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KITRC CIRCUIT & NETWORKS MADE BY AGNA PATEL:130260111011 AESHA CHAUHAN:130260111004 28-Apr-17 1 CONTENTS • Definitions • Kirchoff’s Voltage Law (KVL) • Kirchoff’s Current Law (KCL) • Nodal analysis • Mesh analysis 28-Apr-17 2 BRANCHES AND NODES Branch: elements connected end-to-end, nothing coming off in between (in series) Node: place where elements are joined—entire wire 28-Apr-17 3 NOTATION: NODE VOLTAGES The voltage drop from node X to a reference node (ground) is called the node voltage Vx. Example: a + Va _ b + + _ Vb _ ground 28-Apr-17 4 Linear network Nonlinear network – A circuit or network to A circuit whose parameter parameter are R,L,C are changes there value with always constant change in time, temp irrespective of change in ,voltage is known as non time .voltage, temp., linear parameter. known as linear network. 28-Apr-17 5 Bilateral network – Unilateral network – A circuit characteristic behavior irrespective of direction of current to various element is known as bilateral network 28-Apr-17 A circuit whose operation behavior is depended on direction of current to various element is known as unilateral network 6 Active network Passive network A circuit which contains A circuit which consist of at least one source of energy is called active network. example – voltage source, ac/dc signal, generator , transistor. 28-Apr-17 number any kind of source is called passive network . Example – R,L,C. 7 KIRCHOFF’S VOLTAGE LAW (KVL) The sum of the voltage drops around any closed loop is zero. We must return to the same potential (conservation of energy). Path + V1 “drop” - Path - “rise” or “step up” V2 (negative drop) + Closed loop: Path beginning and ending on the same node Our trick: to sum voltage drops on elements, look at the first sign you encounter on element when tracing path 28-Apr-17 8 KVL EXAMPLE + v2 1 + va b a + vb - v3 2 + Examples of three closed paths: c + vc 3 Path 1: va v 2 vb 0 Path 2: vb v 3 vc 0 Path 3: va v2 v3 vc 0 28-Apr-17 9 KCL [KIRCHOFF CURRENT LAW] “Algebraic sum” of currents entering node = 0 where “algebraic sum” means currents leaving are included with a minus sign “Algebraic sum” of currents leaving node = 0 currents entering are included with a minus sign 28-Apr-17 10 KIRCHHOFF’S CURRENT LAW EXAMPLE 24 A -4 A 10 A i Currents entering the node: 24 A Currents leaving the node: 4 A + 10 A + i Three formulations of KCL: 1: 2: 3: 28-Apr-17 24 4 10 i 24 ( 4) 10 i 0 24 4 10 i 0 i 18 A i 18 A i 18 A 11 Nodal Analysis The selection of mesh current and the application established the mesh analysis For the solution of networks. this was studied in the previous section . In this section ,the same solution Is found by introducing the set of the equation by the application Of kirchoff’s current law . this method is known as nodal analysis. Steps 1. While assuming branch currents, make sure that each unknown branch Current is considered at least once. 2. Convert voltage source present into their equivalent current sources for node analysis where possible. 28-Apr-17 12 3.Follow the same sign convention ,currents entering at node are to be considered positive, while current leaving the node are to be considered as negative. 4.Select the direction of branch current leaving the respective nodes. 28-Apr-17 13 Super node By considering the above figure nodes v2 and v3 are connected directly Through a voltage source without any circuit element . The region surrounding a voltage source which connects two nodes directly is called super node . V2 =v3 + v x 28-Apr-17 14 Super node : circuit 28-Apr-17 15 Nodal Analysis: Concept Illustration: v1 v2 v3 R4 R2 R1 I R3 reference node Figure 6.1: Partial circuit used to illustrate nodal analysis. V1 V2 R2 3 28-Apr-17 V1 R1 V1 R3 V1 V3 R4 I Eq 6.1 16 Nodal Analysis Clearing the previous equation gives, 1 1 1 1 1 1 V1 V2 V3 I R2 R4 R1 R2 R3 R4 Eq 6.2 We would need two additional equations, from the remaining circuit, in order to solve for V1, V2, and V3 4 28-Apr-17 17 Nodal Analysis: Example Given the following circuit. Set-up the equations to solve for V1 and V2. Also solve for the voltage V6. R2 v1 R3 v2 R5 + R1 I1 R4 v6 _ R6 Figure 6.2: Circuit for Example 6.1. 5 28-Apr-17 18 Nodal Analysis: Example : Set up for solution. V1 R1 R2 V2 V1 28-Apr-17 V2 R3 V2 Eq 6.3 Eq 6.4 1 1 1 V1 V2 I 1 R1 R2 R3 R3 Eq 6.5 1 1 1 1 V2 0 V1 R3 R3 R4 R5 R6 Eq 6.6 R4 I1 0 R3 7 V1 V2 R5 R6 19 Mesh Analysis The branch current is resultant current flowing through particular branch of circuit mesh current is fictitious current assumed to be common to all the elements of particular mesh in ckt. Mesh currents are sort of fictitious in that a particular mesh current does not define the current in each branch of the mesh to which it is assigned. I1 28-Apr-17 I2 I3 20 •We assume the current source to be open circuited and create a new mesh called the supermesh . A supermesh constitute by two adjust meshes that have a common current source . We can apply kvl only to those meshes in modified network . This technique is useful as it does not increase the number of eqation to be solved . 28-Apr-17 21 1.certain that n/w is applicable. play n/w or mesh analysis is not 2.Make neat simple circuit diagram indicate all element and source value. 3.Asssuming that the circuit has m mesh assign a clockwise mesh current in each mesh. 28-Apr-17 22 4.If circuit contain only voltage sources apply kvl around each mesh If circuit has only independent voltage source .equate the clockwise sum of all resistance voltage to counter clockwise sum of all source voltage and order the terms of I1 to Im. For each dependent voltage source present ,relate the voltage source and controlling quantity to variable. 5.If circuit contains current source create a super mesh for each one that is common to two mesh by applying kvl around large loop formed by branches not common to mesh.kvl not need be applied to mesh containing current source that lies on perimeter of entire circuit. The assigned mesh current should not be change. 28-Apr-17 23 Basic Circuits Mesh Analysis: R1 + VA V1 R2 _ V2 _ + + _ + I1 VL1 _ Rx + I2 _ VB Figure 7.2: A circuit for illustrating mesh analysis. Around mesh 1: V1 VL1 V A where V1 I1 R1 ; VL1 I1 I 2 RX so, ( R1 RX ) I1 RX I 2 V A 28-Apr-17 Eq 7.1 24 R1 + VA V1 _ + V2 _ + + _ R2 I1 VL1 _ Rx + I2 _ VB Around mesh 2 we have V L1 V2 V B with; V L1 ( I 2 I 1 ) R X ; V2 I 2 R2 Eq 7.2 Eq 7.3 Substituting Eq 7.3 in 7.2 gives, R X I 1 ( R X R2 ) I 2 V B or R X I 1 ( R X R2 ) I 2 V B 28-Apr-17 Eq 7.4 25 We are left with 2 equations: From (7.1) and (7.4) we have, ( R1 RX ) I1 RX I 2 VA Eq 7.5 RX I1 ( RX R2 ) I 2 VB Eq 7.6 We can easily solve these equations for I1 and I2. 28-Apr-17 26 The previous equations can be written in matrix form as: R X I1 V A ( R1 RX ) R ( RX R2 I 2 VB X or Eq (7.7) 1 RX V A I1 ( R1 RX ) I R ( RX R2 VB X 2 28-Apr-17 Eq (7.8) 27 Mesh Analysis Standard form for mesh equations Consider the following: R11 R 21 R31 R12 R22 R32 R13 I1 emfs (1) R23 I 2 emfs ( 2) emfs ( 3) R33 I 3 R11 = of resistance around mesh 1, common to mesh 1 current I1. R22 = of resistance around mesh 2, common to mesh 2 current I2. R33 = of resistance around mesh 3, common to mesh 3 current I3. 28-Apr-17 28 R12 = R21 = - resistance common between mesh 1 and 2 when I1 and I2 are opposite through R1,R2. R13 = R31 = - resistance common between mesh 1 and 3 when I1 and I3 are opposite through R1,R3. R23 = R32 = - resistance common between mesh 2 and 3 when I2 and I3 are opposite through R2,R3. emfs (1) = sum of emf around mesh 1 in the direction of I1. emfs ( 2) = sum of emf around mesh 2 in the direction of I2. emfs ( 3) = sum of emf around mesh 3 in the direction of I3. 28-Apr-17 29 Mesh Analysis: Example Write the mesh equations and solve for the currents I1, and I2. 4 10V + _ 2 7 6 I1 2V +_ I2 _ + 20V Figure 7.2: Circuit for Example 7.1. Mesh 1 Mesh 2 28-Apr-17 4I1 + 6(I1 – I2) = 10 - 2 Eq (7.9) 6(I2 – I1) + 2I2 + 7I2 = 2 + 20 Eq (7.10) 30 Simplifying Eq (7.9) and (7.10) gives, 10I1 – 6I2 = 8 Eq (7.11) -6I1 + 15I2 = 22 Eq (7.12) I1 = 2.2105 I2 = 2.3509 28-Apr-17 31