UNIT = .D.C CIRCUITS _ PART-A SHORT QUESTIONS WITH SOLUTIONS What is a circuit? at. ans Model Papers. Q110) | rout is defined as a clos : aeoeceting wires and Ione Pat that transfers energy from source to load. In gener a eiteuit comprises of ENeTEY ses cls usually. The energy sources ae batteries, generaiors (or) any current supplying elements ie loads may be lights (or sa ie loads may (6r) lamps (or) motors (or) any current absorbing elements. The circuit is shown in figure > Connecting a T Sto > comecting Figure = @% Define the following terms: () Resistor (i) Inductor (iii) Capacitor. Ans: Resistor Resistor is a passive element which absorbs energy whenever a current is passed through it. It is denoted by *R’. The symbolic representation of a resistor is shown in figure (1). R SW Figure (1) fi) Inductor Inductor isa passive element which stores the energy inthe form of magnetic field whenever a curent is passed through it. tis denoted by ‘L’ and its symbolic representation jason in figuie (2), Figure (2) Gi) Capacitor ich stores the energy in the form of electric field whenever a current is'passed through ‘bolic representation is shown in figure (3). Cc — oo Figure (3) DeCTRGH ALL-NFONE JOURNAL POR ENGINEERING STUDENTS Capacitor is a passive element wh it Itis denoted by *C” and its sym SIA GROUP © scanned with OKEN ScannerBASIC El 1.2 ind 3. Classify all the different types of voltage a! current sources. Ans: When ideal source and practical source are considered, then the sources are classified as shown in figure (1). ‘Types of sources — Practical source Ideal source =e Practical zal Ideal Practical ‘ von: ewrent tole te source fouree source source Figure (1) When independent source and dependent source are considered, then the sources are classified as shown in figure (2), Types of sources Independent | I, it, voltage current dependent dependent aan Cimem Curent dependent dependent volage sauce “cure! source Vokage dependent vokage source Vokage dependent ‘current ure (2) Q4. What is source transformation? Ans: Source Transformation Source transformation isa network reduction technique, in which one form of source is replaced with its another ‘equivalent form, Using this technique, a complicated network Can be converted into a simple form and hence, calculation can be made easier. Basically, there are two types of source and the current source. A py always have a resistor in series with i source will have a resistor in transformation technique, a resistor can be converted to the same resistor. soureesi.., the voltage ractical voltage source itand a practical current Parallel with it. Using the source voltage source in series with a @ current source in parallel with LECTRICAL ENGINEERIT SU HYOER AG, Two operations can be performe transformation technique: a given voltage source 1 ey Converting a given VINEE SEE (0a cpg d sing =~ nverting a given cure! urce to a Voltage sy Give the statements of KCL and KyE at Mo tei Curent Law CL Kirchhof"’s Current Law (KCL) states that, the currents entering any node is equal 0th sum gfe Jeaving that node, . ma OR The algebraic sum of currents entering and ey node is zero. Figure (1) From figure (1), Current, (= 1, +1, +1 +1, +1 N+ +141 + haty Kirchhoff’s Voltage Law (KVL) Kirchhof?"s Voltage Law (KVL) sates tha, thay sum of all branch voltages around any closed path in acing is always zero at all instants of time. Figure (2) From figure (2), ' vty, Ee oe Q6. Write down the expressions used in convers#" of Y to A transformations. Ans: model Pape Expression Used in Y to A (Star to Delta) Transforms Le the resistance instar connected network) and R, and the resistances in delta connected networt R, and R, as shown in figure. @ scanned with OKEN ScanneryniT-1_D.C Circuits ‘The expressions used in ¥to Ah ag lO transformation are as p= Relat RaRet RoR, Bake Ry RaRe + RoR + RoRy RytRo+ Q 3. Treannot be applied for networks which contain depend cent sources, 4. Ifthe network contains non-linear elements like tran- sistors, diodes ete, then, superposition theorem is not Q10. State Thevenin’s theorem. Ans: ‘Any linear network having an active vollage and current sources with two terminals 4 and B can be replaced by an ‘equivalent voltage source (V,,) and equivalent resistance (R,) in series combination forming a simple equivalent circuit Where, V, is the open circuit voltage across the te rinals A and B and R,, is the equivalent resistance as seen . from the terminals A and B when the independent sources are oR, +R,+ Rabe deactivated Re Reg G7. Write short note on following, Wage (i) Mesh analysis v, Rt (li) Nodal analysis. ‘Ans: hs (Mesh Analysis ure: Thevenin’s Equivalent Circuit Mesh analysis is also known as oop analysis. tis used | @17- Whatare the limitations of Thevenin's theorem? to determine voltage and currents of planar electrical circuits. This method uses Kirchhoff’s Voltage Law (KVL). ‘Nodal Analysis Nodal analysis is also known as node-voltage analysis (01) branch current method. It is used to find the potential difference between the nodes in terms of branch currents. This analysis uses the Kirchhof?’s Current Law (KCL). 08, State superposition theorem. Ans: theorem states that in a linear network comprising of number of independent sources, the total response in any branch of the network is equal to the algebraic sum of individual response acting alone i.e., considering only one source at atime and making all other independent sources to zer0". However, the dependent sources must be retained in the network. Q9. State superposition theorem limitations. Ans: 1. The first and the foremost drawback of superposition theorem is that it fails to apply if the network contains less than two independent sources. 2. Calculation of power is not possible using superposi- tion theorem as power (which is non-linear quantity) is rroportional to the square of current (or) voltage. Ans: “The following are the limitations of Thevenin’s theorem, 1. Thevenin’s theorem is applicable only for linear circuits. But we know that practically no circuit is 100% linear. Its linear only fora specified range of values. Hence; ‘Thevenin’s theorem is applicable only fora limited range of values 2, The power calculated using Thevenin’s equivalent cir- cuit is not same as that calculated by taking the original network. This is so because the power of any element is proportional to square of current or square of voltage bbut not linearly dependent. 3. The Thevenin’s equivalent circuit has an equivalent V-1 characteristic with respect to load only. Q12. State Norton's theorem. Ans: ‘Model Paper-l, Q1(f) ‘The Norton’s theorem states that any two terminals linear network with current sources, voltage sources and resistances (impedances) can be replaced by an equivalent circuit consisting of a current source in parallel with a resistance (impedance), ‘where the value of the current source is equal to the current passing through the short-circuited terminals and the resistance is equal to the resistance measured between the terminals of the network with all the energy sources replaced by their internal resistance. The Norton’s equivalent circuit is shown in figure, ee eee rir GROUP SPECTROM ALLAN-ONE JOURNAL FOR ENGINEERING STUDENTS @ scanned with OKEN ScannerBe | 14 BASIC ELECTRICAL ENGINEERING [JNTU-HYDERAR apy a 4 % wi " it Figure “ " Q13. List the limitations of Norton's theorem. + Ans: J Figur (i 1 Efficiency of the network cannot be determined by this igure (i theorem The networkin figure (i) epresens the Norton’ ng 2. tis not applicable for magnetically coupled circuits. | lent circuit S_ Nowtinca and unter ccuits cannot be simple | Q46. A D.C voltage of 20 Vis applied ina RL crag © by Norton's theorem where R= 5 Q and L= 10H. Find, 4. Alsoit isnot applicable tothe circuits containing active (i) Tho time constant load (ii) The maximum value of stored energy, Q14. List any three advantages of Thevenin’s and Norton's theorems. Given that, An D.C voltage, ¥ = 20 * Resistance, R= 52 _ The following are the advantages for both Thevenin's Fi ipheieniye aa ren pe sense Assuming the given R-L circuit as a series RL ccuy 1. Reduses the complex networks to simple neworks. | 45 toy nines tele Both the theorems are applicable to linear networks -—w. 50 Both active and bilateral networks can be solved by these 7 theorems. > 4. Inimpedance-matching problems, both the theorems are »vC) 1 very much useful for determination of load r Figure: Geometry of Figure Let / be the steady state current flowing through the How to convert Thevei Model Papers, O19) | circuit, Time Constant ‘The Thevenin’s equivalent circuit isas shown in figure (i). | (i) The time constant T of a series R-L circuit is defined Ry the ratio of inductance L to Tesistance R. . Mathematically, Val Time constant, 7 T=2sec (i) Maximum Value of Stored Energy (W) a Figure @) ‘The steady state current / lowing through te crits | given by, Where, Vv _20 is the Thevenin’s voltage . TRS ,, is the Thevenin’s equivalent resistance * M eran red i g S is, The internal resistance will be same for both Thevenin’s ry Aastha ° and Norton's theorem ive, We 2Lt = yx10x(4)" R=, . => W=805 Look for the SIA GROUP Loco QB on ine TITLE COVER before you buy @ scanned with OKEN ScannerEES INIT-1 D.C Circuits ui “ 1.6 , PART-B ESsay QUESTIONS WITH SOLUTIONS 4.4 ELECTRICAL cir = ‘CUIT ELEMENTS-(R, LAND C), VOLTAGE AND CURRENT SOURCES 17. What are active and passive elements? Ex : plain in detail, The elements of a network are classified j 1. Active elements 2. Passive elements, 1, Active Elements into two types as follows, The elemenis which are capa sive elements. Basically, there one °F Selivering energy to the devices or networks connected across them are known as a ly, there are two kinds of active elements as follows, (i) Independent voltage source : (ii) Independent current source, j) An independent voltage source : © trough ‘88° Source is a source in which the voltage across the terminals is independent of current passing Vv 7 7 Figure (1) Figure (1) shows the circuit representation and V-I characteristics of an ideal source. The voltage source is termed as a D.C voltage source if it has a constant voltage and is represented as in figure (2). igure (2) ‘The V-I characteristics of a practical voltage source is shown in figure (3). Figure (3) is a source which can deliver a constant current independent of the voltage across its jon and the V-I characteristics ofan ideal current source are shown in figure (4). ‘An independent current souree i terminals. The circuit representati : ad : + Figure (4) FOR ENGINEERING STUDENTS Sia GROUP Wi SPECTRUM ALL-IN-ONE JOURNAL @ scanned with OKEN Scanner1.6 Practically the current sources doesnot possess an ideal behaviour. The V-I characteristics of a practical current source is shown in figure (5). ~ Figure 5) 2. Passive Elements ‘The elements which cannot deliver power and can only receive the power are known as passive elements Basically, there are three kinds of passive elements as follows, 7 (Resistor Gi) Inductor ii) Capacitor. Resistor Resistor is’ passive element which absorbs energy whenever a current is passed through it. It is denoted by “R’. The unit of resistor is Ohm. Resistance is defined as the opposition offered by the element forthe flow of current through it ‘The resistance of any material is given by, Reo Where, 4 = Specific resistance I= Length of the material A= Area of cross-section. Consider resistor shown in figure (6). 1 OR i —e —y— Figure (6) Let ‘V" be the voltage applied across it and J be the current through it, Then according to Ohm's law, Vel V=IR v R=—Q 7 ‘Whenever a current is passed through a resistor there is, (@) Avoltage drop across it given by, V=IR (b) A power loss in the form of hat given by, P=PR @) Inductor + Inductor isa passive element which stores the energy in the form of magnetic field whenever a current is pasced through it Relationship of Passive Elements. Teak for the SIA GROUP iodo BASIC ELECTRICAL ENGINEERING [JN | U-F'TVENABAD) a tncone neal, (i). 7 i L TTT —- <——_v— . Figure (7) Let ‘7 be the voltage across it and / be the curren, through it, The voltage across the inductor is given by, di dt From above equation we can see that the voltage across the inductor is a function of time. Hence, the Voltage across the inductor cannot change instantaneously. ‘Whenever & D.C supply is given across the inductor, i acts as short circuit, (ii) Capacitor Capacitor is a passive element which stores the energy in he form of electric field whenever a current is passed through it, tis denoted by Cand the units of capacitance is Farads Pure capacitor (with zero internal resistance) is a non- dissipative passive element which only stores the energy, but practical capacitor (with some internal resistance) Partially stores it and partially dissipates it Consider a capacitor as shown in figure (8). i <—_v— Figure (8) Let V’be the voltage across it and ibe the current through it, Then the current through the capacitor is given by, inc! i From the above equation we can say that the curret through a capacitor isa function of time. Hence, the cu- rent through the capacitor can’t change instantaneous): ‘Whenever a D.C supply is given across a capacitor, it acts as open circuit. Voltage Current Relation of Passive Elements For answer refer Unit-l, Q31, Topic: Voltage-Curett @ scanned with OKEN Scanner‘A passive element iy an element which consume the et amo F enerty mount of energy being delivered by an active element, Some 3 | Active elements supply energy, passive elements stores energy. 2. | Passive elements utilize electrical energy either by 3,| Active elements can supply an ave ‘converting or storing it 3 Tage power : infinite time to external ein Me POM for an | 3, | Passive elements eannot surly vege power 4,| Examples of active elements are, ameter tan 20 fran inft time enters (i Voltage source and 4, | Examples of passive elements are, | | ci) Current source, () Resistor (Voltage Source Gi) Inductor and (ii) Capacitor. @ Resistor ‘ Voltage source is one whi Be source is one which maintains constant " Resistor converts electrical energy into heat voltage at its terminal irrespective ofthe loa, Units are volts (V). ict | | | V. Ww? RQ) (i) Current Source iy Inductor Current source is one which delivers a current Units of inductor is Henry (H) and it denoted by with respect to load. Units are Ampere (A). L. It stores the energy in the form of magnetic fied. | L@) | THe | (iii) Capacitor | |. Units of capacitor is Farad (F) and it is denoted | by C. It stores electrical energy in the form electric field. Cc) Voltage-current relationship for passive elements. Voltage - {Passive Elements aero ene be defined by the way in which the current and voltage are elated for an individual element. The passive elements R, L, : . mnstant fora single clement, then the clement is a resistance R. The resistance © tfthecurrent and voltage V are related by 2 60 R represents the constant of proportionality. arma aera ARUN Eb @ scanned with OKEN ScannerBASIC ELECTRICAL ENGINEERING [JNTU-HYDERABany aw Current, : dr 1 r vonage fides Power, P= Vi= ve dt The units of capacitance C is Farad (F). Figure (1) Voltage, = Rr (Ohm's law) Current, = MR Power, P= Vi =?.R The units of resistance is Ohm (2). ‘An energy source can be divided into two types a, follows, (i) If the current and voltage are related, such that the ‘voltage is the time derivative of current, then the element 1 Ideal source is an inductance L. The inductance L represents the 2 Practical source. constant of proportionality. ¢ Ideal and practical sources are further subdivided intg four types. They are as follows, | (i) Ideal volfage source : L (ii) Practical voltage source (iii) Ideal current source (iv) Practical current source. | (@) Ideal Voltage Source | Figure (2) ‘An ideal voltage source is shown in figures. di eet =L a Ina practical voltage source, the voltage af the depends on the current. The characteristics: of practical vols? Figui source are shown in figure (4) 4 @ scanned with OKEN Scanner Power, P= ‘The units of inductance L is Henry (H). (ili) Ifthe voltage and current are related such that, the current wo" 74 is the time derivative of the voltage, then the element - isa capacitance C. The capacitance Cis the constant of | « a proportionality. ; Figure (2) Figure (3) i | nan ideal voltage source, the voltage across ts teminss is constant always ic, itis independent of eurrent i c ‘The characteristics are shown in figure (3). Inthe s is shown in figure (2) as V(). (ji) Practical Voltage Source terminals # v i : 7 age source, the voltage may vary depending on the time. )D.C Circuits Independent Voltage Source ‘A voltage source which maintains a constant VolLage does not change, when changes takes place in an elect network, 4] () tn ideat independ ge source, voltage (1) 15 independent af eurtent 7) hroweh the source. should have zeto inte ndent voltage source does not exist It indent voltae source have some internal re decreases with an (by Inan id practice, indepet Fesistance due to which voltag increase in current (0) Practical voltage soure internal resistance ‘¢ should possess minjmum Figure (5) _4 ‘The current in an ideal eurre; ure (2) irrent source n zt ce source may vary Figure (2) Independent Current Source ‘Accurrent source which gives constant current that does ' takes place in an electric network. x) Practical Current Source w In practical current source, the current depends on the seminal voltage, not change when change + (@) {In ideal independent current source, current (1) is maigendent of voltage (7) across the souree. It has infin intemal resistance p mies (b) Ideal independent current sources does not exist. Inpracice independent current sources have some internal resistance due to which current decreases with - an inerease in vOllge. Figure (6) we ‘The magnitude of current decreases as voltage across " terminals increases. a M Wi. Differentiate between independent and dependent sources. What is their representation? Ans: Independent Sources Independent sources are of twp types as. follows, (i) Independent voltage source (i) Independent current source. Examples of Independent sources are DC (or) AC Figure (@) fenerators, batteries. (In protical current souees internal resistance should ‘be maximum. Ev + Dependent Sources ). In case of dependent sources, the value of the source : depends on some parameter existing in the same cireuit. The parameter on which the value of source depends can be either : T v f vollage across any element or a current through an element in the same circuit Figure (1) DECTROM @LLIVONE JOURNAL FOR ENGIN EERING STUDENTS SIA GROUP {S @ scanned with OKEN Scannerae BASIC ELECTRICAL FN ee BAD) ‘sare operational amplific pe of dependent variable, (Voltage dependent voltage source i) Current dependent voltage source (iii) Volage depend Gv) Current dependent current source (@ Voltage Dependent Voltage Source Inthis type of dependent source, the val is dependent on the voltage across an element i the voltage source is a function of voltage. Figure (5) shows a vollage dependent voltage source. ¥, isthe dependent variable and &, isa constant. i current source of the sour he value of kV Figure (5) ‘The variable V, will be clearly. defined in the network. If due to some reason, Vis zero then the value of dependent source will also be zero and hence the voltage supplied by the source will be 2er0 and it has to be replaced with a short circuit (i) Current Deper In this type, the value of the dependent source depends on the current through an element ic., the value of the voltage source is a function of current. Figure (6) shows a current dependent voltage source. i, is the dependent variable and &, isa scaling factor having units volts/ampere. lent Voltage Source Figure (6) The variable i, will be clearly defined in the network. If due to some reason, i, is zero, then the value of dependent source will be zero,and hence the voltage supplied by it will be zero and it has to be replaced with a short circuit, i) Voltage Dependent Current Source In this type, the value of the dependent source depends onthe voltage across an element ic. the current supplied by it will be a function of voltage. re (7) shows a vollage dependent current souny ¥, isthe dependent variable and k sa Sealing constant haying units ampere/volts. kV, Figure (7) The variable ¥, will be clearly defined in the network. I¢'due to some reason, V, is zero, then the value of the soure, will be zero and hence the current supplied by it will be ze and now it has to be replaced with an open circuit, ) Current Dependent Current Source In this type, the value of the dependent source depenis ‘on the current through an element i.., the value of the curren, source isa function of current. ki Figure (8) Figure (8) shows a current dependent current source, is the dependent variable and &, is a dimensionless constant The variable i, will be clearly defined in the network. If dueto some reason, i, is zero, then the current supplied by the source Will be zero and hence it has to be replaced with an open ciruit. pent e replaced withenopedicntalt Q22. Explain source transformation with suitable examples. : Ans: Source Transformation For answer refer Unit-I, Q4. 1. Converting a Given Voltage Source to a Currett Source Consider a given voltage source with-magnitude V volts having a resistor ‘R’ in series as shown in figure (8) @ scanned with OKEN Scanner(0). The value of current source is given by ; me R Converting # Given Current Source amp Source to 9 Voltage 7 Consider given cure source with magnitude amperes having resistor “Rin parallel as shown in figure toy Figure (c) Figure id) ‘The current source can be converted intoa voltage source in sees with the same resistor R as shown in figure (d), The value ofvoltage source is given by, Y= IR terminals of the following figure. sa PX 202 20" 102 sal Figure The given circuit is shown in figure (1), 202 10a 100 | Figure (1) To find the equivalent voltage source, replacing the Figure (2) v.20 In Bee dh 200 oh soy > oA 20 Now, the circuit shown in figure (1) can be redrawn as shown as figure (4), \ DeasDsa Ze | 20 Figure (4) Now, replacing 4A. 5 A and SA current source with 2 single current source and single equivalent resistance in parallel. The value of equivalent current source is given by, -4-5 4A = 4A (downwards) ‘The value of equivalent resistance is given by, Via Je ig «5S 10 20 035 a ned SNe” O38 R= 2850 Vohage sources by their respective equivalent current sources o “shown in figures (2) and (3)- ‘rou 2 ANAL FOR ENGINEERING STUDENTS ra A. DECTRUM ALL-IN-ONE JOU @ scanned with OKEN ScannerBASIC ELECTRICAL ENGINEERING (JNTU-HYDERABAp) jreuit reduces as shown in figure (5). ao— 285.2 bo~ Figure (5) valent voltage source, IR 2.85 =1L4v Equivatent voltage source representation ofthe original network is shown in figure (6), 2.85 nav bi Figure (6) Q24. Find the voltage and current source equivalent representation of the following network across ab, as shown in figure. 29 oa + 4 = A)2A eae Figure Ans: The circuit diagram is shown in figure (1). 20 4 2a av a 2A b Figure (1) To find the voltage and current source equivalent, re- placing the voltage source by its equivalent current source as shown in figure (2). Look for the SIA GROUP Loco. Ce 20S ! 0 E Figure (2) a via =2A R 2 Now the cteuit ean be redrawn as shown in fue Oa A)2A 2a(h 22520 —o b Figure (3) Combining the 2A and2A current sources intoa singe current source. 2A+2A=4A ‘Combining the resistances into a single resistane ig, 2x2 29||29= =~ =19 Now, the circuit reduces to as shown in figure (4) ——Oa 4a = 19 ob Figure (4: Equivalent Current Source Representation ‘This is the equivalent current source representation of the original network of figure (1), To obtain the equivalent voltage source representation, replace the current source in figure (4) by its equivalent voligt source as shown in figure (5), 1a o® Zi 5 we : Lo, Lob Figure (6) V=IR=4x1=4V es Equivalent voltage source representation of tie oi network is shown in figure (6). on the TITLE COVER before you buy @ scanned with OKEN ScannerX ywit-1_0¢ Circuits oh rts ea sv Loy Figure (6) 5 Find the voltage across terminals ‘a’and bt i i - @ als ‘a’ andb’ of the circuit as shown in figure using source transformation 2a 10 20 AYIA . Figure ans: ‘The given circuit is as shown in figure (1). 20 1a 3A ° Figure (1) To find, The voltage across ab terminals, V, ‘To find the voltage across a terminals, replacing the voltage source by its respective current source as shown in figure (2)- 29 ‘ —wi-—— 05a wa > an Jee Figure (2) v 17 RR Leosa = to. ‘ “The circuit shown in figure (1 is redrawn as shown in figure (3) 7 19 @ scanned with OKEN Scanner1.14 Tn figure (3), the resistances 2 @ and 2.Q are in parallel, Hence, the equivalent resistance is given by, 2x2 Ray * 342 4 7 =10 ‘Thus, the cireuit can be represented as shown in figure (4). a 1a 0sA@ 12 AA b Figure (4) Replacing 0.5 A, 3A current source witha single current source, we get, 543 =35A Hence, the circuit can be modified as shown in figure (5). Figure (5) The current source can be replaced by its equivalent voltage source as shown in figure (6). 19 2 1a $ Figure (6) Applying KVL to the circuit, we get, M10) +3.5=0 20) =-3.5 = 35 “2 = LISA Voltage across a-b terminals, V=IxR 21752 =35V Bag ces 14.2 KVL AND KCL, ANALYSIS OF SIMPLE oC ITS WITH D.C EXCITATION - ‘3 ian explain Kirchhoff's laws. An: Kirchhoff’s Current Law (KCL) This law states that “the sum of the currents ent into any node is equal to the sum of the currents leaving he node". When two or more branches are interconnected in any parallel circuit, the junction point is called the node. Consider the circuit shown in figure (1). Figure (1) ‘The current / entering the node P is divided into currents Typ Ty 1, and J, which flows out of node P. Therefore, with the Kirchhofl’s current law, we have, I=L +L +1 +Lic. current leaving node P. , current in node P is equal to total Ifnode Q is considered, Kirchhoff’s current law will be same as applied to node P. L+htlti=l Consider the following figure (2). |, Figure (2) Applying Kirchhoff's current law to figure (2), we have, LtL+h=h+l+h, | 1t+L+h-1-1,-1, ‘Therefore, from the above relation it is clear that the algebraic sum of all currents meeting at a node is zero. Kirchhoff’s Voltage Law (KVL) : This law states that “the algebraic sum of all branch ‘Voltages around any closed path.in a circuit is always all instants of time”, : = Look for the SIA GROUP Loco a on the TITLE COVER before you buy @ scanned with OKEN ScannerFigure (3) The voltage drop occurs across each resistor as the arent passes through the circuit, The sum of the voltage drop sud the loop will be equal tothe total voltage in that loop. The volages at points 1, 3, 5, 7 will be more as compared to points 2,4, 6, 8. : ‘Therefore, from the Kirchhoff’s voltage law, we have, VaVAVAVAD, The current supplied by the voltage source can be obtained through Kirchhoff’s voltage law. : Using Ohm’s law, the voltage across each resistor is siven by, V,=IR,, V,= Ry V,=IRy V,=IR, ‘Then from Kirchhoff"s voltage law, we have, VaVtVAVtY, V = IR, +IR, + IR, + IR v 1 BER +R +R, SES veRESECeCoved ea oe Scr reer ereeecceae Q27. What is mesh analysis? Explain the steps involved in it with an example. Ans: Mesh Analysis ‘Mesh analysis is also called as’ “Joop analysis” or’ “mesh Current method”. This method is carried out using Kirchhof?'s Voltage Law (KVL). Mesh analysis is applicable only for planar etworks, a Mesh analysis is used to determine the voltages and cur- Tens ofthe given electrical circuit. KVL is applied for each of the meshes and unknown voltages and currents aré determined. “1.15 ical circuit containing P nodes, B branches, {he number of independent mesh equations is given BY: ~(P=1) -P+1 ' The number of mesh currents is equal to the number of ‘mesh equations, ‘Steps Involved in Mesh Analysis 1, Identify whether the given electrical circuit is planar (a circuit without having crossovers) oF not 2. Ifthe given circuit is planar, then identify the number cof meshes. (A mesh is a loop or closed path without any other loops within it). 3. Assign mesh curregts arbitrarily for the meshes identi- fied. 4. Apply KVL to the meshes identified. which Solve the system of simultaneous equations are obtained by the application of KVL at each of the meshes, Example Using mesh analysis write the mesh current equations and determine the currents for the circuit shown in the figure (1)- 52 Figure (1) Step 1 ‘The given electrical circuit is planar. Hence, mesh analysis can be applied to it. Step 11 . ‘Two meshes M,, M, are present in the given electrical circuit, Step 111 Assign mesh currents J, and J, for the meshes that are identified. sa ov a) 1 Mesh 1 Figure 2) wa wer Sia GROUP Wi SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING ‘STUDENTS = a | @ scanned with OKEN Scanner1.16 BASIC ELECTRICAL ENGINEERING [JNTU-HYDERABAp) Step1V Applying KVL for mesh 1, we get, 1W=S1,+20,-b) = 1, -2h Applying KVL for mesh 2, we get, 2-1) +10,=-50 => ~24, + 12%, ~() 50. = Q) Step V Solving the equations (1) and (2), we get, h- 10 =2h +12 7 =0.25 Amps 4.125 Amps 28. What is nodal analysis? Explain it with an example. Ans: Nodal analysis is also called as “node-voltage analysis” or the “branch current method”. This method is cried ou sg Kirchhoff"s Current Law (KCL). Nodal analysis is used to find out the pote each of the electrical nodes and potential difference between the nodes is determined in terms of branch currents, In an electrical circuit containing P nodes, one out of the P nodes is chosen as the reference node. Hence, for an electia] circuit containing ‘P’ nodes a total of P - I node equations and P ~ 1 node voltages are obtained. difference between the nodes in an electrical circuit. KCL is applied Steps Involved in Nodal Analysis Identify all the nodes (i.e., the points where elements are connected) in a given electrical circuit 1 2. From the identified nodes, select one node as the reference node. 3. Assign the variables to the nodes whose voltages are unknown, 4 5 Apply KCL at each af the unknown node voltages. Solve the system of simultaneous equations which are obtained by the application of KCL at each of the nodes. Example Determine the currents in each branch and also write the node voltage equations for the ‘electrical network shown in figure (1). 30 19 “6 ne +e “G) Figure (1) Step] * In the given electrical network, nodes A, B and C are identified, A 38] ‘B SA t 19a $2 ov “ igure (2) Look for the SIA GROUP LoGo {Qf ch the TITLE COVER before you buy @ scanned with OKEN Scanners the reference node. step Mt Here, the voltages at nodes sun variables and ¥, tether, Sn Fave unknown, Hence, sep Figure (3) Applying KCL at node 4, we get, 2) Applying KCL at node B, we get, Vo-Vs , Va , Ya=10 5 e 7 =) step V Solving the system of simultaneous equations (1) and Q).we get, 13V.4-10%y =150 50) V, =19.89 Volts Vg =10.84 Volts Current in 10 9 branch = 7g 19.89 10 = 1.989 Amps y; Current in $ 9 branch= = 2.168 Amps Varhe Q branch= —“S 19,89-10.84 ea Current in 3 1.47 Vy 10 7 1.4410 1 Current in 1 Q branch = 0.84 Amps. Q29. Determine V, and V, In the circuit shown in below figure. Let I, and I, be the loop currents as shown in igure (2). : 29 > 20 yh © Wy © nv oe ae € h Figure (2) Applying KVL to the loop ab ede a, > 2, +20,-1)+20,-1)=0 =U, +21,-2,+21,-2h,=9 => 61,-21,-2,=9" ~W) “Applying KVL to the loop ed gfe, => Ul,-1)+20,-)= 12 =. U,-2,+2h,- 2-12 (2) -2,+4l,-2h= 12 = 3.01 Amps | rou i @ scanned with OKEN Scanner= ‘Applying KVL to the loop dc hed, = Wi,-1)+2h + 2,1 = U2, +21, +21,-2) = -2),-21,+61,=0 = 4, +2,-61,-0 . By solving equations (1) and (2), we get, 2{61, 21, ~2/,= 0] 6-21, +4, 21, = 12] 0+ (4424), + (- 4-120, = 72 = 20/,-161,=72 = 5i,-41,=18 By solving equations (2) and (3), we gets 21, +41, ~2h, = 12 2,+21,-6l, 61,81, = |, 41, =6 By solving equations (4) and (5), we get, 51, —41, = 18 3h,-4h,= 6 aie 2, +0= 12 A Substituting J, value in equation (5), we get, 3(6)- 41-6 > 4 = = Substituting J, [, 21, + 2(6)- 63) =0 => -2,#12-18=0 = 21,-6=0 6 * 45 =~ h=3A From figures (1) and (2), we get, V=1,x2 = Vi=3x2 = V,=6V And, V,= 1, x2 = V=3x2 ~ V,=6V VW =6V,V, =6V values in equation (3), we get, BASIC ELECTRIGAY =) (4) = (6) G30, Determine the current supplied by battery in the circuit shown in Figure by usiy’ Kirchhoft’s laws. wv a sv T sv 1 Figure Ans: Let the node voltages at nodes-() and (4) be ¥, nay respectively as shown in figure o sa sa Lyk 4a wd . ‘Applying Kirchhoff’s current law at'node (a), we get, L+L+1=0 Y yeh, =v 20 i =) Applying Kirchhoff"s current law at node (b), we get -1+K+1,=0 -5-V,) Vy-5 , VaW30 4 } Se 8 @ scanned with OKEN Scanner4 D.CCircuits unr 1.19 ‘On solving equations (1) and (2), we pel, 2308 a -20 ov Figure (2) Q31. Two resistances when they are in series has an equivalent resistance of 9 ohms and when connected in parallel has an equivalent resistance of 2 ohms. Find the resistances and the ratio of the voltage and current =2.241A sharing between these elements if supply voltage is 50 V. Ya=5-¥y _ 11-722-5~9.064 Ans: iy : Let R,, R, be the two resistances. It is given that when 0.5864 they are connected in series, the equivalent resistance is 9 © and when they are connected in parallel, the equivalent resistance Vi-3 _ 9.0645 is2.0 R Ry Hence, the current supplied by 20 V battery, Le 1.656) = 16564 The current supplied by 5 V battery which is in series wins Oresston % 50V Ry Lo 1, 241A 6 ‘The current supplied by 5 V battery which is in series Figure (b): Parallel Connection With 4 © resistor, 1. Values of Resistances + L The equivalent resistance when connected in series is, given by, : ee Ra Rt R=9 ~() =0.586A ‘The equivalent resistance when connected in parallel is The current supplied by 5 V battery which is in series | given by, With 2 Q resistor, A Ry, = Pi 2 @ , SRR = ‘Substituting equation (1) in equation (2), we get, ‘ 2.032 A RR on The current supplied by 30 V battery, 9 R,R,=18 GB) We know that, 2617A f (a- bP = (a+b) -4ab _ The current supplied by each battery in the circuit is ‘Using the above identity, we have, show in figure (2). RRP HR +RP-ARR, A) SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING STUDENTS Sia Group 25 @ scanned with OKEN ScannerTequation 1) we a Sulbsttuting equations (1) and @) (&, -R, = OF -4(18) = (R,-RYe8I- (R,-RP=9 R-R=3 w= (5) Solving equations (1) and (5), We Bel, +R =9 R-R,=3 2. Ratio of Voltage and Current ‘When Connected in Series 1 R62 R.=32 (a) <—Va, 50 V- Figure c) Let Ip,. Ig, be the current flowing through R, and R, and V,,. Vp, be the voltage across R, and R,. ‘The total current of the circuit shown in figure (c) is, v 30 6! aR 555A 643 In case of series circuit, current remains same hence, Voltage across R,, Vp, = 6 * 5.555 =v Voltage across, Vp, BASIC El LECTRICAL ENGINEERING _ HYDERABAp, o ‘When Connected in ee () y= Ye, SOV ( Volage remains sae case of parallel cir Ya 5 ; ¥, ances are connected in paralle having the ratio of 1:2:3, the total is 100 W whey 40 V is applied to the combinations, find the values of the resistances. Given that, Voltage, ¥= 10V Total power, P= 100 W Ratio of resistances connected in parallel= 123 Let, the three resistances be x, 2x and 3s. ‘The circuit diagram for the given data is shown in gue, RX tt V=10V Figure Let ‘P be the current flowing through the circuit. P=W 100= 107 = = =10A Curent, = 104 Look for the SIA GROUP Loco Qf on the TITLE COVER before jou bly @ scanned with OKEN Scanner{eee Tai the equivalent resistance of tate and R, be“, itee parallel resistors * pquivalent resistance “R, : of the parallel circuit is end Given, Total power = 100 W = PR, = 100 , The values of resistances are, 833. x 1,833 = 3.666 2 x 1,833 =559 Ra 3S ES Qi. Find the power loss in the resistors of the network for the figure shown using nodal analysis. 19 Y 10V 39 4aQ) B20 i 10 O24 Figure Ans: Given circuit is shown in figure, 19 1.21 To determine, Using node-1 analysis, power loss in each resistor. ity Pro P, Pugs P. 2 2 Applying KCL at node-1, we get, Ky Vit-We = > “ Applying KVL at node-2, we eet 1K Va-Vil , Ye 2. Y-W=10 Ve i es 2 = tLe 3¢a) 3 = ° (On solving equations (1) and (2), we get, 26 Volts V,= 6.74 Volts =. Power loss across 2 Q resistor connected at node 1 Pram Bak But, i af ut, ho= > 526 2 . =2.63 Py g= (2.63) * 2 = 13.83 Watts Power loss across | resistor connected between node-I and node-2. =5.26-6.74 =-148A Sia GROUP a @ scanned with OKEN Scanner1 BASIC ELECTRICAL ENGINEERING [JNTUHYDERAB,9, ‘At Node 1 1.22 Here negative sign indicates that the current is in opposite direction, Pia Nin = 1asy x os = 2.19 Watts Power loss across 29 resistor connected across node-2. Page ygYR = 64 = 30 3 tarts 337A | 4 Py G31P*2 = 22.71 Wats Power loss in 3.2 resistor is given by, Pra haY R Vj#10-V, _ 5.26+10-6.74 3 3 84 2.84)? x3 4.19 Watts Q34. By using nodal analysis, find the current flowing through 3 ohms resistor. 3 ohms -() 0 Figure Ans: ie The network given is shown in figure wherein the cor- S responding node voltages are V, and V,, => v ha % = LAB a 12 = -4V,413 ¥, = -4V,413¥,=-15 Solving equations (1) and (2), we get, Figure (Current through 3.Q resistor. 1, V,=-343V Applying Kirchhoft’s current law, V,=-221V Look for the SIA GROUP LOGO {ff on the TITLE COVER before you by @ scanned with OKEN Scannerares 9.c circuits u current through 3 lug 4.3 SUPERPOSITION THEOREM x5, Stéfe and explain the superposition theorem. ie Model Paper, 2a) seatement Foranswer refer Unit-l, Q8. faplanation This theorem is valid only for linear system, Anindependent voltage source can be made inoperative ty replacing it by a short-circuit and an independent current ure can be made inoperative by replacing it by an open- cit. This theorem can be better understood with a numeri- c:leample. Consider the circuit which has two independent sources as follows, ry Th 39 100. -WW\- Figure (1) Sues for Applying Superposition Theorem Step 1 Keep one independent source active and other sources *cinoperative. Obtain the branch response (voltage or current). By superposition theorem, Let, only 20 V source is operative and all other sources ‘inoperative, The circuit now gets modified as, sa 102 1.23 Step 11 Repeat the above procedure for each independent source. Let only current source § A is operative now and all other sources are inoperative. The circuit is modified as, wa oT 30 q SA Figure (3) ia (6) * Gag BISA Step I ‘To determine the net branch response utilising superposi- tion theorem, just add all the responses obtained in step-I and step-Il considering their voltage polarities and current direc- tions. <1. The net response in 32 branch is, 2.5 + 3.125 = 5.625 A Ss Q36. Is superposition valid for power? Substantiate your answer. An: Superposition theorem is valid only for linear systems. It cannot be applied for power because the equation for power is non-linear. Consider the circuit shown below. R R, AWW 1 R Figure (1) ‘When P; is acting, let the current through R, be’. R, WW wiv r R Figure (2) Power = (FR, ‘When ¥, acting, let the current through R, be I". ‘ECTROM @LLIW-ONE JOURNAL FOR ENGINEERING STUDENTS sia Group 3 @ @ scanned with OKEN Scanner1.24 Figure (3) Powers "FR, The total current through R, by superposition is, perer ‘ Power = PR, But, ()'R, + U"PR, #PR, Since, P= ("4 PP = PEP 2 efore, the superpasition theorem is not valid Yor Q37. Find the voltage across the 2.2 resistor by using superposition theorem. k a Figure ‘Model Paper, 02(b) ‘The given circuit is as shown in figure (i) 10a. 20 op ‘NM Zi Esa Ouv 2200 L Hy Cav Figure fi) In Superposition theorem one source is considered at a ‘ime. Consider 10 V voltage source is active and other current ounce (2,A) and voltage source (20 V) are open circuited and short circuited respectively as shown in figure (i) wa © 20 3a Co) Look for the SIA GROUP Loco we on the TITLE COVER before you buy BASIC ELECTRICAL ENGINEERING [JNTU-HYDER, ABA Tet V, be the voltage at node (1), Applying KCL at node (1), we get, a Qa aw v,{0.2928] = os) 3 yesatvew Lith ah ot Sehage acre 2 rear Y ng BAing i= Mae Macaoy Considering 20 V voltage source is only active circuiting the 10 V voltage source and open eircuiting the 7, ‘current source. Now the circuit becomes, 02 © 29 “Wi sa mo I . 20 V Figure (iii) Let V, be the voltage at node (1), Applying KCL at node (1), we get, => 0.2928 V,=2.857 9.7575 Volts ‘The voltage across 2 Q resistor is, ¥n-20 7 2 9.7575~20 To 2.926 Volts we! active toot Now considering only the current souree 1g both the voltage sources as shown ircui @ scanned with OKEN Scannerunt —————_— 100 AWM iw 4 30 200 Zsa 2A Figure iv) Current in 2 Q resistor is, oe 3+ wee +2 Voltage across 2.2 resistor will be, pia Is2 =073*2 1.46 Vols . I=2x IBA, cording to Supesposition theorem the voltage across 20 resistor in presence ofall source is sum ofthe voltage produced ty each individual source. [ ‘Voltage across 2 © resistor is, Wawra rh = 0.972.926 + 1.46 =~ 0.496 volts “The negative sign represents that the voltage at 4 is negative. 4.4 THEVENIN THEOREM 038,Explain the steps for solving a network problem using Thevenin’s theorem. Ans: “The following are the steps involved in order to apply Thevenin's theorem. Step 1 Open circuit the terminals of the load if given i. be determined, Step 2 _ remove the element (by open circuiting) for which the response is to Determine the voltage across the open terminals using any of the network techniques. This voltage is nothing but V,,. Step 3 In the given network, eplace all the voltage source by short-circuit and al Step 4 I the current source by open circuit Determine the equivalent resistance as seen through the open terminals (oad terminals). This resistance i nothing but R, 4 Yoo % : Ifthe circuit contains any dependent source then R,,, is given by, Ry, = a = a , Where Jy is the short-circuit current isc Isc : ‘through the load terminals. . “Sia GROUP i SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING STUDENTS, @ scanned with OKEN Scanner1.26 Step 6 In order to determine Jy. with a short-circuit. Determine terminals using any network technique ina race the open oa terminal eee ough he sored Step 7 Replace the original network with the Thevenin’ lent circuit shown in figure. Rov Vou. "sequiva- Figure 98. State and explain Thevenin’s theorem. Model Papers, 2(a) Ans: ‘Thevenin’s Theorem For answer refer Unit-I, QUO, Procedure to Calculate Thevenin’s Equivalent Circuit Consider a network as shown in figure (2). R a oR & @) B Figure (2) The circuit consists of two voltage sources V, and V, and three resistances R,, R, and R,. t Steps to Calculate Thevenin’s Voltage (V,,) ___ In order to obtain Thevenin’s voltage, open circuit the resistance (R,). The circuit becomes as shown in figure (3).. rR oa R, BASIC ELECTRICAL ENGI einectclnteth ntti ing KVL to the cir By app! UR, +R)+¥,-V,=0 fee SCC ing equation (1) in equation 2) yy calculate the value of Vy & Steps to Caleulate Ry, In order io calculate Thevenin’s resistance (g voltage sources are short circuited and current Sout aege circuited ‘Therefore, can be found by shor cteuitng heyy, age sources V, and V,. The circuit becomes, R, A R Ay CT | 1 [= R, B Te Figure (4) Thevenin’s resistance, From figure it is observed that both the resistancesag in parallel ‘Therefore, the Thevenin’s equivalent LM circuit tained from derived parameters is as fallows, Ru A Figure (5) Q40. In the given figure, find the current flow through R, using Thevenin's theorem. R=20 @ scanned with OKEN Scanner| ue | sme given circuit is as shown in figure (1). | | igure (1) In order to determine the current through R, resistance cenin’s theorem, first remove 20 resistance as shown ties ving THEY infigure 2) Env rewa(f)in3a N. Figure (2) =43- Sx =4x3- $93 =12-6 = 6V with B at higher potential ‘Thevenin’s Resistance, Ry, Inorder to determine Ryn the voltage source is replaced with short circuit and the current source 1s replaced with open circuit as shown in figure (3). A B Rao Raa ' 2 ANW- N Figure (3) LR, 4 =72 Hence, the Thevenin’s equivalent circuit can be drawn rr yuit-t_ 0.6 Circuits Ryn Current through Resistor R, Ete Var 6) Figure (4) ‘Now, connecting the resistor R, with thevenin’s equivalent circuit as shown in figure 6). Rye 70 ~ ee Figure (5) Q41. Deter terminals A and B as shown in figure. 15 Ohms B 5 Ohms Ww 5Ohms = 20V =~ .5 Ohms 5Ohms =~10V A 15Wv . Figure Model Papert, 2{b) “The given circuit is as shown in figure (1), 1IsQ B 52 NNW ANN eo wvVa- «5a 5a 10V A BV Figure (1) The given cigcuit can be redrawn as shown in figure (2) I8v 159 5Q SV 20v> 52, “shown in figure (4). SPECTRUM ALL-IN-ONE 1OURNAL FOR ENGINEERING STUDENTS Figure (2) Sia GROUP @ scanned with OKEN Scanner1.28 ‘The two closed loops (1) and (2)are independent of each other, et J, nd J, be the currents in loops (1) and (2) respec tively Applying KVL to loop (1), we get, 20= 181, +5/, 201,~20 eta Applying KVL to loop (2), we get, loss To,= 10 Lela «Sh, Voltage drop 5 Q resistance in loop (1) is, =1A) Voltage drop across 5 Q resistance in loop (2), 1x5 xs v [F h= 1A} ‘The voltage between points A and B is sum of the volt- ages as shown in figure (3). ‘The equivalent resistance seen into terminals AB is, Ryg= (15 15) + 5) 16 +5) _ [(1sxs ze ia 4 1140) = (8.75) || (10) 666 2 ‘The Thevenin’s equivalent circuit is as shown in figure (4). Ryy = 4.666 0 A Vua=5¥ Figure (4) Look for the SUA GROUP Loco a on the TITLE COVER before you buy BASIC ELECTRICAL ENGINEERING LINTU-HYDERABAD | _, 1,5 NORTON THEOREM 42. State and explain Norton's theorem, ins: Statement ‘te Norton's theorem tts that ay two ening network with curent soutees, volage sources and revs (Gmpedances) can be replaced by an equivalent circutegne of a curent source in parallel witha resistance (imped. ‘where the value ofthe current source is equal tothe cee passing through the short-circuited terminals and theres equal to the resistance measured between the teminalyep network with all the energy soures replaced by ther ina resistance In shor, this theorem is used, where itis easier plify a network in terms of current instead of volta theorem reduces a normally complicated networks t parallel circuit consisting of, ‘osin, BES. Thi, 102 simple (@) Anideal eurent source /, of infinite interna esisane and (b) resistance R, in parallel with current source showy in figure (2). -————_3 Network |» B Figure (1): Normally Complicated Circuit : A wy B Figure (2): Norton's Equivalent _ Where, /,, isthe current which flows through a shor: circuit placed across terminals 4 and B. Rg is the circuit resistance looking from the open 4-3 terminal pene eee 43. By using Norton’s theorem determine the current through 5 © resistor (all resistances are in 2) as shown in figure. 20 3a 4 7 wove 2 oe Figure @ scanned with OKEN Scannereee ane ‘The given circuit is shown in figure (1) a 3a wv re (1) Current through 5 Q resistor In order to determine the current through 5 @ resis- tor using Nortons’s theorem, we have to first find Norton’s equivalent citeuit which consists of current source /, in paral- lel with resistance R,, where, 1, is the current through the short ciuited terminals 4 and B and R, is the equivalent resistance seen through the terminals 4 and B. (i) Norton’s Current Remove 5 Q resistor and replace it with a short circuit as shown in figure (2). AAW AWN a ly : Figure (2) Equivalent resistance of the circuit shown in figure (2) isgiven by, R,=2Q+(12Q||3Q) 36, 15 =2+—a2+ 1243 =24+24=449 Total current supplied by the source is, ~ LL Meama Ry 44 ‘Norton’s current is given by, 12 12 y= (| = 2.272x 75= 181BA Ty =1818A SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING STUDENTS (i) Norton’s Resistance In order to determine R,, voltage source is replaced by 4 short circuit, 32 Figure (3) Norton’s resistance R,= QQI\I22)+32 Pays 4 2x12 435 2412 =1.714+3=4.714Q ‘The Norton’s equivalent circuit can be drawn as shown in figure (4). Figure (4) “+ Current through 5 © resistor is given as, x | Rw v' [Ry +5] * i 4714 | y= le 1g | 4714 471445 818 0.485 Sia GRouP 25 @ scanned with OKEN Scanner1.30 | 1.6 TIME-DOMAIN ANALYSIS OF FIRST- “ORDER RL AND RC CIRCUITS Q44, Derive the expression for tho response ina R-L circuit for D.C. excitation. Define time constant. OR Derive the expression for I(t) of a R-L series circult when D.C. voltage Is applied toitatt=0 by closing the switch. Draw the response curve i(t) Vs t. Define time constant of R-L circult. Model Papers, 02(e) Ans: Consider the RL circuit shown in figure (1). Applying KVL, 180 cearer + Figure (1) General solution of this differential equation is given as, v i= Bt Kew Since,inductor behaves as an open-circuit at switching, 40) =0 Therefore, CO} i= Ru-e*") Then the voltage across the resistor and inductor is givenas, V0 (R= Ree (1) V4) = Ve) 10-140. Look for the SLA GROUP Loco {Qf on the TITLE COVER before you buy BASIC ELECTRICAL ENGINEERING ER ABA | Vin ver™ @ — Atr=0, i) =O and V0 {+ Lacts as open circuit ty =“ (i) Atte, i= and ¥,()=0 [L acts a short circuitatr= v,Q=0 Aus Ser i= R0-e)=0632 and ¥, (9 = Ve!= 0.368 V ¥,() = 0632V ‘The values of V, (#) and V,(t) are shown in figures @ and @). 7=UR Figure (2): For itt) Figure 3): For V, (and V, (0 Time Constant t= fis known asthe time constant ofthe ct er is defined as the interval after which currént or volta 63.2% of its total change. © scanned with OKEN ScanneryNiT-1_ D.C Circuits Gis. Explain the transient analysi circuit having D.C excitation (first order circuit). OR Derive the expression for i(t) and voltage across capacitance V,(t) forseries R-C circuit with D.C. voltage applied to it at t= 0. Explain about the time constant of R-C circuit. Ans: nc series Circuit RC circuit with D.C excitation is shown in figure (1). cr Nef} ’ + ) Figure (1): R-C circuit with D.C. Excitation “Applying KVL, we get, nio* Zfuou=v Differentiating dil) “ae respect to fon both sides, we get, i C. 0 d 1 = 7+ ZGil=0 1 = (p+2-}an- 7 ( + x} M)=0 a ‘The solution for equation (1) is, (= kee (2) Where, k is a constant and it can be calculated using initial conditions. ‘The circuit for calculating i(0*) is shown in figure (2). R v Aura 0" iO) = Substituting the value of k in equation (2), we get, VL ae vee i “This ype of equation is kriown as an &x shown in figure (3). The plot shows the transition period i ©. final from its initial value of tothe which the current adjus value zero, the steady state. i Figure (3) v 1 2L) IC When B is posttive, Applying KVL to the circuit, we get, Therefore, when P is positive, the roots are real ang did Ve RisL e+e fide (1) | unequal. Now equation (3) can be written as, Differentiating equation (1) with réspect to f, we get, Im CaeR tage, 1 dt ai ai + [eeee For this type of roots we will have over damped response and the current curve for overdamped case is shown in figure 2, +2) ‘Therefore, equation (2) represents a second order linear Lorrenns 0. Calculate ifor t= 2s and t= 5s. sg, t=0 42 62 Modet Papers, 22(b) To determine, iy fort Oand ‘i) for ¢= 2s and ¢= 5s In order to find i) considering three time intervals i.¢.. 150,015 4 and ¢2 4 separately. The switches 5, and S, are ‘opened for <0 and current through the inductor is zero i.e. 1 = 0, because inductor doesn’t allow the sudden changes in current, Thus, 40) =1()=i(')=0 For 0 < 1< 4, switch S, is closed and 5, is stil open Then, 40 _ 40 )~ F46 10 io) =4 Now, Equivalent resistance, R= 4+ 6= 100 Time constant, += Thus, (0) = Ho) + (0) —il)Je = 44 [0-d]e 5 =4-4e% i()=4- eA, Fort $ 4.switch S, is closed. Since the inductor doesn’t allow the sudden changes in current, the initial current at = 4 Osrs4 is, i(a)= (= 4 (Ie) =4(l-e) = 4(1-3.35 * 10) = 4(0.99) . i@=4a ‘To find i(«), assume the voltage ‘17 at node P. Now, applying KCL at node P, we get, o-¥ 40-¥ fo-V @ scanned with OKEN Scanner1.34 i(o)=2.727A ‘At the inductor terminal, the equivalent resistance is, R=4)/246 _ 4x2 “442 8 yg 8436 4 “60° 6 6 2 And, bes 5 BT 7333706818 Hence, i) = 1 () + A)", 14 ‘The term (¢—4) in the exponential is due to time delay. “Thus, substituting (0), (4) and t values in the above equation, wwe get, i 727 + [4-2.72T]eU 0818 =2.727 + [1.273]e tM) i ()= 2.727 1.273e CM), > 4 Hence, 0, 10 i=} M-e, osrs4 2.727 +1.273¢° 4660-9, 124 Now, ALI=2s, iQ)=4 (le) =4(Le) =4(1- 0.0183) ‘ = 40.98) y= Look for the SIA GROUP Loco {ion the TITLE COVER before you buy BASIC ELECTRICAL ENGINEERING MNT RUNDERABAD) , Atr= 5s, {(5)=2.727 4 12TBetHe- 2.7274 127i 2.727 + 1.273(0.2308) = 2.727 +0.293 3.020 i) is in position 1 for along time and broughs’® position 2 at time t = 0. Determine the ey current. rei s0v— 7 jure ‘The given circuit is shown in figure, 5a 1 x Hey 2H Figure Let, current i be flowing through the circuit when the switches are close. In figure, switch ‘S? is at position-1 for steady state condition. AL L=0 :. Curren through the inductor is given as, 4 R Dawa 5 The inductor will not allow to change the cure through it, i(0") =i") =10A It is required to obtain the current response afte He switch is moved to position 2. ‘The current in the circuit is, * i@=q+ree + sn) We know that, 0A Atr=0, @ scanned with OKEN ScannerUNIT-1. D.C Circuits . 1.35 ‘Substituting the values in equation (I) | 5 wets ae | : “® | Now, switch is moved to position 2 ic., at 1.00 -10 a ‘Substituting the values in equation (1), we have, 2 =¢,+¢,0) oe? Substituting ¢, in equation (2), we get, ete, =10 10-2 (+ 4e%%) Q49, Att=0, switch 1 in figure is closed, and switch 2is closed 4s later. Find i(t) for t > 0. Calculate ifort = 2s and t=5s. ao Si {7° 60 aov( * Ans: ‘The given circuit is shown in figure, 4g f°? 40Vv Figure To determine, (0) for > 0 and i( for t= 2s and 1= 5s SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING STGDENTS Th order to find A) considering three time intervals He 10,0514 and 12 4separately. The switches S, and Safe opened for £ <0 and current through the inductor is 2¢r0 1 = 0, because inductor doesn’t allow the sudden changes "2 current Thus, §(0)=1)=10)=0 For 0 <5 4, switch S, is closed and S, is still open- ‘Then, Now, Equivalent resistance, R= 4+ 6= 100 Time constant, === e constant, ae at yo" 2 ‘Thus, oA (0) = Heo) + L(0)— Hele * =44 [0-4}e 5 =4-4e% 1@=A(- eA, For t < 4,switch S, is closed. Since the inductor doesn’t allow the sudden changes in current, the initial current at = 4 is, osts4 (ey =4(-e*) =4(1-3.35 x 104) = 4(0.99) i(4)=3.96=4 i@)=4A To find i(o0), assume the voltage *V” at node P. Nov applying KCL at node P, we get, Sia GRouP @ @ scanned with OKEN Scannerane. BASIC ELECTRICAL ENGINEERING [INTU-HYDERABay, ‘Thus, At the inductor terminal, the equivalent resistance is, R=4j)2+6 42 eae 3c. 8436 _ 44 Fer onne a6 2 R= 2o-r330 Ai, eaeaes Bn Typ 7 0.6818 Hence, HG) = 1 () + [H(4) (eye, 124 The term (#~ 4) in the exponential is due to time delay. Thus, substitutin we get, FO = 2.727 + [42,727] UH 0.6H18 = 2.727 + [L273]e-t-4"466) = 2.727 127A, 3g Hence, 150 i= osts4 2.727+1.273e 146-4 Now, AUr=2s, iQ)=4 (1%) =4(ee4) =4(1- 0.0183) =4 (0.98) iQ)=3.92. Atr= 5s, 15) = 2.727 + 1.273 e1ams-0 2727+ 1.271 Look for the SIA GROUP Loco {on the TITLE COVER before you buy a i (©), (4) and + values in the above equation, @ scanned with OKEN Scanner. UNIT-1. D.C Cirouits = 2.727 + 1.27 0.2308) = 2.727 + 0,293 HS) = 3.020 A G50. A constant voltage Is applied to a RL circuit at t = 0. The voltage across the inductor at t= 3.46 ms is 20 Vand 5 V at t= 25 ms. Obtain Rif =2H. 1.37 Ans: Given that, 20V WjasVv A6ms = 3.46 * 10” see = 25 ms = 25 « 10° see We know that, Under transient analysis, the voltage across the inductor in a series RL circuit is given as, yevelt)! Now, at time, 1, 3.46ms, we have, M, vel! ) Substituting values, ,2,f, we get, (fos? : = 20=Ve O= Vetere v=20¢'™0"R ~() And also, At time, 1, = 25 ms, we have, ",, velit) Substituting values V;, ,L,f we get, o sere Ve BSR 5g S078 --(Q) From equations (1) and (2), we get, ae! 07h 592 5008 (2.52407 eR = ous a el BXO™ AR 3 > 1.07 R SIA GROUP Zh. @ scanned with OKEN ScannerBASIC ELECTRICAY = — Applying natural log, on both sides, we get ea > 07710 R tog. 4 log. e' toga} : => [(10.77%10%)xR}loge=log.4 [+ loga”™ => 510.77 10°) R (1) = 1.3862 [slogee=H] 3862 10.7710 1.3862 “10.77 => R=0.1287« 10° R x10? = 128.710 @ scanned with OKEN Scanner