Loop analysis or Mesh Analysis

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This is forth in series of the tutorials on circuit analysis. It has been derived by the basic Kirchoffs voltage law. Mesh analysis performs the calculation of values of current, voltage and power of any branch / mesh of the circuit/network by simply applying KVL to that loop. The method involves assumption of a loop current corresponding to a particular mesh. This method is applied to resistive circuits involving independent voltage/current sources. However when we do involve dependent source than modified version named Supermesh is discussed. The procedure is simplified by illustrative examples. To get the realtime knowhow of this method, that is, for a live interaction, learners are directed to contact me at parag.vlsi@gmail.com.

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Mesh or Loop Analysis : Mesh or Loop Analysis Presented By : Parag Parandkar Email: parag.vlsi@gmail.com Contact: +919826139931

PowerPoint Presentation : The second systematic technique to determine all currents and voltages in a circuit. It is dual to node analysis - it first determines all currents in a circuit And then it uses ohm’s law to compute necessary voltages There are situation where node analysis is not an efficient technique And where the number of equations required by this new method is Significantly smaller Loop analysis

PowerPoint Presentation : Apply node analysis to this circuit There are 4 non reference nodes There is one super node There is one node connected to the Reference through a voltage source We need three equations to compute All node voltages …But there is only one current flowing through all components and if That current is determined all voltages can be computed with ohm’s law i Strategy: 1. Apply kvl (Sum of voltage drops =0) 2. Use ohm’s law to express Voltages in terms of the “loop current.” RESULT IS ONE EQUATION IN THE LOOP CURRENT!!! SHORTCUT Skip this equation Write this one Directly

PowerPoint Presentation : Loops, meshes and loop currents Each component Is characterized By its voltage Across and its Current through A loop is a closed path that does not Go twice over any node. This circuit has three loops FABCDEF a mesh is a loop that does not enclose any other loop. fabef , ebcde are meshes A loop current is a ( ficticious ) current That is assumed to flow around a loop FABEF EBCDE A mesh current is a loop current Associated to a mesh. I1, i2 are mesh Currents Claim: in a circuit, the current through Any component can be expressed in terms Of the loop currents Fact: not every loop current is required To compute all the currents through Components The direction of the loop Currents is significant For every circuit there is a minimum Number of loop currents that are Necessary to compute every current In the circuit. Such a collection is called a minimal Set (of loop currents).

PowerPoint Presentation : For a given circuit let B number of branches N number of nodes The minimum required number of Loop currents is Mesh currents are always independent An example Two loop currents are Required. The currents shown are Mesh currents. Hence They are independent and Form a minimal set Kvl on left mesh Replacing and rearranging These are loop equations for the Circuit Determination of loop currents Kvl on right mesh Using ohm’s law In matrix form

PowerPoint Presentation : Write kvl on each mesh Write the mesh equations Identify all voltage drops Bookkeeping Branches = 8 Nodes = 7 Loop currents needed = 2 Use ohm’s law And we are told to Use mesh currents! This defines the loop Currents to be used

PowerPoint Presentation : Draw the mesh currents. Orientation Can be arbitrary. But by convention They are defined clockwise Now write kvl for each mesh and apply Ohm’s law to every resistor. At each loop follow the passive sign Convention using loop current reference Direction Developing a shortcut Whenever an element Has more than one Loop current flowing Through it we compute Net current in the Direction of travel

PowerPoint Presentation : Shortcut: polarities are not needed. Apply ohm’s law to each element as kvl Is being written Rearrange Express variable of interest as function Of loop currents This selection is more efficient Rearrange learning example: find io using loop analysii An alternative selection of loop currents

PowerPoint Presentation : If the circuit contains only independent Source the mesh equations can be written “By inspection” The right hand side is the algebraic sum Of voltage sources around the loop (Voltage rises - voltage drops) Must have all mesh currents with the Same orientation In loop k the coefficent of ik is the sum of resistances around the loop. the coefficient of ij is the sum of resistances common to both k and j and with a negative sign. Loop 1 Loop 2 A practice example Loop 1 Loop 2 loop 3

PowerPoint Presentation : 1. Draw the mesh currents 2. Write mesh equations Mesh 1 Mesh 2 divide by 1k. get numbers for coefficients on the left and ma on the rhs Learning Extension 3. Solve equations

PowerPoint Presentation : Bookkeeping: b = 7, n = 4 1. Draw mesh currents 2. Write mesh equations. Use kvl Equations by inspection Choose your favorite technique To solve the system of equations

PowerPoint Presentation : Circuits with independent current sources There is no relationship between v1 and The source current! However ... Mesh 1 current is constrained Mesh 1 equation Mesh 2 “By inspection” Current sources that are not shared By other meshes (or loops) serve to Define a mesh (loop) current and Reduce the number of required equations To obtain v1 apply kvl to any closed Path that includes v1 Kvl

PowerPoint Presentation : Two mesh currents are defined by current Sources Mesh 3 “By inspection” kvl for vo use kvl to compute vo Learning example

PowerPoint Presentation : We actually need the current on the Right mesh. Hence, use mesh analysis Mesh 1: Mesh 2: Mesh 1: Mesh 2: Learning extensions

PowerPoint Presentation : 1. select loop currents. in this case we use meshes. we note that the current source could define one mesh. 2. write loop equations. loop 1 loop 2 loop 3 since we need to compute vo it is efficient to solve for i3 only. hint: divide the loop equations by 1k. coefficients become numbers and voltage source becomes ma. 3 non-reference nodes. 3 meshes one current source, one super node Selecting the solution method Both approaches seem comparable. Choose Loop analysis we use the fact that i1 = is loop 2 loop 3

PowerPoint Presentation : Current sources shared by loops - the supermesh approach 1. Select mesh currents 2. Write constraint equation due to Mesh currents sharing current sources 3. Write equations for the other meshes 4. Define a supermesh by (mentally) Removing the shared current source Supermesh 5. Write kvl for the supermesh Now we have three equations in three Unknowns. The model is complete

PowerPoint Presentation : Current sources shared by meshes - the general loop approach The strategy is to define loop currents That do not share current sources - Even if it means abandoning meshes For convenience start using mesh currents Until reaching a shared source. At that Point define a new loop. In order to guarantee that if gives an Independent equation one must make sure That the loop includes components that Are not part of previously defined loops A possible strategy is to create a loop By opening the current source The loop equations for the loops with Current sources are The loop equation for the third loop is The mesh currents obtained with this Method are different from the ones Obtained with a supermesh . Even for Those defined using meshes.

PowerPoint Presentation : Three independent current sources. Four meshes. One current source shared by two Meshes. For loop analysis we notice... Careful choice of loop currents Should make only one loop equation Necessary. Three loop currents can Be chosen using meshes and not Sharing any source. SOLVE FOR THE CURRENT I4. USE OHM’S LAW TO C0MPUTE REQUIRED VOLTAGES Now we need a loop current that does Not go over any current source and Passes through all unused components. HINT: IF ALL CURRENT SOURCES ARE REMOVED THERE IS ONLY ONE LOOP LEFT MESH EQUATIONS FOR LOOPS WITH CURRENT SOURCES KVL OF REMAINING LOOP

PowerPoint Presentation : A comment on method selection The same problem can be solved by node analysis But it requires 3 equations

PowerPoint Presentation : Circuits with dependent sources Treat the dependent source as though It were independent. Add one equation for the controlling Variable Combine equations. Divide by 1k

PowerPoint Presentation : SOLVE USING MATLAB PUT IN MATRIX FORM » R=[1,0,0,0; %FIRST ROW 1,1, -1, 0; %SECOND ROW 0,1,3,-2; %THIRD ROW 0,-1,-1,2] %FOURTH ROW R = 1 0 0 0 1 1 -1 0 0 1 3 -2 0 -1 -1 2 DEFINE THE MATRIX DEFINE THE RIGHT HAND SIDE VECTOR » V=[4;0;8;12] V = 4 0 8 -12 Since we divided by 1k the RHS is mA and all the coefficients are numbers SOLVE AND GET THE ANSWER » I=R\V I = 4 -6 -2 -10 The answers are in mA >> is the MATLAB prompt. What follows is the command entered

PowerPoint Presentation : learning extension: dependent sources we treat the dependent source as one more voltage source AND SOLVE... the selection of loop currents simplifies expression for vx and computation of vo. Using loop currents Using mesh currents Mesh 1 Loop 1 Mesh 2 Loop 2 Now we express the controlling variable in terms of the loop currents Replace and rearrange Solutions Notice the difference between mesh Current i1 and loop current i1 even Though they are associated to the Same path find Vo

PowerPoint Presentation : Dependent current source. Current sources not shared by meshes we treat the dependent source as a conventional source then kvl on the remaining loop(s) and express the controlling variable, vx , in terms of loop currents equations for meshes with current sources we are asked for vo. we only need to solve for i3 Replace and rearrange

PowerPoint Presentation : Draw mesh currents Write mesh equations. Controlling variable in terms of Loop currents Replace and rearrange Solve for i2

PowerPoint Presentation : In the following we shall solve using loop analysis two circuits that had previously been solved using node analysis This is one circuit. we recap first the node analysis approach and then we solve using loop analysis

PowerPoint Presentation : Learning example find the voltage vo Identify nodes –and super nodes At super node Controlling variable Solve equations now Variable of interest Recap of node analysis

PowerPoint Presentation : USING LOOP ANALYSIS Determine Vo SELECT LOOP CURRENTS START SELECTION USING MESHES SELECT A GENERAL LOOP TO AVOID SHARING A CURRENT SOURCE Write loop equations Controlling variable: Variable of interest

PowerPoint Presentation : Learning example find the current io Find nodes – and super nodes 7 eqs in 7 variables Variable of interest Recap of node analysis

PowerPoint Presentation : Find the current io using mesh analysis Select mesh currents Write loop/mesh equations 8 eqs in 8 unknowns Variable of interest:

Parag Parandkar
Assistant Professor in Electronics
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