energy (or) can able to generate electric energy is called active elements. e.g., (a) to (d). Passive element: Elements which consumes electrical energy or store energy is called passive elements. e.g., (e) to (g). 1.5 Network Terminology 6 DC Circuit: The closed path followed by an electric current (DC) is called an electrical circuit. The essential parts of an electrical circuit are the source of power (battery, generator, etc.) the conductor to carry the current and the loads. Load: The device that utilses the electrical energy, e.g., lamp, motor, heater, etc. Load may be a part of an electrical circuit (or) may not. Parameters: The various elements of an electric circuit are called its parameter (i.e., resistance, capacitance, inductance, etc.). They may be in groups (or) separated in a circuit. Node: Node is a junction where two or more circuit elements can connect together. Minor node Point at which two elements are connected. Major node Point at which three or more element are connected. Branch: It is a part of a circuit which lies between two junction points. Loop: A loop is any closed path of a circuit. Mesh: A mesh is the most elementary form of a loop and cannot be further divided into other loops. Problems 1.5 The resistance of a 230 V lamp is 270. Calculate the current taken by the lamp. Solution: Given: Resistance of the lamp, R  270. Voltage of the lamp, V  230V. To find: Current of the lamp, I = ? By Ohm’s law, we know that VI R I V R  230  0.85 Amps. 270  The current taken by the lamp = 0.85 Amps. 7 1.6 An electric lamp consumes 100 watts of power. The supply Voltage is 230 V. Determine a) the current flowing through the filament b) its resistance c) the energy consumed in 45 minutes. Solution: Given: Power consumed by electric lamp, P = 100 watts. Supply Voltage, V = 230 V. Time, T = 45 minutes. a) Current flowing, I P from [P  V  I ] V 100 I = 230  I  0.434 Amps. b) Resistance of filament, R V from [P  V  I ] I 230 R= 0.434  R  529.95. c) Energy consumed in 45 min, E  P T  100  45 60 E  75 watts (or) 0.07 kw. 1.6 Kirchoff’s Law: Gustov Kirchoff a German scientist gave his findings with electrical circuit in a set of two laws. 1. Kirchoff’s Current law or First law, 2. Kirchoff’s Voltage law or Second law. 8 1.6.1 Kirchoff’s Current law (KCL) Statement: “The algebraic sum of currents meeting a junction or node in an electrical circuit is zero”. I = 0 Explanation and Velocity Consider the figure shown. In KCL the sign of the quantity is taken into account. From the figure, there are five currents I1, I2, I3, I4 and I5 meeting at a point or node “0”. Assume, the current flowing towards “0” is positive and the current flowing away from “0” is negative. Applying Kirchoff’s Current law at the junction “0” we get, (I1 )  (I2 )  (I3 )  (I4 )  (I5 )  0 I1  I2  I3  I4  I5  0 I1  I3  I5  I2  I4 From the above equation, the sum of incoming current is equal to the sum of the outgoing current. Now the statement of KCL can also be written as “the sum of the currents flowing towards any junction in an electrical circuit is equal to the sum of the current flowing away from the junction”. Validity of KCL KCL is true because electric current is merely flow of electrons and they cannot accumulate at any point in the circuit. 1.6.2 Kirchoff’s Voltage Law Statement: “In a closed circuit the sum of the potential drops is equal to the sum of the potential rises” (or) “In a closed circuit or mesh, the algebraic sum of all the emf’s plus the algebraic sum of all the voltage drops is zero”.  Algebraic Sum   0. i.e., Algebraic Sum    of all emf of all Voltage drops      Mathematically,  E  V  0. 9 Consider the below circuit, Consider the entire loop ABCDA, Sum of emf = E the potential drop across AB, VAB = I . R1 the potential drop across BC, VBC = I . R2 the potential drop across CD, VCD = I . R3 Sum of potential drops = I . R1 + I . R2 + I . R3 By KVL,  emf   Voltage drop = 0 E  VAB  VBC  VCD  0 i.e., E  IR1  IR2  IR3  0  E  IR1  IR2  IR3. i.e., The KVL can also be stated as “In a closed circuit or mesh, the algebraic sum of all emfs is equal to the algebraic sum of all voltage drops”. While applying KVL, algebraic sum are involved. So it is necessary to assign proper signs to the emfs and to the voltage drops. 1.7 Sign Conversion: Sign conversions are very important while considering Kirchoff’s Voltage Law. It is very important to assign proper signs to emfs and voltage drop. In order to assign proper signs both loop current and branch current direction are considered. 1.7.1 sign of emf: if we go from negative terminal of the battery or source to the positive terminal then there is rise in potential and so the emf should be given positive sign. Emf is either  denoted as     E or V  10 Therefore + E (or) + V. If we go from positive terminal of the battery (or) source to the negative terminal, then there is fall in potential and so the emf should be assigned negative sign.  (– ve) = – E (or) – V. 1.7.2 Sign of Voltage drops: When the loop current and branch current are flowing in the same direction through a resistor then there is a fall in potential. So the sign of the voltage drop across the resistor is negative. Fall in voltage or fall in potential. When the loop current and branch current are flowing opposite to each other through a resistor then there is a rise in potential. So the sign of the voltage drop across the resistor is positive. Rise in voltage or rise in potential. Illustration – 1: Using KVL solve the circuit below:- Step1:- Mark the branch current direction, 11 Note:Always the branch current will flow from the positive terminal of the battery or source. Step2:- Mark the loop current direction, Note:Always assume the loop current in clock wise direction. Step 3: Apply KVL for ABCDA Vs  I1R1  I 1 R2  0 Vs  I1 (R1  R2 )  0 Vs  I1 (R1  R2 ) Illustration 2 Apply KVL for the above circuit and assume the loop current direction in anti-clockwise direction. Step1 Mark the nodes as ABCDA, Mark the branch current direction. 12 Step2 Mark the loop current direction i.e., in anti-clockwise direction as given in questions. Step3 Vs  I1R1  I 1 R2  0 Vs  I1 (R1  R2 )  0 Vs  I1 (R1  R2 ) Simple problems on KVL and KCL 1.7 For the given circuit apply Kirchoff’s law and find the current flowing through it? Step1 Mark the nodes as ABCDA, Mark the branch and loop current direction. Step2 Apply KVL, Vs  I1R1  I1R2  0 Vs  I1 (R1  R2 )  0 10  I1 (2  6)  0 8I1  10  I1  1.25Amps 13  Current through each element:𝐼2Ω = 𝐼6Ω = 𝐼1 = 1.25 Amps  Voltage drop across each element 𝑉2Ω = 𝐼2Ω. 𝑅2Ω = 1.25 ∗ 2 = 2.5 𝑉 𝑉6Ω = 𝐼6Ω. 𝑅2Ω = 1.25 ∗ 6 = 7.5 𝑉 1.8 Find the current through 2Ω resistor? Also find the current in each branch and current through 35 V and 40 V source [NOV - 2009] Solution Step1 Mark the current as ABCDEFA. Mark the branch and loop current. Step2 Since the above circuit has two loops (i.e), ABCFA & CDEFC we can apply KVL for each loop or mesh. Apply KVL for the mesh ABCFA, 14 35 − 3𝐼1 − 2(𝐼1 + 𝐼2) = 0 35 − 3𝐼1 − 2𝐼1 − 2𝐼2 = 0 35 − 5𝐼1 − 2𝐼2 = 0 −5𝐼1 − 2𝐼2 = −35 ∴ 5𝐼1 + 2𝐼2 = 35 −−−−1 Apply KVL for the mesh CDEFC, −40 + 4𝐼2 + 2(𝐼1 + 𝐼2) = 0 −40 + 4𝐼2 + 2𝐼1 + 2𝐼2 = 0 −40 + 2𝐼1 + 6𝐼2 = 0 2𝐼1 + 6𝐼2 = 40 − − − − − − − 2 Step3 Solving the equation 1 and 2 by matrix method 5𝐼1 + 2𝐼2 = 35 −−−−1 2𝐼1 + 6𝐼2 = 40 − − − − − − − 2 [𝑅][𝐼] = [𝑉] 5 2 𝐼1 35 [ ] [𝐼 ] = [40] 2 65 2 2 ∆=| | = 26 2 6 35 2 ∆𝐼1 = | | = 130 40 6 5 35 ∆𝐼2 = | | = 130 2 40 15 ∴ 𝐼1 = ∴ 𝐼2 = ∆𝐼1 ∆ ∆𝐼2 ∆ = = 130 26 130 26 =5A =5A Step4 Current through 2Ω resistor, 𝐼2Ω = 𝐼1 + 𝐼2 =5+5 𝐼2Ω = 10 Amps Step5 Current through each branch, In branch AB = 𝐼1 = 5 A BC = 𝐼2 = 5 A CD = 𝐼2 = 5 A DE = 𝐼2 = 5 A EF = 𝐼2 = 5 A FA = 𝐼1 = 5 A CF = 𝐼1 + 𝐼2 = 10 Amps Step6 Current through the source 35V and 40 V 𝐼35 = 𝐼1 = 5 A 𝐼40 = 𝐼2 = 5 A 1.9 A battery of emf 40 V and the internal resistance of 2 Ω is connected in parallel with a second battery of 44 V and internal resistance of 4 Ω. A load resistance of 6 Ω is connected across the ends of the parallel circuit. Calculate the current in each battery and in the load? [April/May 2011; April 2010] Solution The circuit is, 16 Step1 Mark the circuit as ABCDEFA, Mark the branch and loop currents [No. of Mesh = No. of Currents] Step2 Apply KVL for the ABCFA, 40 − 2 𝐼1 − 6 (𝐼1 + 𝐼2) = 0 40 − 2 𝐼1 − 6𝐼1 − 6𝐼2 = 0 40 − 8 𝐼1 − 6𝐼2 = 0 −8 𝐼1 − 6𝐼2 = 0 ∴ 8𝐼1 + 6𝐼2 = 40 (1) Apply KVL for CDEFC, −44 + 4𝐼2 + 6(𝐼1 + 𝐼2) = 0 −44 + 4𝐼2 + 6𝐼1 + 6𝐼2 = 0 6𝐼1 + 10 𝐼2 = 44 17 (2) Step3 Solve the equation 1 and 2 by matrix method and find the value of the Currents I1and I2. 𝐼 40 [8 6 1 ] [ ] =[ ] 6 10 𝐼2 44 8 6 ∆= | | = 80 − 36 = 44 6 10 40 6 ∆𝐼 = | | = 400 − 264 = 136 1 44 10 8 40 ∆𝐼2= | | = 352 − 240 = 112 6 44 ∆𝐼1 = 3.09 A ∴ ∆𝐼1 = ∆ ∆𝐼2 = 2.54 A ∴ ∆𝐼2 = ∆ Step4 To find the current through 40 V and 44 V battery, 𝐼40v = 𝐼1 = 3.09 A 𝐼44v = 𝐼2 = 2.54 A Step5 Current through the 6Ω load, 𝐼6Ω = 𝐼1 + 𝐼2 = 3.09 + 2.54 ∴ 𝐼6Ω = 5.63 Amps 1.10 Find the loop currents, current through each element and current across the voltage source, power delivered by 100 V and 40 V source. [April 2011; Nov 2011] Step1 Mark the circuit as ABCDEFGHA, Mark the loop currents and branch current directions. No. of mesh = No. of currents No. of mesh = 3 i.e., (ABCHA; CDGHC; DEFGD) No. of currents = I1, I2 and I3. 18 Step2 1) Apply KVL for the loop ABCHA, 100 − 8𝐼1 − 4(𝐼1 − 𝐼2) = 0 100 − 8𝐼1 − 4𝐼1 + 4𝐼2 = 0 100 − 12𝐼1 + 4𝐼2 = 0 −12𝐼1 + 4𝐼2 = −100 ∴ 12𝐼1 − 4𝐼2 = 100 2. Apply KVL for the loop ABCHA, −10𝐼2 − 10(𝐼2 + 𝐼3) + 4(𝐼1 − 𝐼2) = 0 −10𝐼2 − 10𝐼2 − 10𝐼3 + 4𝐼1 − 4𝐼2 = 0 4𝐼1 − 24𝐼2 − 10𝐼3 = 0 (2) 3. Apply KVL for the loop DEFGD, 19 (1) −40 + 4𝐼3 + 10(𝐼2 + 𝐼3) = 0 −40 + 4𝐼3 + 10𝐼2 + 10𝐼3 = 0 (3) 10𝐼2 + 14𝐼3 = 40 Step3 Solve the equation 1, 2 and 3 by matrix method, [𝑅][𝐼] = [𝑉] 12 −4 0 𝐼1 100 [4 − 24 − 10] [𝐼2] = [ 0 ] 0 10 14 𝐼3 40 12 −4 0 ∆ = |4 − 24 − 10| = −2608. 0 10 14 100 −4 0 ∆𝐼1 = | 0 − 24 − 10 | = −2200. 40 10 14 12 100 0 ∆𝐼2 = | 4 0 − 10 | = −800 0 40 14 12 − 4 100 ∆𝐼3 = | 4 − 24 0 | = −6880 0 10 40 ∆𝐼1 = 8.4 A 𝐼1 = ∆ ∆𝐼2 = 0.306 A 𝐼2 = ∆ ∆𝐼3 = 2.63 A 𝐼3 = ∆ ∴ Loop Currents are 𝐼1 = 8.4 A 𝐼2 = 0.306 A 𝐼3 = 2.63 A Step4 (i) Current through each element, 𝐼8Ω = 𝐼1 = 8.4 A 𝐼4Ω = 𝐼1 − 𝐼2 = 8.4 − 0.306 = 8.094 A 𝐼10Ω = 𝐼2 = 0.306 A 𝐼10Ω = 𝐼2 + 𝐼3 = 0.306 + 2.63 = 2.936 A 20 (ii) Current through the voltage sources 𝐼100 = 𝐼1 = 8.4 A 𝐼40 = 𝐼3 = 2.36 A (iii) Power delivered by 100 V and 40 V source, 𝑃100 = 𝑉1 ∗ 𝐼100𝑉 = 100 ∗ 8.4 = 840 watts 𝑃40 = 𝑉2 ∗ 𝐼40𝑉 = 40 ∗ 2.63 = 105.2 watts 1.11 Find the current through each element and also find the current and power developed from the battery. Solution: Step1 Mark the circuit or ABCDEFGHA, Mark the loop and branch current directions No. of mesh = 3 No. of currents = 𝐼1, 𝐼2 & 𝐼3 i.e, No. of mesh = No. of Currents Step2 Apply KVL, (i) For mesh ABCHA, 21 90 − 30𝐼1 − 30 (𝐼1 − 𝐼2) = 0 90 − 30𝐼1 − 30𝐼1 + 30𝐼2 = 0 90 − 60𝐼1 + 30𝐼2 = 0 −60𝐼1 + 30𝐼2 = −90 ∴ 60𝐼1 − 30𝐼2 = 90 (1) (ii) Apply KVL for CDGHC, −10𝐼2 − 30(𝐼2 − 𝐼3) + 30 (𝐼1 − 𝐼2) = 0 −10𝐼2 − 30𝐼2 + 30𝐼3 + 30𝐼1 − 30𝐼2 = 0 30𝐼1 − 70𝐼2 + 30𝐼3 = 0 (2) (iii) Apply KVL for DEFGD, −30𝐼3 − 30𝐼3 + 30(𝐼2 − 𝐼3) = 0 −30𝐼3 − 30𝐼3 + 30𝐼2 − 30 𝐼3 = 0 −90𝐼3 + 30𝐼2 = 0 ∴ 30𝐼2 − 90𝐼3 = 0 (3) Step3 Find the currents 𝐼1, 𝐼2 & 𝐼3 by using matrix, [𝑅][𝐼] = [𝑉] 90 60 − 30 0 𝐼1 𝐼 [ 30 ] [ ] = [ − 70 30 0] 2 0 30 − 90 𝐼3 0 60 − 30 0 △= | 30 − 70 30 | = 243000 0 30 − 90 90 − 30 0 △ 𝐼1 = | 0 − 70 30 | = 486000 0 30 − 90 60 90 0 △ 𝐼2 = | 30 0 30 | = 243000 0 0 − 90 60 − 30 90 △ 𝐼3 = |30 − 70 0| = 81000 0 30 0 ∆𝐼1 486000 = = 2 Amps ∴ 𝐼1 = ∆ 243000 ∆𝐼2 243000 = = 1 Amps 𝐼2 = ∆ 243000 ∆𝐼3 81000 = = 0.333 Amps 𝐼3 = ∆ 243000 Step4 (i) Current supplied through the 90 V battery, 𝐼90 = 𝐼1 = 2 Amps (ii) Power generated through the 90 V battery, 22 𝑃90𝑉 = 𝑉. 𝐼 = 90 ∗ 2 = 180 watts (iii) Current through each element, 𝐼30Ω = 𝐼1 = 2 Amps (BC Branch) 𝐼30Ω = 𝐼1 − 𝐼2 = 2 − 1 = 1 Amps (CH Branch) 𝐼10Ω = 𝐼2 = 1 Amps (CD Branch) 𝐼30Ω = 𝐼2 − 𝐼3 = 1 − 0.333 = 0.667 Amps (DG Branch) 𝐼30Ω = 𝐼3 = 0.333 Amps (DE Branch) 𝐼30Ω = 𝐼3 = 0.333 Amps (EF Branch) 1.12 Find the current through each element? Solution Step1 Name the circuit and draw the loop and branch current directions, No. of loops = No. of currents = 3 From the fig, in order to mark the current through the branch HG, 𝐼1 − 𝐼3 + 𝐼2 + 𝐼3 = 0 𝐼1 + 𝐼2 = 0 ∴ The current through HG branch = 𝐼1 + 𝐼2. Step2 Apply KVL in all the 3 mesh, 1. Apply KVL in ABHGA, 23 10 − 0.4 (𝐼1 − 𝐼3) − 1.1(𝐼1 + 𝐼2) = 0 10 − 0.4 𝐼1 + 0.4𝐼3 − 1.1𝐼1 − 1.1𝐼2 = 0 10 − 1.5 𝐼1 − 1.1𝐼2 + 0.4𝐼3 = 0 −1.5 𝐼1 − 1.1𝐼2 + 0.4𝐼3 = − 10 1.5 𝐼1 + 1.1𝐼2 − 0.4𝐼3 = 10 (1) 2. Apply KVL in HEFGH, −16 + 0.4 (𝐼2 + 𝐼3) + 1.1(𝐼1 − 𝐼3) = 0 −16 + 0.4𝐼2 + 0.4𝐼3 + 1.1𝐼1 + 1.1𝐼2 = 0 (2) 1.1𝐼1 + 1.5𝐼2 + 0.4𝐼3 = 16 3. Apply KVL in BCDEHB, −2𝐼3 − 0.4(𝐼2 + 𝐼3) + 0.4(𝐼1 − 𝐼3) = 0 −2𝐼3 − 0.4𝐼2 − 0.4𝐼3 + 0.4𝐼1 − 0.4𝐼3 = 0 (3) 0.4𝐼1 − 0.4𝐼2 − 2.8𝐼3 = 0 Step3 Solving the equation 1, 2 and 3 by using matrix 1.5𝐼1 + 1.1𝐼2 − 0.4𝐼3 = 10 1.1𝐼1 + 1.5𝐼2 − 0.4𝐼3 = 16 0.4𝐼1 − 0.4𝐼2 − 2.8𝐼3 = 0 [𝑅][𝐼] = [𝑉] 10 1.5 1.1 − 0.4 𝐼1 𝐼 [ 1.1 1.5 0.4 ] [ 2] = [16] 0.4 − 0.4 − 2.8 𝐼3 0 1.5 1.1 − 0.4 ∆= | 1.1 1.5 0.4 | = −2.08 0.4 − 0.4 − 2.8 24 (1) (2) (3) 10 1.1 − 0.4 ∆𝐼1 = | 16 1.5 0.4 | = 11.44 0 − 0.4 − 2.8 1.5 10 − 0.4 ∆𝐼2 = | 1.1 16 0.4 | = −32.24 0.4 0 − 2.8 1.5 1.1 10 ∆𝐼3 = |1.1 1.5 16| = 6.24 0.4 − 0.4 0 ∆𝐼1 11.44 = = −5.5 Amps ∴ 𝐼1 = ∆ −2.08 ∆𝐼2 −32.24 = = 15.5 Amps 𝐼2 = ∆ −2.08 ∆𝐼3 6.24 = = −3 Amps 𝐼3 = ∆ −2.08 Step4 Current through each element, 𝐼0.4Ω = 𝐼1 − 𝐼3 = −5.5— 3 = −5.5 + 3 = −2.5 Amps 𝐼1.1Ω = 𝐼1 + 𝐼2 = −5.5 − 15.5 = 10 Amps 𝐼0.4Ω = 𝐼2 + 𝐼3 = 15.5 − 3 = 12.5 Amps 𝐼2Ω = 𝐼3 = −3 Amps 1.13 Find using Kirchoff’s law, the voltage drop across the middle resistance? + + - - Solution Step1 Name the nodes or joints Mark the branch and loop current directions. 25 No. of Mesh = No. of Current = 2 (i.e., I1 and I2) Step 2 i. Apply KVL for ABCFA, 100 10I1  2(I1  I2 )  0 10I1  2I1  2I2  100 12I1  2I2  100 12I1  2I2  100 (1) ii. Apply KVL for CDEFC, 50  5I2  2(I1  I2 )  0 5I2  2I1  2I2  50 2I1  7I 2  50 (2) Step 3 Solving equations 1 and 2 by matrix, RI   V  12 2  I1  100 2 7  I   50     2    12 2   84  4  80 2 7 I1  100 2  700 100  600 50 7 12 100  600  200  400 2 50 600 I  7.5 Amps 1 80 400 I  5 Amps 2 80 I2  Step 4 Current and voltage drop through middle resistance i.e., 2 I2  I1  I2  7.5  5  12.5 Amps V2  I2  R2  12.5 2  25 V. 1.14 In the bridge network AB = 6; BC = 12; CD = 10; DA = 3; BD = 20  and as an emf. of 2V. Calculate the I delivered to the battery? 26 i) Apply KVL for loop ABDA, 6I1  20I3  3I2  0 6I1  3I2  20I3  0 (1) ii) Apply KVL for loop BCDB, iii) 20I3 12(I1  I3 ) 10(I2  I3 )  0 20I3 12I1 12I3 10I2 10I3  0 12I1 10I2  42I3  0 (2) Apply KVL for the loop ABCFA, 6I1 12(I1  I3 )  2  0 6I1 12I1 12I3  2  0 18I1 12I3  2 (3) Solve the equation 1, 2 and 3 by matrix method, 27 6 3  20 I1  0  12 10 42  I  = 0     2    12   I 3   2  18 0 6 3  20  = 12 0 42 =  6156 18 0 12 0 3  20 42  652 12 I1  0 10 2 0 6 0  20 I2  12 0 18 0 42  984 12 1.8 Resistance in Series and Voltage Division Technique The circuit in which resistances are connected end to end so that there is one path for the current flow is called series circuit. Fig. 1 Fig. 2 In series circuit, the current through all the three resistors are same but the voltage drop across the each is different. Let three resistances R1, R2 and R3 are connected in series across a battery of V volts. Let V1, V2 and V3 be the voltage drop across R1, R2 and R3. By Ohm’s Law V1 = Voltage drop across R1 = I . R1 V2 = Voltage drop across R2 = I . R2 (1) V3 = Voltage drop across R3 = I . R3 Therefore the total voltage drop across the circuit, V  V1 V2 V3 (2) Sub. 1 in 2, V  I R1  I R2  I R3 By ohm’s law, the total voltage V, 28 (3) V  I RT (4) Equate equation 3 and 4 we get, I RT  I R1  I R2  I R3 I RT  I (R1  R2  R3 )  RT  R1  R2  R3 Voltage Division Rule: From the above circuit, V  I RT (By Ohm's law) i.e., V  I (R1  R2  R3 ) V  I  (1) R1  R2  R3 Voltage drop across the R1 , V1  I R1 (2) Substitute equation 1 in 2 V1  V R1 R1  R2  R 3 Similarly, V2    V R2 R1  R2  R 3 V R R1  R2  R 3 3 Therefore the equations can be written in words as, Applied Voltage Voltage drop across a resistor =  Corresponding Resistance Total Resistance V3  Concept of Series circuit 1. The current flowing in all the parts of the circuit is same. 2. Voltage across the different elements will depend upon the resistance of elements. 3. Voltage drops are additive. 4. Resistances are additive. 5. Powers are additive. 6. Applied voltage equals to the sum of different voltage drops. Disadvantages of Series circuit 1. If break occurs at any point in the circuit, no current will flow and the entire circuit becomes useless. 2. It is not applicable for practical lighting circuits, since the supply voltage increases depending upon the number of lamps (or) elements connected. 3. Since electrical devices have different current ratings, they are not connected in series for efficient operations. 29 1.9 Resistance in Parallel and Current Division Technique Resistance in Parallel Combination If one end of all the resistors is joined to a common point and the other ends are joined to another common point, the combination is said to be a parallel combination between two common points. Let three resistances R1, R2 and R3 are connected in parallel across a battery of V volts as shown below in fig (a). The total current I divide into three parts, I1 flowing through R1 I2 flowing through R2 I3 flowing through R3 and The voltage drop across each resistance is same as that of supply voltage. By ohm’s law, V 1I  R1 V (1) 2I  R2 V 3I  R3 Total Current, I  I1  I 2  I 3 (2) Substitute 1 in 2, V V V   R1 R2 R3 1 1 1 I  V R  R  R   1 2 3  1 1 I 1     V R1 R2 R3 I By Ohm’s law, from fig. (b), 30 (3) I  1  V (4) RT Equating equation 3 and 4, we get, 1 RT  1 1 1   R1 R2 R3  The total resistance, 1 1 1    R R R   1 2 3  RT   1 Note If there are two resistance R1 & R2 , then RT  1 1 1  R1 R2 1 = R1 R2 R1R2  RT  R1R2 R R 1 2 Product of Resistances  R T Sum of Resistance Current Division Rule: Supply voltage  Voltage across R1 = Voltage R2 i.e.,V  I1R1  I2 R2  I1R1  I2 R2 IR (1)  I2  1 1 R2 Total Current, I  I1  I2 (2) 31 Substitute equation 1 in 2, I1R1  I  I1  R 2 I1R2  I1R1 = R2 I (R  R2 ) I 1 1 R2   Current through R1 , (3)  I.R2 I1  R  R 1 i.e., I1  2 Supply Current  Opposite Resistance Sum of Resistance Similarly, I2  I. R1 R1  R2 Concepts of Parallel Circuits: i. Voltage is same across all the elements. ii. Current through the various resistors are different. iii. Total resistances of a parallel circuit is always less than the smallest of the resistances. iv. Powers are additive. v. Source current is the sum of branch currents. vi. Conductance’s are additive. Advantages of parallel circuits: i. Electrical appliances of different power ratings may be rated for the same voltage. Example: In a house, tube lights, bulbs, fridge, fans, motor are all rated for 230 V. But their power ratings are different. In such cases, if they are all connected in parallel the performance of other devices will not be affected. ii. In case a break occurs in any of the branch circuits, it will not affect the other branch circuits. Comparison between series and parallel circuits Series Circuit 1. Current is same through all the elements. 2. Voltage is distributed. It is proportional to the resistance. Parallel Circuit Current is divided inversely proportional to resistance Voltage is same across each element in the parallel combination. 3. Total (or) equal resistance is equal to the sum of individual resistances. Reciprocal of equal resistance is equal to the sum of reciprocals of individual 32 i.e., R  R1  R2  R3 resistances. 1 1 1 1 i.e.,    R R1 R2 R3 4. There is only one path for flow of Current There are more than one path for the flow of current Illustration of how to solve series and Parallel circuits: Illustration 1:   The given circuit is a series circuit. The total resistance, RT  R1  R2  Because R1 and R2 are connected in series. Illustration 2:  It is a parallel circuit since R1 is connected in parallel with R2 RR  RT  1 2 R1  R2 Illustration 3:  It is a series circuit, since R1 is in series with R2  RT  R1  R2       Illustration 4:    It is a parallel circuit R1 is in parallel with R2 RR RT  1 2 R1  R2 Illustration 5: 33  It is a combination of parallel and series circuit. Hence we need to reduce the circuit in order to find the value of RT. Here R2 is in series with R3,  R2 , R3  R2  R3 Therefore the above circuit can be reduced as,  Here R1 and R2,3 are in parallel R R  RT  1 2,3 R1  R2,3 Therefore, the reduced circuit is, Illustration 6: R23  R2  R3 Since R2 and R3 are in series 34 The above circuit has two resistances R1 and R2,3 , which are connected in parallel with each other.  RT  R1 R2,3 R1  R2,3  Illustration 7:   It is a parallel circuit. R R RT  1 2 R1  R2 Problems on Series Circuit 1.15 In the circuit shown below, find the current, the voltage drop across each resistor and the power dissipated in resistor. Given R1  2; R2  6; R3  7. Solution: Total Resistance, R  R1  R2  R3  R  2  6  7  15 Supply voltage  V  45V V 45 Circuit current I    3A R 15 I  3A Voltage drop across 2 resistor V1  IR1  3 2  6V Voltage drop across 6 resistor V2  IR2  3 6  18V Voltage drop across 7 resistor V3  IR3  3 7  21V Power dissipated in R1 is P1  I 2 R1  32  2  18W Power dissipated in R is P  I 2 R  32 6  54W 2 2 2 Power dissipated in R3 is P3  I R3  32  7  63W 2 35 1.16 Three resistors R1, R2 and R3 are connected in series across a constant voltage of “V” volts. The voltage drop across R1 is 20 Volts. The power loss in R2 is 25 Watts and R3 has a resistance of 2. Find the voltage “V” if the current is 5 Amps. Solution Voltage across R1 is V1  20V Current, I  5A V1 20  1 I 5 R1  4 R Power loss in R2 is P2  25W i.e., P  I 2 R 2 R  2 2 P2 I 2  25 52 R2  1 Total resistance, RT  R1  R2  R3  4 1 2 RT  7 Supply voltage, V  IR  5 7 V  35Volts 1.17 A lamp can work on 50 volt mains taking 2 amps. What value of resistance must be connected in series such that it can be operated from 200 V mains giving the same power. 36 Solution Given Voltage, V  50V Current, I  2Amps To find r  ? Resistance of the lamp R  V I 50   25 2 R  25 Resistance connected in series with lamp  r Supply voltage  200V Circuit current, I  2 A V 200 Total resistance, RT    100 I 2 i.e., RT  R  r 100  25  r  r 100  25 r  75  R1 , R2 and R3 are connected in series. The power dissipated in P1 is 50 w. The resistance of R2 is 5. The current through the series circuit is 5 A. When 1.18 Three resistors the supply voltage is 100 v, determine the power dissipated in circuit and voltage R. across 3 Solution Given P1  50w R2  5 I  5A V  100volts To find, Power dissipated = ? Voltage across R3 = ?  Total Power dissipated  VI  I 2TR P 1005  500w 37 Power Dissipated across R1 , P1  I 2 R1  50w (given) 50 50 50   I 2 52 25 R1  2 R1  By Ohm’s law, V R   100  20 T I 5 RT  20 w.k.t., RT  R1  R2  R3  R3  RT  (R1  R2 )  20  (2  5)  13 V3  IR3  5(13)  65volts Problems on Parallel circuit 1.19 Three resistors of 6, 9 and 15 are connected in parallel to a 18 v supply, calculate (a) the current in each branch of the network, (b) the supply current, (c) the total resistance of the network. Solution By Ohm’s law, I  (i)  V R Current through 6 resistor is I 6  V 18  6  3A R 1 Current through 9 resistor is 9I  V 18   2A R2 9 Current through 15 resistor is I15  V 18   1.2A R2 15 (ii) By KCL, 38  V   R   1  Supply Current = Source Current I  3  2 1.2 I  6.2A (iii) By Ohm’s law, Total resistance, V 18 I 6.2  R  2.9 R   1.20 Three resistors 4, 12 and 6 are connected in parallel. The total current is 12 A find the voltage across the parallel circuit and current across each branch. 1 1 1 1    RT 4 12 6 6 = 12  RT  2 Potential difference across the parallel circuit, V  IR = 12  2 V  24volts Current through 4  I1   V R1 24  6A 4 39 V R2 Current through 12  I2   Current through 6  I3  24  2A 12 V R3 24  4A 6 1.21 Two resistors are connected in parallel and a voltage of 200 volts is applied to the terminals. The total current taken is 25A and the power dissipated in one of the resistors is 1500 w. What is the resistance of each element.  Solution Given V  200 Volts I  25 A Power dissipated in resistor R1 Say P1  1500 watts V2 But R1  P1     V 2    P  R V 2 (200)2   R1  P1 1500 40000   26.67 1500 R1  26.67 V R   200  8 Total 25 I resistor W.K.T., RR R  1 2  8 R1  R2 8  R1R2  R R  8(R  R ) 1 2 1 2 R1  R2 40 R1R2  8R1  8R2 R1R2  8R2  8R1 R2  R1  8  8R1 R2 8R1 R1  8 8(26.67) 213.36   26.67  8 18.67  R1  26.67 R2  11.43 R  26.67 1  Note R2 Can be calculated by the following method also. V I   200  7.5A 1 R 26.67  I  I1  I2 I2  I  I1  25  7.5 I2  17.5A V 200  R2    11.43 I 2 17.5 R2  11.43  1.22 Calculate the current supplied by the battery in the given circuit shown below Solution For easy understanding the given circuit can be re-drawn as shown below. There is no change in the connections. 41 R1 and R2 are in parallel across the voltage of 48 volts. R1R2 R1  R2 816  8 16 16 R   3  Equivalent circuit is The total resistance R   Current from battery by Ohm’s Law, I V R   48  16     3  I  9A Note From the circuit below, by applying ohm’s law to R1 and R2 , we get I1 and I2 V 48   6A R1 8 V 48   3A 2I  R2 16  I  I1  I2  6  3  9A I  9A I 1 1.11 Series-Parallel Circuits As the name suggests, this circuit is a combination of series and parallel circuits. A simple example of such a circuit is shown in fig. below. Note that R1 and R2 are connected in parallel with each other and that is together connected in series with R1 . One simple rule to solve such circuits is to first reduce the parallel branches to an equivalent series branch and then solve the circuit as a simple series circuit. 42 Referring to the above series-parallel circuit, R2 R3 RPfor parallel combination  R 1 RR 2 3 Total circuit resistance,  R1  R2 R3 RR 2 3 Voltage across parallel combination  I  R2 R3 R2  R3 Problems on Series-Parallel Combination 1.23 Find the equivalent resistance of the circuit shown below: Solution First reduce the last parallel combination 8 || 12 .  812 96   4.8 8 12 20  The circuit becomes 5 and 4.8 are in series.  5  4.8  9.8 [Eqn. Series Combination] 43 The circuit becomes Again the parallel combination is reduced as 6  9.8 58.8    3.7 6  9.8 15.8 R eqn = 3.7 1.24 A circuit consists of 2 parallel resistors, having resistances of 20  and 30 respectively, connected in series with 15. If current through resistor is 3 A, Find i) the current through 20  and 30 resistor, ii) the voltage across the whole circuit, iii) total power. Solution: (i) The total current 3A is divided between the 20  and 30 resistors as shown below. 30 Current through 20 , I1  3 (20  30) I1  1.8A Current through 30 , I  3 2 20 (20  30) I2  1.2A (ii) For parallel circuit, 20  30 600 RP  20  30 50 RP  12  The circuit becomes Total resistance 15 12 R eqn = RT  27  Supply voltage V  3 27 (iii)  81V Total power  VI or I 2 R  81 3  243 watts 44 1.25 Calculate the equivalent resistance of the following combination of resistors and also the source current Solution Start simplifying from inner combination Step1: Replace parallel combination of 12 and 6  to get the following circuit 12 ||16  12 parallel with 16  12 6 72   4 12  6 18 Step2: (4 + 16) and 40 are in parallel Their equivalent resistance is 20  40 20  40 Step3: 13.33 and 4 are in series, 45  800 60  13.3  RT  13.33  4 R T  17.33   Step4: I  Source Current V   50  2.885A RT 17.33 I  2.885A 1.26 Find the R equivalent and current flowing due to a source of 12v for the circuit shown. Solution By reducing the last parallel combination 510 50 R   3.33 P 5 10 15 RP  3.33  The circuit becomes, Now, 100  and 3.33 are in series, 100  3.33  103.33 46 Then the circuit becomes This parallel combination can be reduced by 50103.33   50 ||el 103.33  (50 103.33)  33.69  The two resistances are in series.  Rs  R1  R2  25  33.69 Req = Rs  58.69  V 12  0.204A R 58.69 I  0.204A Current I  1.27  Two resistors R1 and R2 are connected in parallel with total resistance of Obtain the expression for the currents in the individual resistor. If R2  0.05 and It  50A , Find I1 and I2 Solution Give data IT  50A R1  1.95 R2  0.05 To find I1 and I2 Total current It  I1  I2 R2 R 1 R 2   = 50   0.05  (1.95  0.05)    I1  It    47 It . R1 1.95 , 0.05  = 50    2  = 50  0.025 I1 = 1.25 R1 R 1 R 2   = 50   1.95  (1.95  0.05)    = 50  0.975  48.75A I2  48.75A I2  IT    1.28 What is the equivalent resistance of the network. Solution Since the last three resistances (1, 6 and 1) one series, we can redraw the circuit as shown by adding the 3 resistances. Now the two 8 are in parallel, Hence, the equivalent resistance of the parallel combination is 88 64   RP  8  8 16 = 4 48 Now the resistors 2, 4 and 2 are in series RP  8 8  4 88 Total Resistance = 1 + 4 + 1 = 6     Problems on Series and Parallel Circuits 1.29 Find the resistance across AB for the network Solution 49 RAB  RARB  RB RC  RC RA (9  9)  (9  9)  (9  9)  RC 9 RAB  27 RBC  27 RCA  27 Thus connecting star connection to delta 27 || 9 R1R2 R1  R2 27  9  27  9 RP  6.75 RP    RS  R1  R2 = 6.75+6.75 =13.5  27 ||13.5 27 13.5 364.5   9 27 13.5 40.5 Resistance across AB for the network given is 9. RAB  1.30 For the circuit shown below, find the current flowing through the 10  resistor. Solution: By applying -Y transformation, the circuit is modified as 𝑅𝐴𝐵𝑅𝐶𝐴 𝑅𝐴 = 𝑅𝐴𝐵 + 𝑅𝐵𝐶 + 𝑅𝐶𝐴 50 2 ×2 = 16 = 4 16 = 1 4 Ω 1 𝑅 = Ω 4 𝑅𝐶𝐴𝑅𝐴𝐵 𝑅𝐵 = = 𝑅𝐴𝐵 + 𝑅𝐵𝐶 + 𝑅𝐶𝐴 2 × 12 = 16 24 16 = 3 2 Ω 3 𝑅 = Ω 2 𝑅𝐶𝐴𝑅𝐵𝐶 𝑅𝐶 = = 𝑅𝐴𝐵 + 𝑅𝐵𝐶 + 𝑅𝐶𝐴 2 × 12 = 16 24 16 = 3 𝑅𝐶 = Ω 2 The circuit becomes, 10Ω and 1⁄4 Ω resistors one in series 10 + 1 = 4 = 40 + 1 41 4 51 4 3 2 Ω Similarly 1⁄4 Ω and 2 are in series 3 2 +2= 7 2 Ω 7⁄ Ω and 7⁄ are in parallel 2 2 49 7 7 49 2 49 7  2 24    14 7 7 4 14 28 4 2 2 2 40 7 47    11.75 Req = 4 4 4 V I R 40 I  3.4 A 11.75 1.31 Find the current supplied by the source in the circuit shown in the figure below. Also determine the power in each resistance. Solution: The circuit diagram is simplified in the following steps. Step1:1Ω + 1Ω = 2Ω Step2:- 52 The last 2 resistors are in parallel [2 || 2] 2×2 (2 + 2) = Step3:3 and 1 are in series ∴3+1=4Ω Step4: 4 and 4 are in parallel [4 || 4] = 4 ×4 4+4 = 16 8 = 2Ω Step5: 20 = 3.33A 𝐼1 = 8 𝑉 20 = = 2.5A 𝐼2 = 8 𝑅2 ∴ = 𝐼1 + 𝐼2 = 3.33 + 2.5 = 5.83𝐴 Power dissipated in 6 Ω = I12 × 6 = (3.33)2 × 6 = 66.53 w 1.12 Star and Delta Transformation: 53 4 4 = 1Ω (i) Star Connected Circuits: The linear lumped passive three terminal circuit consisting of three resistance 𝑅1, 𝑅2 and 𝑅3 in the form of Star, Y or T connection as shown in the figure are calles as Star-Connected Circuits. They are characterised by a common node. Star-Connected Circuit (ii) Delta-Connected Circuits The linear lumped passive three-terminal circuit consisting of three resistances 𝑅1, 𝑅2 and 𝑅3 in the form of delta,  or  is called as Delta-Connected Circuits. 1.13 Delta - Star Transformation:  Referring to the Delta network, Resistance between A & 𝐵 =  Referring to the Star network, Resistance between A & 𝐵 = 𝑅𝐴𝐵 ||(𝑅𝐵𝐶 + 𝑅𝐶𝐴) 𝑅𝐴𝐵 + 𝑅𝐵𝐶 + 𝑅𝐶𝐴 𝐴 + 𝑅𝐵 Since the two networks are electrically equivalent, equate 1 and 2, 54 (1) (2) 𝑅 𝐴 + 𝑅𝐵 = 𝑅𝐴𝐵 ||(𝑅𝐵𝐶 + 𝑅𝐶𝐴) (3) 𝑅𝐴𝐵 + 𝑅𝐵𝐶 + 𝑅𝐶𝐴 Similarly at terminals BC and CA, 𝑅𝐵𝐶 ||(𝑅𝐴𝐵 + 𝑅𝐶𝐴) 𝑅 𝐵 + 𝑅𝐶 = 𝑅𝐴𝐵 + 𝑅𝐵𝐶 + 𝑅𝐶𝐴 𝑅𝐶𝐴 ||(𝑅𝐴𝐵 + 𝑅𝐵𝐶) 𝑅𝐶 + 𝑅𝐴 = 𝑅𝐴𝐵 + 𝑅𝐵𝐶 + 𝑅𝐶𝐴 Subtract equation 3 and 4, (𝑅𝐴 + 𝑅𝐵) − (𝑅𝐵 + 𝑅𝐶) = 𝑅𝐴𝐵 ||(𝑅𝐵𝐶 + 𝑅𝐶𝐴) 𝑅𝐴𝐵 + 𝑅𝐵𝐶 + 𝑅𝐶𝐴 (4) (5) − 𝑅𝐵𝐶 ||(𝑅𝐴𝐵 + 𝑅𝐶𝐴) 𝑅𝐴𝐵 + 𝑅𝐵𝐶 + 𝑅𝐶𝐴 𝑅𝐴𝐵 𝑅𝐵𝐶 + 𝑅𝐴𝐵 𝑅𝐶𝐴 − 𝑅𝐵𝐶 𝑅𝐴𝐵 − 𝑅𝐵𝐶 𝑅𝐶𝐴 𝑅𝐴 + 𝑅𝐵 − 𝑅𝐵 − 𝑅𝐶 = 𝑅𝐴𝐵 + 𝑅𝐵𝐶 + 𝑅𝐶𝐴 𝑅𝐴𝐵 𝑅𝐶𝐴 − 𝑅𝐵𝐶 𝑅𝐶𝐴 𝑅𝐴 − 𝑅𝐶 = 𝑅𝐴 − 𝑅𝐶 = 𝑅𝐶𝐴(𝑅𝐴𝐵 𝑅𝐴𝐵 + 𝑅𝐵𝐶 + 𝑅𝐶𝐴 − 𝑅𝐵𝐶 ) (6) 𝑅𝐴𝐵 + 𝑅𝐵𝐶 + 𝑅𝐶𝐴 Adding equation 5 and 6, 𝑅𝐶 + 𝑅𝐴 + 𝑅𝐴 − 𝑅𝐶 = 𝑅𝐶𝐴 ||(𝑅𝐴𝐵 + 𝑅𝐵𝐶) + 𝑅𝐶𝐴(𝑅𝐴𝐵 − 𝑅𝐵𝐶 ) 𝑅𝐴𝐵 + 𝑅𝐵𝐶 + 𝑅𝐶𝐴 𝑅𝐴𝐵 + 𝑅𝐵𝐶 + 𝑅𝐶𝐴 𝑅𝐶𝐴 𝑅𝐴𝐵 + 𝑅𝐶𝐴𝑅𝐵𝐶 + 𝑅𝐶𝐴𝑅𝐴𝐵 − 𝑅𝐶𝐴𝑅𝐵𝐶 2𝑅𝐴 = 𝑅𝐴𝐵 + 𝑅𝐵𝐶 + 𝑅𝐶𝐴 2𝑅𝐶𝐴 𝑅𝐴𝐵 2𝑅𝐴 = 𝑅𝐴𝐵 + 𝑅𝐵𝐶 + 𝑅𝐶𝐴 𝑅𝐶𝐴 𝑅𝐴𝐵 𝑅𝐴 = (7) 𝑅𝐴𝐵 + 𝑅𝐵𝐶 + 𝑅𝐶𝐴 Similarly, 𝑅𝐴𝐵 𝑅𝐵𝐶 𝑅𝐵 = 𝑅𝐶 = (8) 𝑅𝐴𝐵 + 𝑅𝐵𝐶 + 𝑅𝐶𝐴 𝑅𝐴𝐶 𝑅𝐵𝐶 (9) 𝑅𝐴𝐵 + 𝑅𝐵𝐶 + 𝑅𝐶𝐴 Therefore, 55 𝑅𝐴𝐵 𝑅𝐶𝐴 𝑅𝐴 = 𝑅𝐵 = 𝑅𝐶 = Any arm of Star connection = 𝑅𝐴𝐵 + 𝑅𝐵𝐶 + 𝑅𝐶𝐴 𝑅𝐴𝐵 𝑅𝐵𝐶 𝑅𝐴𝐵 + 𝑅𝐵𝐶 + 𝑅𝐶𝐴 𝑅𝐴𝐶 𝑅𝐵𝐶 𝑅𝐴𝐵 + 𝑅𝐵𝐶 + 𝑅𝐶𝐴 Prod. of two adjacent arms of Delta network Sum of arms of Delta network 1.14 Star – Delta Transformation [April - 2009] We know that, 𝑅𝐴𝐵 + 𝑅𝐶𝐴 𝑅𝐴 = 𝑅𝐴𝐵 + 𝑅𝐵𝐶 + 𝑅𝐶𝐴 𝑅𝐵𝐶 + 𝑅𝐴𝐵 𝑅𝐵 = 𝑅𝐶 = 𝑅𝐴𝐵 + 𝑅𝐵𝐶 + 𝑅𝐶𝐴 𝑅𝐶𝐴 𝑅𝐵𝐶 𝑅𝐴𝐵 + 𝑅𝐵𝐶 + 𝑅𝐶𝐴 (1) (2) (3) Multiply equation 1 and 2, 𝑅𝐴 . 𝑅𝐵 = 𝑅𝐵𝐶 . 𝑅𝐴𝐵 𝑅𝐴𝐵. 𝑅𝐶𝐴  𝑅𝐴𝐵 + 𝑅𝐵𝐶 + 𝑅𝐶𝐴 𝑅𝐴𝐵 + 𝑅𝐵𝐶 + 𝑅𝐶𝐴 𝑅𝐴𝐵2𝑅𝐵𝐶 𝑅𝐶𝐴 𝑅𝐴 . 𝑅𝐵 = (𝑅𝐴𝐵 + 𝑅𝐵𝐶 + 𝑅𝐶𝐴 )2 56 (4) Multiply equation 2 and 3, 𝑅𝐵𝐶 + 𝑅𝐴𝐵 𝑅 𝐵. 𝑅 𝐶 = 𝑅𝐴𝐵 + 𝑅𝐵𝐶 + 𝑅𝐶𝐴 ∗ 𝑅𝐶𝐴 𝑅𝐵𝐶 𝑅𝐴𝐵 + 𝑅𝐵𝐶 + 𝑅𝐶𝐴 𝑅𝐵𝐶 2𝑅𝐴𝐵 𝑅𝐴𝐶 ∴ 𝑅𝐵 . 𝑅𝐶 = (5) (𝑅 𝐴𝐵 + 𝑅𝐵𝐶 + 𝑅𝐶𝐴 )2 Multiply equation 1 and 3, 𝑅𝐴𝐵 + 𝑅𝐴𝐶 𝑅𝐴. 𝑅𝐶 = 𝑅𝐴𝐵 + 𝑅𝐵𝐶 + 𝑅𝐶𝐴 ∗ 𝑅𝐴𝐶 𝑅𝐵𝐶 𝑅𝐴𝐵 + 𝑅𝐵𝐶 + 𝑅𝐶𝐴 𝑅𝐴𝐶 2𝑅𝐴𝐵 𝑅𝐵𝐶 ∴ 𝑅𝐴 . 𝑅𝐶 = (𝑅𝐴𝐵 + 𝑅𝐵𝐶 + 𝑅𝐶𝐴 )2 (6) Add equation 4, 5 and 6, 𝑅𝐴 . 𝑅𝐵 + 𝑅𝐵 . 𝑅𝐶 + 𝑅𝐴. 𝑅𝐶 = = = 𝑅𝐴𝐵 2𝑅𝐵𝐶 𝑅𝐶𝐴 + 𝑅𝐵𝐶 2𝑅𝐴𝐵 𝑅𝐴𝐶 + 𝑅𝐴𝐶 2𝑅𝐴𝐵 𝑅𝐵𝐶 (𝑅𝐴𝐵 + 𝑅𝐵𝐶 + 𝑅𝐶𝐴)2 𝑅𝐴𝐵𝑅𝐵𝐶 𝑅𝐶𝐴(𝑅𝐴𝐵 + 𝑅𝐵𝐶 + 𝑅𝐶𝐴) (𝑅𝐴𝐵 + 𝑅𝐵𝐶 + 𝑅𝐶𝐴 )2 𝑅𝐴𝐵𝑅𝐵𝐶 𝑅𝐶𝐴 (7) (𝑅𝐴𝐵 + 𝑅𝐵𝐶 + 𝑅𝐶𝐴) We know that, 𝑅𝐵 = 𝑅𝐴𝐵 + 𝑅𝐵𝐶 (8) 𝑅𝐴𝐵 + 𝑅𝐵𝐶 + 𝑅𝐶𝐴 Sub. Equation 8 in 7, 𝑅𝐴𝑅𝐵 + 𝑅𝐵𝑅𝐶 + 𝑅𝐶𝑅𝐴 = 𝑅𝐵 . 𝑅𝐶𝐴 ∴ 𝑅𝐶𝐴 = 𝑅𝐶𝐴 = 𝑅𝐴𝑅𝐵 + 𝑅𝐵𝑅𝐶 + 𝑅𝐶𝑅𝐴 𝑅𝐵 𝑅𝐵[𝑅𝐴 + 𝑅𝐶] 𝑅 𝐵 ∴ 𝑅𝐶𝐴 = 𝑅𝐴 + 𝑅𝐶 + 57 + 𝑅 𝐶𝑅 𝐴 𝑅𝐵 𝑅 𝐶𝑅 𝐴 𝑅𝐵 Similarly, 𝑅𝐴𝐵 = 𝑅𝐴 + 𝑅𝐵 + 𝑅𝐴𝑅𝐴𝐵 𝑅𝐶 𝑅𝐵𝐶 = 𝑅𝐵 + 𝑅𝐶𝐵 + 𝑅 𝐵𝑅 𝐶 𝑅𝐴 1.15 Difference between Star and Delta Connection Delta Connection (i) If the Delta network, effective conductance between A and B is 𝐺𝐴𝐵 = 𝐺𝐴𝐵 + 𝐺𝐶𝐴 Star Connection In Star network conductance is 𝐺𝐴𝐵(𝐺𝐵 + 𝐺𝐶) 𝐺𝐴𝐵 = 𝐺+𝐺+𝐺 (ii) it is also known as  - network it is also known as Y - network (iii) in delta connection, three resistances are connected end to end. (iv) it is of delta shape, so it is known as delta connection. (v) it is represented as 𝐴 𝐵 𝐶 in star connection, one end of each resistance is connected at a point called Star-point. it appears as a star so it is known as Star connection. it is represented as 1.32 Calculate the equal resistance across AB of the circuit by using Star and Delta transformation [Nov. 2011] Step1 58 The above circuit is a combination of Star and Delta network. Always star reducing the circuit from the inner side.  Take the star network and convert it into  - network. 𝑅 =6Ω 𝑅 =6Ω 𝑅 =6Ω (𝑅𝐴𝑅𝐵) 𝑅𝐶 (6 ∗ 6) =6+6+ 6 =6+6+6 𝑅𝐴 = 18 Ω (𝑅𝐵𝑅𝐶) 𝑅𝐵𝐶 = 𝑅𝐵 + 𝑅𝐶 + 𝑅𝐴 𝑅𝐵𝐶 = 18 Ω (𝑅𝐶𝑅𝐴) 𝑅𝐶𝐴 = 𝑅𝐶 + 𝑅𝐴 + 𝑅𝐵 𝑅𝐵 = 18 Ω 𝑅𝐴𝐵 = 𝑅𝐴 + 𝑅𝐵 +  The star network is converted to delta network.  The resultant circuit is Step2 In the above circuit, the resistance connected across AB i.e. 4 and 18 are in parallel connection. 59 Similarly the resistance across BC and CA are also parallel  The above circuit can be reduced as follow, 𝑅𝐴𝐵 = 4 ||18 4 ∗ 18 𝑅𝐴 = = 3.27Ω 4 + 18 4 ∗ 18 𝑅𝐵 = = 3.27Ω 4 + 18 4 ∗ 18 𝑅𝐶 = = 3.27Ω 4 + 18  The above circuit is reduced and the resultant circuit is, Step3: The resistance connected across the AC and BC are in series, ∴ = 3.27 + 3.27 𝑅𝐴𝐶 & 𝐵 = 6.54 Ω 𝐴𝐶 & 𝐵𝐶 The resultant circuit, Step4: The resistance across A and B are parallel 3.27 ∗ 6.54 ∴ 𝑅𝐴𝐵 = 3.27 + 6.54 ∴ 𝐴𝐵 = 2.18 Ω ∴ The effective Resistance = 2.18 Ω 1.33 Calculate the equivalent resistance across AB 60 Step1 Reduce the 6  -connected network into a star network. 𝑅𝐴𝐵. 𝑅𝐶𝐴 𝑅𝐴 = 𝑅𝐴𝐵 + 𝑅𝐵𝐶 + 𝑅𝐶𝐴 = ∴ 6∗6 6+6+6 𝐴 = 2Ω 𝑅𝐵𝐶 . 𝑅𝐴𝐵 𝑅𝐵 = 𝑅𝐴𝐵 + 𝑅𝐵𝐶 + 𝑅𝐶𝐴 ∴ 𝐵 = 2Ω 𝑅𝐶𝐴. 𝑅𝐵𝐶 𝑅𝐶 = 𝑅𝐴𝐵 + 𝑅𝐵𝐶 + 𝑅𝐶𝐴 ∴ 𝑅𝐶 = 2Ω  The resultant network, Step2 61 In the above circuit obtained from step 1. Reduce the inner star connected network.  2 and 2 are in series. 𝑅 = 2 + 2 = 4Ω 𝑅 = 2 + 2 = 4Ω 𝑅 = 2 + 2 = 4Ω   The resultant circuit of step 2, Step3 Reduce the inner star network to delta network, 𝑅𝐴 . 𝑅𝐵 𝑅𝐶 4∗4 𝑅𝐴 = 4 + 4 + 4 ∴ 𝑅𝐴 = 12 Ω 𝑅𝐵 . 𝑅𝐶 𝑅𝐵𝐶 = 𝑅𝐵 + 𝑅𝐶 + 𝑅𝐴 4∗4 𝑅𝐵 = 4 + 4 + 4 ∴ 𝑅𝐵 = 12 Ω 𝑅𝐴𝐵 = 𝑅𝐴 + 𝑅𝐵 + 62 𝑅𝐶 . 𝑅𝐴 𝑅𝐵 4∗4 𝑅𝐶 = 4 + 4 + 4 ∴ 𝑅𝐶 = 12 Ω 𝑅𝐶𝐴 = 𝑅𝐶 + 𝑅𝐴 +  The resultant circuit of step 3, Step4 The 4  and 12  are in parallel in all the AB, BC, CA arms. 4 ∗ 12 = 3Ω ∴ 𝑅𝐴𝐵 = 4 ||12 = 4 + 12 4 ∗ 12 = 3Ω 𝑅𝐵 = 4 ||12 = 4 + 12 4 ∗ 12 = 3Ω 𝑅𝐶 = 4 ||12 = 4 + 12  The resultant circuit, Step5 Therefore the effective resistance across AB, 𝑅𝐴𝐵(𝑅𝐵𝐶 + 𝑅𝐶𝐴) 𝑅𝐴𝐵 + 𝑅𝐵𝐶 + 𝑅𝐶𝐴 3(3 + 3) Reff of 𝐴𝐵 = = 3+3+3 Reff of 𝐴 = 2 Ω 1.16 Source Transformation:  For mesh analysis, the source should be a voltage source. 63  For nodal analysis, the source should be a current source. i. Current to Voltage Transformation: The current source present in the mesh analysis circuit can be transformed into a voltage source by using current to voltage transformation. i.e., consider the circuit. 𝑉 = 𝐼𝑅 = 50 ∗ 5 = 250𝑉 Always a current source is accompanied by resistance R which is always in parallel with the current source. By using ohm’s law, the voltage is found by using the formula 𝑉 = 𝐼𝑅. While transforming a current source to voltage source, the resistance R which is parallel with the current source will now become series with the voltage source. i.e. A voltage source is always accompanied with a series resistance. ii. Voltage to current source transformation 𝑉 𝑅 230 𝐼= 10 = 23 Amp 𝐼= 1.17 Mesh and Nodal Analysis for DC circuit 1.17.1 Maxwell’s Mesh Analysis (Current Method) In this method, mesh or loop currents are taken instead of branch currents. The following steps are taken while solving a network by this method:- 64 1. The whole network is divided into number of mesh. Each mesh is assigned a current having a continuous path. (Current is not splitted at the junction). These mesh currents are drawn in the clockwise direction. The common branch carries the algebraic sum of the mesh current flowing through it. 2. Write Kirchoff’s voltage law equation for mesh using the same signs as applied to Kirchoff’s law. 3. Number of equations must be equal to the number of unknown quantities. Solve the equations and determine the mesh currents. 4. From the mesh current, determine the branch currents. Generalisation of Mesh Analysis: Consider the network, No. of Mesh = 3(ABGHA, BCFGB, CDEFC) Step1: Apply KVL loop 1, [leading current is 𝐼1] 𝐸 − 𝐼1𝑅1 − (𝐼1 − 𝐼2)𝑅4 = 0 𝐸 − 𝐼1𝑅1 − 𝐼1𝑅4 + 𝐼2𝑅4 = 0 −𝐼1𝑅1 − 1𝑅4 + 𝐼2𝑅4 = − 𝐸𝐴 −𝐼1(𝑅1 + 𝑅4) + 𝐼2𝑅4 = − 𝐸𝐴 𝐼1(𝑅1 + 𝑅4) − 𝐼2𝑅4 = 𝐸𝐴 (1) Apply KVL for loop 2, [leading current is 𝐼2] + (𝐼1 − 𝐼2)4 − 𝐼2𝑅2 − (𝐼2 − 𝐼3)𝑅5 = 0 𝐼1𝑅4 − 𝐼2𝑅4 − 𝐼2𝑅2 − 𝐼2𝑅5 + 𝐼3𝑅5 = 0 𝐼1𝑅4 − (𝑅4 + 𝑅2 + 𝑅5)2 + 𝐼3𝑅5 = 0 ∴ − 𝐼1𝑅4 + 𝐼2(𝑅4 + 𝑅2 + 𝑅5) − 𝐼3𝑅5 = 0 (2) Apply KVL for loop 3, [leading current is 𝐼3] +𝑅5(𝐼2 − 𝐼3) − 𝐼3𝑅3 − 𝐸𝐵 = 0 𝐼2𝑅5 − 𝐼3𝑅5 − 𝐼3𝑅3 − 𝐸 = 0 𝐼2𝑅5 − 𝐼3(𝑅3 + 𝑅5) = 𝐸𝐵 −𝐼2𝑅5 + 𝐼3(𝑅3 + 𝑅5) = − 𝐸𝐵 65 (3) The equation 1, 2 and 3 can be written in matrix form, 0  I1   E A  R1  R4  R4  R R R R R I   0  4 2 5 5   2    4 0  R5 R3  R5   I 3   EB   The above matrix can be written as, R  11 R  21  R31 R12 R 22 R32 R13  I1  V1   R I   V  23   2   2  R33   I 3  V3   It can be generalized as, RI   V   9.1 Calculate the current through the circuit using Mesh analysis? [May 2010] No. of Mesh = 3 No. of unknown current  3(I1, I2 and I3 )  The size of R-matrix  33  The matrix is, R11 R12 R13  I1  V1   R R R  I     V  21 22 23   2   2   R31 R32 R33   I 3  V3  R11  3  8  5  16 R12  5 R13  8 R21  5 66 R22  5  4  9 R23  4 R31  8 R32  4 R33  8  6  4  18  V1  20V V2  40V V3  80V  The matrix, 16  5 5 9   8  4  8  I1   4  I     2 18   I 3  16  5   5 9 8  4 20  5 8  4  990 18 8 I1  40 9 80  4  4  15160 18 16 8  4  19720 18 20 I2  5 40 8 80 16  5  20 40   80 20 I3  5 9 40  15520 8  4 80 I I  1  15.21A 1   I I  2  19.91A 2  I I  3  15.67A 3  1.35 Solve the network and find the current in each branch? [April/May 2009] 67 R11  5  8 10  28 R12  8 R22  10  4  6  8  28 R21  8 28  8 I1  0  (15 15)   8   I  40 (25 15) 28    2       8  720 28    8 0 288   I   320 1 40 28  28 0   1120 I2      I1840 I  0.444 A 1  I I  2  1.555A 2   The current in each branch, AB  I1  0.444 A BC  I2  1.555A AF  I1  0.444 A CD  I2  1.555A ED  I2  1.555A EB  I2  I1  1.555  0.444  1.111A 68 1.36 Calculate the current in each mesh using mesh analysis [April 2008] No. of Mesh = 3 No. of unknown current = 3(I1, I2, I3) Size of Matrix = 3 × 3 R11 R12 R R 22  21  R31 R32 R13  I1  V1  R  I   V  23   2   2  R33   I 3  V3  R11  0.4 1.1  1.5 R12  1.1 R13  0.4 R21  1.1 R22  1.1 0.4  1.5 R23  0.4 R31  0.4 R32  0.4 R33  2  0.4  0.4  2.8 V1  10V V2  16V V3  0 1.5 1.1  0.4  I1  10  1.1 1.5  0.4  I   16    2    0.4  0.4     2.8   I 3  0  1.5 1.1  0.4   1.1 1.5  0.4 = 2.08 0.4  0.4 2.8 69 10 I1  16 0 1.5 I2  1.1 0.4  1.1 1.5  0.4  0.4  0.4 = 92.24 2.8 10 16 0 1.5 1.1 I2  1.1 1.5 0.4 0.4 I1  44.34 A  I1   I I  2  46.65A 2  I I  3  12.99 A 3   0.4  0.4 = 97.04 2.8 10 16 = 27.04 0 1.17.2 Nodal Analysis [April 2011] In this method, one of the nodes is taken as the reference node and the other node as independent nodes. The voltage at the different independent nodes is assumed and the equations are written as per the KCL for each node. After solving these equations, the node voltage are determined. Consider the circuit,  In the circuit Mark the nodes as 1 and 2.  Apply KCL for each node. Step1:- Apply KCL for node 1. 70 Sum of Incoming Current = Sum of the Outgoing Current I1  I 2  I 4 (1) On applying Ohm’s law, I1 I 2 I 4 I 4 VA V1     (2) R1 V1 V2 (3) R2 V1 VR  VR  0  Since it is a reference node. R4 V1 R4 (4) Substitute equation 2, 3 and 4 in equation 1,  VA V1  V1 V2  V1 R1 R2 R4 VA V1 V1 V2 V1     R1 R1 R2 R2 R4 1  1 1  V  1   VA 1   V R R R R  2   R 1  1 2 4  2    Apply KCL at Node 2 71  (5) Sum of Incoming Current = Sum of the Outgoing Current I2  I3  I5 (6) On applying Ohm’s law, I 2 I 3 I 5 V1 V2 (7) R2 VB V2 (8) R3 V2 VR = R5 Substitute equation 7, 8 and 9 in equation 6, V2 (9) R5 V1 V2 VB V2 V2   R2 R3 R5 V1 V2 VB V2 V2     R2 R2 R3 R3 R5 VB V2 V2 V2 V1     R3 R5 R3 R2 R2 VB  1    V2  1 1  1    V R R R R  1   R 3  3 5 2  2        (10)  By Matrix Method,  1 1  1  1  V1   V1   R R R R     R      1  1 2 4 2  1  1 1   1    VB  R R R R     R   2 2 3 5  V2   3  1 (conductance) G R G11 G12  V1  I1  G G  V    I   21 22   2   2           In general, GV   I  1.37 Calculate the node voltage for the given circuit? 72   1 1 G11 = +  0.36 6 5 1 G12 =   0.2 5 G13 =0 1 G21 =   0.2 5 1 1 1 G22 = + +  1.03 5 2 3 1 G23 =   0.5 2 G31 =0 1 G32 =   0.5 2 1 1 1 G33 = +   0.75 2 4 3 Sum of Incoming Current = Sum of the Outgoing Current  I1  4  3  1A I2  3  2  1A I3  2  4  2 A By matrix, 0.36 0.2  0  0.2 1.03  0.5 0  V1  1   0.5  V   1    2    0.75 V3   2 73 0.36  0.2   0.2 0 1.03  0.5 1 V1  0.2 2  0.2 1.05  0.5 0.36 1 V2  0.2 0 1 2 0.36  0.2 V3  0.2 0 V1   V1 1.03  0.5 0  0.5 = 0.1581 0.75 0  0.5 = 0.47 0.75 0  0.5 = 0.06 0.75 1 1 =  0.3816 2  2.98V  V2 V  0.37V 2  V V  3  2.41V 3  1.38 Calculate the voltage at the node A and B in the given circuit? [Jan 2010]   Mark the nodes as A and B The voltage source in the circuit must be transformed into a current source. 74 I V R  10  5A 2  The resultant circuit, Apply KCL at node A, I1  25  5  20 A 1 1 G    0.3 11 10 5 1 G12    0.2 5 Apply KCL at node B, 75 I2  Incoming  Outgoing current = (5  5)  6 = 10  6 I2  4 A 1 G21    0.2 5 1 1 G22    0.7 5 2  0.2 V1  20  0.3 0.2 0.7 V  4      2     0.2  0.17 0.7  0.2  14.8 0.7 0.3 0.2 V1  20 4 V2  0.3 20 0.2 4  5.2 V1  87.05V 1  V V  2  30.5V 2  V 1.39 Use the Nodal Voltage method and hence find the power dissipated in 10  resistor?  The voltage source 40 V must be transformed into a current source. I V R  40 = 10A 4 76  The resultant circuit, Apply KCL at Node 1, 2, 3, I1 = Sum of Incoming Current - Outgoing Current  10 10 I1  0 A I2  0 I3  15 10 I3  25A 77 1 1 1 G11 = + + =0.575 4 8 5 1 G12 =  =  0.2 5 G13 = 0 1 G21 =  =  0.2 5 1 1 1 G22 = +  = 0.5 5 5 10 1 G23 =  =  0.2 5 G31 =0 1 G32 =  =  0.2 5 1 1 1 G33 = + + = 0.575 4 5 8  0.2 0 0.575 0.2 0.5  0.2   0  0.2 0.575  V1  0   V   0    2     V3   25 0.57  0.2 0   0.2 0.5  0.2  0.2 0.57  0.2 0 0 0 V1  0 25 0.5  0.2  0.2  1 0.57 0.57 0 0 V2  0.2 0  0.2 0 25  0.1193  2.875 0.57 0.57  0.2 V3  0.2 0.5 0 0  6.1875  0.2 25 V2  24.09V V10  V2   Power Delivered By 0 10, P10  V10 * I10 (24.09)2 V2  P10  10   58.03W R10 10 78