BEEE UNIT-1 DC CIRCUITS Topics- Electrical circuit elements (R, L and C), voltage and current sources, Kirchhoff’s current law, Kirchhoff’s voltage laws, Analysis of simple circuits with dc excitation, Superposition Theorem, Thevenin’s Theorems, Norton’s Theorems. Time-domain analysis of first-order RL circuits. 1.1 Electric Current Flow of free electrons is called electric current. When electric pressure is applied to a copper strip, free electrons being negatively charged will start moving towards positive terminal round the circuit. This directed flow of electron is called electric current. The convention current flows from positive terminal of source to negative terminal of source. (Opposite to the flow of electrons). The strength of electric current I is the rate of flow of electrons i.e., charge flowing per second. πΆπ’πππππ‘ (πΌ) = (π (πππ’ππππ))/(π‘ (π ππ)) Unit of current is ampere i.e. coulomb/second One Ampere of current is set to flow through a wire if at any section one coulomb of charge flows in one second. 1.2 Electric Potential & Potential Difference Electric Potential: The charged body has the capacity to do work by moving other charges either by attraction or by repulsion. This ability of the charged body to do work is called electric potential. πΈππππ‘πππ πππ‘πππ‘πππ (π) = (ππππ ππππ)/πΆβππππ = (π (π½ππ’ππ))/(π (πΆππ’ππππ)) Unit of electric potential is volt, joule/coulomb Potential Difference: The difference in the potentials of two charged bodies is called potential difference. 1.3 Resistance The opposition offered by a substance to the flow of electric current is called resistance. The practical unit of resistance is ohm and is represented by the symbol Ω. It is defined as under : A wire is said to have a resistance of 1 ohm if a potential difference of 1 volt across its ends causes 1 ampere to flow through it The resistance of a conductor has following characteristics: a) It is directly proportional to the length of the conductor. b) It is inversely proportional to the area of cross section of the conductor. c) It depends on the nature of the material of the conductor. d) It depends on the temperature of the conductor. 1|P a ge BEEE UNIT-1 DC CIRCUITS Hence if R is resistance of a conductor of length l, cross-sectional area A and π is the resistivity of the material. π ππ ππ π‘ππππ (π ) = π π/π΄ πβπ Resistors are used for dividing voltage, current limiting etc. 1.4 Electric Power The rate at which work is done in an electric circuit is called electric power, πΈππππ‘πππ πππ€ππ (π) = (ππππππππ ππ πππππππ πππππ’ππ‘)/ππππ π = ππΌ = πΌ^2 π = π^2/π 1.5 Inductor An Inductor is a passive electrical component consisting of a coil of wire which is designed to take advantage of the relationship between magnetism and electricity as a result of an electric current passing through the coil. In its most basic form, an Inductor is nothing more than a coil of wire wound around a central core. For most coils the current, ( i ) flowing through the coil produces a magnetic flux, ( NΦ ) around it that is proportional to this flow of electrical current. Inductors are formed with wire tightly wrapped around a solid central core which can be either a straight cylindrical rod or a continuous loop or ring to concentrate their magnetic flux. The schematic symbol for inductor is that of a coil of wire so therefore, a coil of wire can also be called an Inductor. Inductors usually are categorised according to the type of inner core they are wound around, for example, hollow core (free air), solid iron core or soft ferrite core with the different core types being distinguished by adding continuous or dotted parallel lines next to the wire coil as shown below. 2|P a ge BEEE UNIT-1 DC CIRCUITS An inductor opposes the rate of change of current flowing through it due to the build up of selfinduced emf. In other words, inductors resist or oppose changes of current but will easily pass a steady state DC current. This ability of an inductor to resist changes in current and which also relates current, i with its magnetic flux linkage, NΦ as a constant of proportionality is called Inductance which is given the symbol L with units of Henry, (H) after Joseph Henry. An inductor coil has a central core area, (A) with a constant number of turns of wire per unit length, (l). So if a coil of N turns is linked by an amount of magnetic flux, Φ then the coil has a flux linkage of NΦ and any current, ( i ) that flows through the coil will produce an induced magnetic flux in the opposite direction to the flow of current. Then according to Faraday’s Law, any change in this magnetic flux linkage produces a self-induced voltage in the single coil of: π½π³ = π΅ π ∅ ππ΅π π¨ π π = π π π π π Where: ο· N is the number of turns ο· A is the cross-sectional Area in m2 ο· Φ is the amount of flux in Webers ο· μ is the Permeability of the core material ο· l is the Length of the coil in meters ο· di/dt is the Currents rate of change in amps/second L=μN2A/l The relation between the flux in the inductor and the current flowing through the inductor is given as: NΦ = Li. 1.6 Capacitors A capacitance consist of two conducting surfaces separated by a layer of insulating medium called dielectric. The dielectric can be paper, glass, ceramic, air etc. Capacitance is the electrical property of a capacitor and is the measure of a ability to store electric charge. The capacitance may be defined as the amount of charge required to create a unit potential difference between its plates. C = Q / V = Coulombs / Volts 1 Farad = 1 Coulombs / Volt Capacitance is proportional to the area (A) of one of the plates and inversely proportional to the separation between the plates (d). C = εA/d Where ε is the permittivity of the dielectric medium. 1.7 Voltage and Current sources 3|P a ge BEEE UNIT-1 DC CIRCUITS Ideal Voltage Source It is that voltage source whose output voltage remains constant irrespective of the change in load current. It has zero internal resistance. In practice, ideal voltage source is not available, and every voltage source has some internal resistance. Smaller the resistance of a voltage source, more it will approach to the ideal voltage source. The ideal voltage source can be represented by either of the symbols shown in the figure below. Ideal Current Source It is that current source whose output current remains constant irrespective of the change in load resistance. Its internal resistance is infinity. At any load resistance, it supplies the constant current. In practice, ideal current source has very high resistance. Higher the resistance of a current source, more it will approach to the ideal source. The ideal current source can be represented by the symbol shown in the figure below. Source Transformation The following points may be noted about a source transformation: Case (i) A voltage source with a series resistance can be converted into (or replaced by) an equivalent current source with parallel resistance. Example: It may be noted that if the polarity of a voltage source changes, accordingly the direction of the equivalent current source also changes. 4|P a ge BEEE UNIT-1 DC CIRCUITS Case (ii) A current source with a parallel resistance can be converted into (or replaced by) an equivalent voltage source with series resistance as shown below: Example: It may be noted that if the direction of a current source changes, accordingly the polarity of the equivalent voltage source also changes. 1.8 Kirchhoff’s Laws These laws are used for solving electrical networks Kirchhoff’s laws, two in number, are particularly useful (a) in determining the equivalent resistance of a complicated network and (b) for calculating the currents flowing in the various conductors. The two-laws are: 1. Kirchhoff’s Current Law (KCL) Statement: In any electrical network, the algebraic sum of the currents meeting at a point (or junction) is zero. It simply means that the total current leaving a junction is equal to the total current entering that junction. It is obviously true because there is no accumulation of charge at the junction of the network. Assuming the incoming currents to be positive and the outgoing currents negative, we have I1 + (-I2) + (-I3) + (+ I4) + (-I5) =0 I1 + I4 – I2 – I3 – I5 = 0 or I1 + I4 = I2 + I3 + I5 or incoming currents = outgoing currents Similarly, in Fig. 18(b) for node A + I + (-I1) + (-I2) + (-I3) + (-I4) = 0 or I = I1 + I2 + I3 + I4 We can express the above conclusion thus: ΣI = 0 at a junction 5|P a ge BEEE UNIT-1 DC CIRCUITS 2. Kirchhoff's Voltage Law (KVL) Statement: The algebraic sum of the products of currents and resistances in each of the conductors in any closed path (or mesh) in a network plus the algebraic sum of the e.m.f.s in that path is zero. In other words, ΣIR + Σe.m.f. = 0 ...round a mesh Determination of Voltage Sign- In applying Kirchhoff's laws to specific problems, particular attention should be paid to the algebraic signs of voltage drops and e.m.fs., otherwise results will come out to be wrong. Following sign conventions is suggested: Sign Conventions used in KVL While applying Kirchhof’s voltage law to a closed circuit, algebraic sums are considered. Therefore, it is very important to assign proper signs to emf's and voltage drops in the closed circuit. The following sign convention may be followed. A rise in potential can be assumed positive while a fall in potential can be considered negative. The reverse is also possible and both conventions will give the same result. a) If we go from positive terminal of the battery or source to negative terminal, there is a fall in potential and so, the emf should be assigned negative sign. If we go from negative terminal of the battery or source to positive terminal, there is a rise in potential and so, the emf should be given positive sign. b) When current flows through a resistor, there is a voltage drop across it. If we go through the resistance in the same direction as the current, there is a fall in the potential and so, the sign of this voltage drop is negative. If we go opposite to the direction of current flow, there is a rise in potential and hence, this voltage drop should be given positive sign. 6|P a ge BEEE UNIT-1 DC CIRCUITS 1.9 DC Circuit The closed path followed by a direct current (dc) is called a dc circuit. DC circuits can be classified as, a) Series Circuit b) Parallel Circuit c) Series – Parallel Circuit 1.9.1 Series Circuit The circuit in which resistances are connected end to end so that there is only one path for the current to flow is called as series circuit. Consider three resistances of R1, R2 and R3 ohm connected in series across a battery of V volt as shown in Fig. Obviously, there is only one path for the current I, i.e., current is same throughout the circuit. By Ohm's law, Voltage drop across the resistance R1, V1=IR1 Voltage drop across the resistance R2, V2=IR2 Voltage drop across the resistance R3, V3=IR3 Now, for a series circuit, sum of voltage drops is equal to the applied voltage. So, V = V1 + V2 + V3 =IR1 + IR2 + IR3 =I(R1 + R2 + R3) V/I= R1 + R2 + R3 But V/I is the total resistance R between the points A and B. R is called the total or equivalent resistance of the three resistances. So, R= R1 + R2 + R3 Hence, when a number of resistances are connected in series, the total resistance is equal to the sum of the individual resistances. The following points may be noted about a series circuit: 7|P a ge BEEE UNIT-1 DC CIRCUITS (i) The current flowing through each resistance is same. (ii) The applied voltage equals the sum of different voltage drops. (iii) The total power consumed in the circuit is equal to the sum of the powers consumed by the individual resistances. (iv) Every resistor of the circuit has its own voltage drop. Voltage Divider Rule in a series circuit A series circuit acts as a voltage divider as it divides the total supply voltage into different voltages across the circuit elements. Figure below shows a voltage divider circuit in which the total supply voltage Vin has been divided into voltages V1 and V2 across two resistances R1 and R2. The current through both resistances is same, i.e., I. According to Ohm’s law, V1 = IR1 and V2 = IR2 Let R is the total resistance of the circuit, and it is given by, R = R1+R2 Also, from the circuit, we have, V = IR = I(R1+R2) But I = V1R1 = V2R2 Therefore from the above equations, Vin = V1 (R1+R2) /R1 ∴ V1 = VinR1/(R1+R2) 8|P a ge BEEE UNIT-1 DC CIRCUITS Similarly, Vin = V2 (R1+R2)/ R2 ∴ V2 = VinR2/(R1+R2) 1.9.2 Parallel Circuit The circuit in which one end of each resistance is joined to a common point and the other end of each resistance is joined to another common point, so that there are as many paths for current flow as the number of resistances, is called a parallel circuit. Consider three resistances of R1, R2 and R3 ohm connected in parallel across a battery of V volt as shown in Fig. The total current Is divide into three parts: I1 flowing through R1, I2 flowing through R2 and I3 flowing through R3. Obviously, the voltage across each resistance is the same (i.e., V volt in this case) and there are as many current paths as the number of resistances. Hence, when a number of resistances are connected in parallel, the reciprocal of the total resistance is equal to the sum of reciprocals ofthe individual resistances. The following points may be noted about a parallel circuit: (i) The voltage drop across each resistance is same. (ii) The total current equals the sum of the branch currents. (iii) The total power consumed in the circuit is equal to the sum of the powers consumed by the individual resistances. (iv) Every resistor has its own current. 9|P a ge BEEE UNIT-1 DC CIRCUITS Current divider rule used in parallel circuits A parallel circuit acts as a current divider as it divides the total circuit current in its all branches. Figure1 shows a current divider circuit in which the total circuit current I has been divided into currents I1 and I2 in two parallel branches with resistances R1 and R2. Although, we can notice that the voltage drop across both resistances is same, i.e., V. According to Ohm’s law, I1 = V/R1 and I2 = V/R2 Let R is the equivalent resistance of the circuit, and it is given by, R = R1R2/(R1+R2) Also, from the circuit, we get, I = V/R = V×((R1+R2)/R1R2) But, we know that the voltage across both resistances is same. V = I1R1 = I2R2 Hence, from equations (2) & (3), we finally get, I = I1R1((R1+R2)/R1R2) = I1((R1+R2)/R2) ∴ I1 = IR2/(R1+R2) Similarly, I = I2R2((R1+R2)/R1R2) = I2((R1+R2)/R1) ∴ I2 = IR1/(R1+R2) 10 | P a g e BEEE UNIT-1 DC CIRCUITS Numericals: 1. Calculate the effective resistance of the circuit of Fig. and the current through 8β¦ resistance, when potential difference of 60 V is applied between the points A and B. Solution 11 | P a g e BEEE UNIT-1 DC CIRCUITS 2. What is reading of the ammeter shown in the circuit shown? Solution 12 | P a g e BEEE UNIT-1 DC CIRCUITS By Ohm’s law, π 100 Total current, πΌ = π = 2 = 50π΄ ππ 2 By current division rule, πΌ2 = 50 × 2+2 = 25π΄ Hence, the ammeter reading is 25A. 3. Calculate the effective resistance between the points A and B in the circuit shown. Solution 13 | P a g e BEEE UNIT-1 DC CIRCUITS Thus, the equivalent resistance between the terminals A and B, RAB = 3.43 β¦ 1.10 Junction, Node, Loop and Mesh Node- Point in the network where two or more circuit elements meet together. Junction- Point in the network where three or more branches meet together. Loop- A loop is any closed path through a circuit where no node is encountered more than once. Mesh- A mesh is a closed path during a circuit with no other paths inside it. A mesh is additionally a loop but a loop may or might not be a mesh. In the above figure, Node is A and C, Junction is B and E, Branch is AB, BC, CD, BE and AF, Loop is ABCDEFA, Mesh is ABEFA, BCDEB. 14 | P a g e BEEE UNIT-1 DC CIRCUITS 1.11 Maxwell's Mesh Current Method In this method, Kirchhoff's voltage law is applied to each mesh in terms of mesh currents instead of branch currents. Each mesh is assigned a separate mesh current. This mesh current is assumed to flow in clockwise direction around the perimeter of the mesh without splitting at a junction into branch currents. Kirchhoff's voltage law is applied to write equation in terms of unknown mesh currents. Once the mesh currents are known, the branch currents can be easily determined. Maxwell's mesh current method consists of the following steps: 1. Each mesh is assigned a separate mesh current. For convenience, all mesh currents are assumed to flow in clockwise direction. For example, in Fig, meshes ABDA and BCDB have been assigned mesh currents I1 and 12 respectively. 2. If two mesh currents are flowing through a circuit element, the actual current in the circuit element is the algebraic sum of two. Thus, in Fig, there are two mesh currents I 1 and 12 flowing in R2. If we go from B to D, current is (I 1 – I2) and if we go in the other direction (i.e. from D to B), current is (I2 – I1). 3. Kirchhoff"s voltage law is applied to write equation for each mesh in terms of unknown mesh currents. 4. If the value of any mesh current comes out to be negative in the solution, it means that true direction of that mesh current is anticlockwise, i.e., opposite to the assumed clockwise direction. Consider a circuit as shown in Fig. Applying Kirchhoff's voltage law to mesh ABDA, −πΌ1 π 1 − (πΌ1 − πΌ2 )π 2 + πΈ1 = 0 πΌ1 (π 1 + π 2 ) − πΌ2 π 2 = πΈ1 Applying Kirchhoff's voltage law to mesh BCDB, −πΌ2 π 3 − (πΌ2 − πΌ1 )π 2 − πΈ2 = 0 − πΌ1 π 2 + (π 2 + π 3 )πΌ2 = −πΈ2 Solving equations simultaneously, mesh currents I1 and I2 can be calculated. Once the mesh currents are known, the branch current can be readily obtained. Note: Branch currents are the real currents because they actually flow in the branches and can be measured. However, mesh currents are fictitious quantities and cannot be measured directly. Hence, mesh current is concept rather than a reality. 15 | P a g e BEEE UNIT-1 DC CIRCUITS Numericals: 1. With the help of mesh current method, find the magnitude and direction of the current flowing through the 1β¦ resistor in the network. Solution Marking the different nodes and assigning the separate mesh current for each mesh, we have 16 | P a g e BEEE UNIT-1 DC CIRCUITS 2. By mesh analysis, find mesh currents I1, I2, and I3 in the network. Solution Marking the different nodes, we get the circuit as shown Applying KVL to mesh ABEDCA, −2 − 6πΌ1 + 19 − 1(πΌ1 − πΌ2 ) = 0 7πΌ1 − πΌ2 = 17 … … … … … … … … … … … … … … … … . . (π) Applying KVL to mesh CDGFC, − (πΌ2 − πΌ1 ) − 3(πΌ2 − πΌ3 ) + 25 − 2πΌ2 = 0 πΌ1 − 6πΌ2 + 3πΌ3 = −25 (ππ) Applying KVL to mesh CDGFC, − 19 − πΌ3 − 3(πΌ3 − πΌ2 ) = 0 3πΌ2 − 4πΌ3 = 19 (ππ) The values of I1, I2 and I3 can be found by solving the above three simultaneous equations, I1 = 2.95A I2 = 3.65A I3 = -2.01A 17 | P a g e BEEE UNIT-1 DC CIRCUITS 1.11 Nodal Analysis (Node Voltage Method) This method is based on Kirchhoff's current law (KCL). Normally, this analysis is carried out to determine voltages of different nodes with respective to reference node. However, after determination of node voltage, currents in all branches can be determined. This method is useful where number of loops is large and hence, mesh analysis becomes lengthy. Nodal analysis also has advantage that a minimum number of equations need to be written to determine the unknown quantities. Following steps are to be taken while solving a problem by nodal analysis. Consider the circuit shown. Step I: Mark all nodes. Every junction of the network where three or more branches meet is regarded as a node. In a circuit, there are four nodes (marked by bold points). But lower two nodes are same, and by joining them, we get only three nodes as shown. Step II: Select one of the nodes as reference node. Normally, for convenience, choose that node as reference where maximum elements are connected or maximum branches are meeting. Obviously, node C is selected as reference node. Reference node is also called zero potential node or datum/ground node. Step III: Assign the unknown potentials of all nodes with respect to the reference node. For example, at nodes A and B, let the potentials are V A and VB with respect to the reference node. Step IV: At each node (excluding reference node), assume the unknown currents and mark their directions (choose the current directions arbitrarily). 18 | P a g e BEEE UNIT-1 DC CIRCUITS Step V: Apply the KCL at each node and write the equations in terms of node voltages. By solving the equations, determine the node voltages. From node voltages, current in any branch can be determined. Numericals: Type I: Simple circuit (Only Voltage sources) 1. By node voltage method, find the current through 15β¦ resistor in the circuit. Solution Marking the different nodes and assigning the unknown currents, we obtain the following circuit: Applying KCL to node 1, πΌ1 = πΌ2 + πΌ3 100 − π1 π1 − 0 π1 − π2 = + 20 10 15 13π1 − 4π2 = 300 (π) Applying KCL at node 2, πΌ3 = πΌ4 + πΌ5 π1 − π2 π2 − 0 π2 − (−80) = + 15 10 10 π1 − 4π2 = 120 (ππ) From equation’s (i) and (ii), π1 = 15 π, π2 = −26.25 π 19 | P a g e BEEE UNIT-1 DC CIRCUITS Hence, π°ππβ¦ = π°π = π½π − π½π = π. ππ π¨ (→) ππ Type II: Branch currents given 2. By node voltage method, find the currents I 1, I2 and I3 in the circuit given. Solution Marking the different nodes, we get Applying KCL at node 1, πΌ1 = 3 + πΌ2 120 − π1 π1 − π2 = 3+ 0.2 0.3 5π1 − 2π2 = 358.2 (π) Applying KCL at node 2, πΌ2 + πΌ3 = 20 π1 − π2 110 − π2 + = 20 0.3 0.1 π1 − 4π2 = −324 (ππ) From equation’s (i) and (ii), π1 = 115.6 π, π2 = 109.9 π Hence, π°π = πππ − π½π π½π − π½π = ππ π¨, π°π = = ππ π¨ π. π π. π 20 | P a g e BEEE UNIT-1 DC CIRCUITS π°π = 1.11 πππ − π½π = ππ¨ π. π Superposition Theorem This theorem is applicable for linear and bilateral networks. According to this theorem, if there are number of sources acting simultaneously in any linear bilateral networks, then each source acts independently of the others, i.e., as if other source did not exit. Statement- This theorem may be stated as in a linear network containing more than one source, the resulted current in any branch is the algebraic sum of the currents that would be produced by each source, acting alone, all other sources of emf being replaced by their respective internal resistances. Illustration Consider a network shown in Fig, having two voltage sources V1 and V2. Let us calculate the current in branch AB of the network by using superposition theorem. Case (i) According to superposition theorem, each source acts independently. Consider source V1 acting independently. At this time, other sources must be replaced by internal resistances. But as internal resistance of V2 is not given, i.e., it is assumed to be zero, V2 must be replaced by short circuit. Thus, the circuit becomes as shown in Fig (a). Using any of the techniques, obtain the current through branch AB, i.e., I AB due to source V1 alone. Case (ii) now, consider source V2 alone, with V1 replaced by a short circuit, to obtain the current through branch AB. The corresponding circuit is shown in Fig (b). Obtain IAB due to V2 alone by using any of the techniques such as mesh analysis, nodal analysis, and source transformation. 21 | P a g e BEEE UNIT-1 DC CIRCUITS Case (iii) According to superposition theorem, the resultant current through branch AB is the sum of the currents through branch AB produced by each source acting independently. Hence, IAB = IAB Current due to V1 + IAB Current due to V2 Numericals: 1. In the circuit shown, find the current flowing through different resistors by using super position theorem. Solution Step I Considering 35 V source acting alone, replacing 40 V source by short circuit, we get the circuit as shown. In circuit, as single source is acting, the actual directions of currents are marked. These branch currents can be calculated as follows: The equivalent resistance across the source, 2 ×4 π ππ = 3 + (2 β₯ 4) = 3 + = 4.33 β¦ 2+4 By ohm’s law, total circuit current, π 35 πΌ= = = 8.08 π΄ π ππ 4.33 Hence, current in 3β¦ resistor, I3β¦ = 8.08 A(→) By current division rule, current in 4β¦ resistor, 2 πΌ4β¦ = 8.08 × = 2.69 π΄ (→) 2+4 22 | P a g e BEEE UNIT-1 DC CIRCUITS Current in 2β¦ resistor, I2β¦ = 8.08 – 2.69 = 5.39 A (↓) Step II Considering 40 V source acting alone, replacing 35 V source by short circuit, we get the circuit as shown. In circuit, as single source is acting, the actual directions of current are marked. These branch currents can be calculated as follows, The equivalent resistance across the source, 2 ×3 π ππ = 4 + (2 β₯ 3) = 3 + = 5.2 β¦ 2+3 By ohm’s law, total circuit current, π 40 πΌ= = = 7.69 π΄ π ππ 5.2 Hence, current in 4β¦ resistor, I4β¦ = 7.69 A(←) By current division rule, current in 3β¦ resistor, 2 πΌ3β¦ = 7.69 × = 3.08 π΄ (←) 2+3 Current in 2β¦ resistor, I2β¦ = 7.69 – 3.08 = 4.61 A (↓) Step III By principle of superposition, the resultant branch currents can be calculated as follows: Current in 3β¦ resistor, π°πβ¦ = π. ππ π¨(→) + π. ππ π¨ (←) = π π¨ (→) Current in 2β¦ resistor, π°πβ¦ = π. ππ π¨(↓) + π. ππ π¨ (↓) = ππ π¨ (↓) Current in 4β¦ resistor, π°πβ¦ = π. ππ π¨(→) + π. ππ π¨ (←) = π π¨ (←) 2. Determine the current in 20β¦ resistor in the network shown in circuit below by superposition theorem. Solution Step I: Considering 10 V source acting alone, replacing 8 V source and 12 V source by short circuit, we obtain circuit as shown. 23 | P a g e BEEE UNIT-1 DC CIRCUITS In the circuit above, the 8 β¦ resistor gets short circuited, i.e., I8β¦ =0A, for circuit simplification, the 8 β¦ resistor can be removed. By removing 8β¦ resistor, we get the circuit as shown: By Ohm’s law 10 πΌ20β¦ = = 0.476 π΄(↓) 1 + 20 Step II: Considering 12 V source acting alone, replacing 8 V source and 10 V source by short circuit, we get In the circuit, the 10 β¦ resistor gets short circuited, i.e., I10β¦ = 0A, for circuit simplification, the resistor 10 β¦ can be removed. By removing 10 β¦ resistors, we get the circuit as shown in Fig. (a) Now, replacing series combination (1β¦ + 20β¦), we get the circuit as shown in Fig. (b) By Ohm’s law 12 πΌ20β¦ = = 0.571 π΄(↑) 1 + 20 Step III: Considering 8V source acting alone, replacing 10V source and 12V source by short circuit, we get the circuit beside. In the circuit of Fig. 1.233, the resistors 10 β¦ and 8 β¦ get short circuited. For circuit simplification, the resistors 10β¦ and 8β¦ can be removed. By removing 10 β¦ and 8 β¦ Q resistors, we get the circuit as shown in Fig. By Ohm’s law, 8 πΌ20β¦ = = 0.381 π΄(↑) 1 + 20 Step IV: According to superposition theorem, resultant current through 20 β¦ resistor is algebraic sum of currents due to individual sources acting alone. Hence, π°ππβ¦ = π. πππ π¨(↓) + π. πππ π¨(↑) + π. πππ π¨(↑) = π. πππ π¨(↑) 24 | P a g e BEEE UNIT-1 DC CIRCUITS 3. Determine the current in 1β¦ resistor in the network shown in Fig. below by superposition theorem. Solution Step I: Considering 4V source acting alone, replacing 1A source and 3A source by open circuit, we get circuit (a) and (b): By ohm’s law, πΌπ΄π΅ = πΌ1β¦ = 4 = 1.33 π΄(↓) 2+1 Step II: Considering 3A source acting alone, replacing 4V source by short circuit and 1A source by open circuit, we obtain the circuit shown beside, In Fig. above, current flowing through 3β¦ resistor I3β¦ = 3A (←). This current divides at node A. Resistors 2β¦ and 1β¦ are in parallel. By current division rule, 2 πΌπ΄π΅ = πΌ1β¦ = 3 × = 2 π΄(↓) 2+1 Step II: Considering 1A source acting alone, replacing 4V source by short circuit and 3A source by open circuit, we obtain the circuit shown beside, In Fig. above, current flowing through 3β¦ resistor I3β¦ = 1A (←). This current divides at node A. Resistors 2β¦ and 1β¦ are in parallel. By current division rule, 2 πΌπ΄π΅ = πΌ1β¦ = 1 × = 0.67 π΄(↓) 2+1 Step IV: According to superposition theorem, resultant current through 1β¦ resistor is algebraic sum of currents due to individual sources acting alone. Hence, π°πβ¦ = π. ππ π¨(↓) + π π¨(↓) + π. ππ π¨(↓) = π π¨(↓) 25 | P a g e BEEE UNIT-1 DC CIRCUITS 4. Determine the current in 10β¦ resistor in the network shown in Fig. below by superposition theorem. Solution Step I Considering 80V source acting alone, replacing 20A source by open circuit and 20V source by short circuit, we get circuit as shown. In fig above resistors 20β¦ and 50 β¦ are in parallel. By circuit reduction technique, we obtain the simplified circuit as shown By ohm’s law, πΌ10β¦ = 80 = 2.73 π΄(→) 5 + 10 + 14.29 Step II Considering 20V source acting alone, replacing 20A source by open circuit and 80V source by short circuit, we get circuit as shown. By source transformation, i.e., converting series combination of voltage source of 20 V and resistor of 20β¦ into equivalent parallel combination of current source and resistor, we get the circuit as shown in Fig. By circuit reduction technique, we obtain the simplified circuit as shown By current division rule, 14.29 πΌ10β¦ = 1 × = 0.49 π΄(←) 5 + 10 + 14.29 Step III Consider 20A source acting alone, replacing 80V source and 20V source by short circuit we get circuit as shown 26 | P a g e BEEE UNIT-1 DC CIRCUITS The circuit can be re drawn as follows By circuit reduction technique, we obtain the simplified circuit as shown By current division rule, πΌ10β¦ = 20 × 5 = 3.41 π΄(←) 5 + 10 + 14.29 Step IV: According to superposition theorem, resultant current through 1β¦ resistor is algebraic sum of currents due to individual sources acting alone. Hence, π°ππβ¦ = π. ππ π¨(→) + π. πππ π¨(←) + π. ππ π¨(←) = π. ππ π¨(←) 27 | P a g e BEEE UNIT-1 DC CIRCUITS 1.12 Thevenin's Theorem Thevenin's theorem is a powerful tool in the hands of engineers to simplify a complex problem and obtain the circuit solution quickly. It reduces the complex circuit to a simple circuit. This theorem is particularly useful to find the current in a particular branch of a network as the resistance of that branch is varied while all other resistances and sources remain constant. This theorem was first stated by French engineer M.L. Thevenin in 1883. According to this theorem, any two terminal networks, however complex, can be replaced by a single source of emf VTH (called Thevenin voltage) in series with a single resistance RTH (called Thevenin resistance). Figure (a) shows a complex network enclosed in a box with two terminals A and B brought out. The network in the box may consist of any number of resistors and emf sources connected in any manner. But according to Thevenin, the entire circuit behind terminals A and B can be replaced by a single source of emf V TH in series with a single resistance RTH as shown in Fig (b). The voltage VTH is the voltage that appears across terminals A and B with load removed. The resistance RTH is the resistance obtained with load removed and looking back into the terminals A and B when all the sources in the circuit are replaced by their internal resistances. Once Thevenin's equivalent circuit is obtained, current through any load RL connected across AB can be readily obtained. Statement- Thevenin's theorem as applied to dc circuits may be stated as under: Any network having terminals A and B can be replaced by a single source of emf V TH (called Thevenin voltage) in series with a single resistance R TH (called Thevenin resistance). (i) The emf VTH is the voltage obtained across terminals A and B with load, if any, removed, i.e., it is open-circuited voltage between A and B. (ii) The resistance RTH is the resistance of the network measured between A and B with load removed and replacing all the voltage/current sources by their internal resistances. Illustration The concept of Thevenin's equivalent circuit across the load terminals can be explained by considering the circuit shown in Fig. (a). The Thevenin's equivalent circuit is shown in Fig. (b). 28 | P a g e BEEE UNIT-1 DC CIRCUITS Note: The internal resistance of an ideal voltage source is zero and that of an ideal current source is infinite. Hence, while finding out RTH, voltage sources are replaced by short circuits and current sources by open circuits. The voltage VTH is obtained across the terminals A and B with RL removed. Hence, VTH is also called open circuit Thevenin's voltage. The circuit to be used to calculate V TH is shown in Fig. (a). While RTH is the equivalent resistance obtained as viewed through the terminals A and B with RL removed and replacing voltage source short circuit and current source by open circuit as shown in Fig. (b). While obtaining VTH, any of the network simplification techniques can be used. When the circuit is replaced by Thevenin's equivalent across the load resistance, the load current can be obtained as πππ» πΌπΏ = π πΏ + π ππ» Steps to apply for Thevenin's Theorem Step 1:Remove the branch resistance through which current is to be calculated. Step 2:Calculate the voltage across these open circuited terminals, by using any one of the network simplification techniques. This is VTH. Step 3:Calculate RTH as viewed through the two terminals of the branch from which current is to be calculated by removing that branch resistance and replacing all sources by their internal resistances. 29 | P a g e BEEE UNIT-1 DC CIRCUITS Step 4:Draw the Thevenin's equivalent circuit showing source VTH with the resistance RTH in series with it. Step 5:Reconnect the branch resistance. Let it be RL. The required current through the branch is given by πΌπΏ = πππ» π πΏ + π ππ» Numericals: 1. Determine the current through 5β¦ resistor in the network by Thevenin's theorem. Solution Current through 5β¦ resistor is required. This resistance can be called load resistance RL. Its terminals A and B are called load terminals. Step I: Calculation of VTH Removing the load resistance from the network, we get the following network: In Fig., voltage appears across the load terminals A and B, which is called Thevenin's voltage (VTH). The voltage VTH, i.e., VAB, can be calculated as follows: Select any path from A to B and marked the selected path by dotted line as shown in Fig. By using any circuit simplification techniques, calculate the current through all the resistances present in the selected path. Then travel through the selected path from B to A and add all voltage drops and emf's algebraically. 30 | P a g e BEEE UNIT-1 DC CIRCUITS In Fig., 3β¦ resistor is present in the selected path, and for calculation of V TH, current through the 3β¦ resistor is required. By using mesh analysis, I 3β¦ can be calculated. Applying KVL to mesh 1 - 6I1 – (I1 – I2) + 10 = 0 7I1 – I2 = 10 Applying KVL to mesh 2 - 2I2 – 3I2 – (I2 – I1) = 0 I1 – 6I2 = 0 Solving (i) and (ii), (i) (ii) I2 = 0.244 A Hence, I3β¦ = 0.244 A (↓) Thus, VTH = VAB = 20 + 3I2 = 20 + 3(0.244) = 20.732 V Step II: Calculation of RTH Removing the load resistance from the network and replacing the voltage sources by short circuit, we get the following network: In the circuit equivalent resistance across the load terminals A and B is called Thevenin’s resistance RTH. By series parallel circuit reduction techniques, we obtain the circuit as shown below. Thus, RTH = RAB = 1.463 β¦ Step III: Calculation of load current Thevenin’s equivalent circuit can be drawn as follows: 31 | P a g e BEEE UNIT-1 DC CIRCUITS By Ohm’s law, πΌπΏ = πππ» π πΏ + π ππ» Hence, π°π³ = π°πβ¦ = ππ. πππ = π. ππ π¨ (↓) π + π. πππ 2. Determine the current through 1.5 β¦ resistor by Thevenin's theorem. Solution Current through 1.5 β¦ resistor is required. This resistance can be called load resistance RL. It’s terminals A and B are called load terminals. Step I: Calculation of VTH Removing the load resistance from the network, we get the circuit as shown in Fig. In Fig., voltage appears across the load terminals A and B, which is called Thevenin's voltage VTH. For calculation of VTH, i.e., VAB, the selected path from A to B is marked by dotted line in Fig. As this path contains the resistors 5 β¦ and 6 β¦ currents through these resistances are required. By using mesh analysis, these required currents can be calculated. Applying KVL to mesh 1, -9I1 – 7.5(I1 – I2) – 6I1 = 0 -22.5I1 + 7.5I2 = 0 (i) Applying KVL to mesh 2, -7.5(I2 – I1) – 30 -5I2 = 0 7.5I1 – 12.5I2 = 30 (ii) Solving (i) and (ii), I1 = - 1 A, I2 = - 3 a Hence, VTH = VAB = - 6 – 5I2 – 6I1 = - 6 -5(-3) – 6(-1) = 15 V 32 | P a g e BEEE UNIT-1 DC CIRCUITS Step II: Calculation of RTH Removing the load resistance from the network and replacing the voltage sources by short circuits, we get the circuit as shown in Fig (a). In Fig (a)., equivalent resistance across the load terminals A and B is called Thevenin's resistance R TH. The circuit can be redrawn as shown in Fig (b). Below, By series parallel circuit reduction techniques, we obtain the circuit as shown below. Thus, RTH = RAB = 4.5 β¦ Step III: Calculation of load current Thevenin’s equivalent circuit can be drawn as follows: By Ohm’s law, πππ» πΌπΏ = π πΏ + π ππ» Hence, π°π³ = π°π.πβ¦ = ππ = π. π π¨ (↓) π. π + π. π 33 | P a g e BEEE UNIT-1 DC CIRCUITS 3. Determine the current through 8 β¦ resistor in the network by Thevenin's theorem. Solution Current through 8 β¦ resistor is required. This resistance can be called load resistance RL. It’s terminals A and B are called load terminals. Step I: Calculation of VTH Removing the load resistance from the network, we get the following network: In fig. beside, voltage appears across the load terminals A and B, which is called Thevenin's voltage VTH. For calculation of VTH, i.e., VAB, the selected path from A to B is marked by dotted line in Fig. As this path contains the resistor 5β¦, current through this resistance is required. By using mesh analysis, this required current can be calculated as follows. 34 | P a g e BEEE UNIT-1 DC CIRCUITS Step II: Calculation of RTH In Fig. above, equivalent resistance across the load terminals A and B is called Thevenin's resistance RTH. By series-parallel circuit reduction techniques, we modify the circuit as shown below Thus, RTH = RAB = 3.636 β¦ Step III: Calculation of load current Thevenin’s equivalent circuit can be drawn as follows: By Ohm’s law, πππ» πΌπΏ = π πΏ + π ππ» Hence, π°π³ = π°πβ¦ = π. πππ = π. πππ π¨ (↓) π + π. πππ . 35 | P a g e BEEE UNIT-1 DC CIRCUITS 1.13 Norton’s Theorem Norton's theorem is converse of Thevenin's theorem. Norton's equivalent circuit uses a current source instead of voltage source and a resistance RN (which is same as RTH) in parallel with the source instead of being in series with it. E.L. Norton, an engineer employed by the Bell Laboratory, USA, first developed this theorem. According to this theorem, any two-terminal network can be replaced by a single current source of magnitude IN (called Norton current) in parallel with a single resistance RN (called Norton resistance). Figure shows a complex network enclosed in a box with two terminals A and B brought out. The network in the box may contain any number of resistors and sources connected in any fashion. But according to Norton, the entire circuit behind terminals AB can be replaced by a current source I N in parallel with a resistance RN as shown in Fig. The current IN is equal to the current that would flow when terminals A and B are short circuited, i.e., IN, is equal to the current flowing through short-circuited terminals AB. The resistance RN is the same as Thevenin's resistance RTH, i.e., RN is the resistance measured at AB with load removed and replacing all sources by their internal resistances. Statement- Norton's theorem as applied to dc circuits may be stated as under: Any complex network having two terminals A and B can be replaced by a current source of current output IN in parallel with a resistance RN. i) The output IN of the current source is equal to the current that would flow through AB when A and B are short circuited. ii) The resistance RN is the resistance of the network measured between A and B with load removed and replacing the source with their internal resistances. Steps to apply Norton's theorem Step 1: Short the branch resistance through which current is to be calculated. Step 2:Obtain the current through this short-circuited branch, using any of the networksimplification techniques. This current is Norton's current IN. Step 3:Calculate RN as viewed through the two terminals of the branch from which current is to be calculated by removing that branch resistance and replacing all sources by their internal resistances. Step 4:Draw the Norton's equivalent circuit showing current source I N, with the resistance RN in parallel with it. Step 5:Reconnect the branch resistance. Let it be RL. The required current through the branch is given by 36 | P a g e BEEE UNIT-1 DC CIRCUITS πΌπΏ = πΌπ × π π π πΏ + π π Numericals: 1. By Norton's theorem, find the current in 20 β¦ resistor in the network shown in Fig. Solution Current through 20 β¦ resistor is required. This resistance can be called load resistance RL. Its terminals A and B are called load terminals. Step I: Calculation of IN Removing the load resistance from the network and short circuiting the load terminals, we get the modified network as shown in Fig. In Fig., current flowing through the short circuit placed across the load terminals A and B is called Norton current IN. This current can be calculated by mesh analysis as shown below. Applying the KVL to mesh l, -10I1 – 10(I1 – I2) + 100 = 0 -20I1 + 10I2 = -100 Applying the KVL to mesh 2, -15I2 – 10(I2 – I1) = 0 10I1 – 25I2 = 0 Solving equation (i) and (ii), I2 = 2.5 A Hence, IN = 2.5 A, from A to B Step II: Calculation of RN Removing the load resistance from the network replacing the voltage source by short circuit, we get the network as shown in Fig. In fig equivalent resistance across the load terminals A and B is called Norton’s resistance RN. Thus, RN = RAB = (10 || 10) + 15 = 20 β¦ (i) (ii) and 37 | P a g e BEEE UNIT-1 DC CIRCUITS Step III: Calculation of load current Norton’s equivalent circuit can be drawn as shown in fig By current division rule, π°π³ = π°ππβ¦ = π. π × ππ = π. ππ π¨ (↓) ππ + ππ 2. Obtain Norton's equivalent circuit across A and B as shown in Fig. Solution For simplicity, the circuit shown in Fig. above can be redrawn as shown in Fig. below Step I: Calculation of IN In Fig. above, current flowing through the short circuit placed across the load terminals A and B is called Norton's current IN In the circuit, there are two combinations as follows: (i) Series combination of voltage source of 10 V and resistor of 5β¦ (ii) Series combination of voltage source of 5 V and resistor of 4β¦ Converting the above combinations into equivalent combinations, we get the simplified circuit as follows: 38 | P a g e BEEE UNIT-1 DC CIRCUITS In Fig. above, all elements are in parallel. The resistors 5β¦ and 2β¦ are in parallel. Similarly resistors 1β¦ and 4β¦ are in parallel. In the circuit of Fig. above, there are two combinations as follows: (i) Parallel combination of current source of 2 A and resistor of 1.43β¦ (ii) Parallel combination of current source of I .25 A and resistor of 0.8β¦ Converting the above combinations into equivalent combinations, we get the simplified circuit as follows: By applying KVL to above circuit 2.86 + 1.43IN + 0.8IN + 1 = 0 IN = - 1.73 A Step II: Calculation of RN Replacing the voltage sources by short circuit, we get the network as shown in Fig. Below In circuit, equivalent resistance across load terminals A and B is called Norton's resistance R N. In Fig. above, resistors 5β¦ and 2β¦ are in parallel. Also resistors 1β¦ and 4β¦ are in parallel. 39 | P a g e BEEE UNIT-1 DC CIRCUITS Thus, RN = RAB = 1.43 + 0.8 = 2.23 β¦ Step III: Norton's equivalent circuit Norton's equivalent circuit can be drawn as follows: 1.15 Growth and decay of current in L-R circuit The current growth and decay in RL circuits can be understood by considering a series RL circuit consisting of a resistor, an inductor, a constant source of emf, and two switches. When the first switch is closed, the circuit is equivalent to a single-loop circuit consisting of a resistor and an inductor connected to a source of emf. In this case, the source of emf produces a current in the circuit. If there were no self-inductance in the circuit, the current would rise immediately to a steady value of ε/R. However, from Faraday's law, the increasing current produces an emf across the inductor, which has opposite polarity. In accordance with Lenz’s law, the induced emf counteracts the increase in the current. As a result, the current starts at zero and increases asymptotically to its final value. Thus, as the current approaches the maximum current ε/R, the stored energy in the inductor increases from zero and asymptotically approaches a maximum value. The growth of current with time is given by 40 | P a g e BEEE UNIT-1 DC CIRCUITS When the first switch is opened, and the second switch is closed, the circuit again becomes a single-loop circuit but with only a resistor and an inductor. Now, the initial current in the circuit is ε/R. The current starts from ε/R and decreases exponentially with time as the energy stored in the inductor is depleted. The decay of current with time is given by the relation The quantity inductance over resistance is given by measures how quickly the current builds toward its final value; this quantity is called the time constant for the circuit. When the current is plotted against time, It grows from zero and approaches ε/R asymptotically. At a time equal to time constant, the current rises to about 63%, of its final value, but during decaying, at the same time constant, it decreases to about 37%, of its original value. 41 | P a g e
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