L-04(GDR)(ET) ((EE)NPTEL)

Module

2

DC Circuit

Lesson

4

Loop Analysis of

resistive circuit in the context of dc voltages

and currents

Objectives

Meaning of circuit analysis; distinguish between the terms mesh and loop.

To provide more general and powerful circuit analysis tool based on Kirchhoff’s voltage law (KVL) only.

L.4.1 Introduction

The Series-parallel reduction technique that we learned in lesson-3 for analyzing DC circuits simplifies every step logically from the preceding step and leads on logically to the next step. Unfortunately, if the circuit is complicated, this method (the simplify and reconstruct) becomes mathematically laborious, time consuming and likely to produce mistake in calculations. In fact, to elevate these difficulties, some methods are available which do not require much thought at all and we need only to follow a well-defined faithful procedure. One most popular technique will be discussed in this lesson is known as ‘mesh or loop’ analysis method that based on the fundamental principles of circuits laws, namely, Ohm’s law and Kirchhoff’s voltage law. Some simple circuit problems will be analyzed by hand calculation to understand the procedure that involve in mesh or loop current analysis.

L.4.1.1 Meaning of circuit analysis

The method by which one can determine a variable (either a voltage or a current) of a circuit is called analysis. Basic difference between ‘mesh’ and ‘loop’ is discussed in lesson-3 with an example. A ‘mesh’ is any closed path in a given circuit that does not have any element (or branch) inside it. A mesh has the properties that (i) every node in the closed path is exactly formed with two branches (ii) no other branches are enclosed by the closed path. Meshes can be thought of a resembling window partitions. On the other hand, ‘loop’ is also a closed path but inside the closed path there may be one or more than one branches or elements.

L.4.2 Solution of Electric Circuit Based on Mesh (Loop)

Current Method

Let us consider a simple dc network as shown in Figure 4.1 to find the currents through different branches using Mesh (Loop) current method.

Applying KVL around mesh (loop)-1:(note in mesh-1, I1 is known as local current and

other mesh currents I2&I3 are known as foreign currents.)

Va Vc (I1 I3)R2 (I1 I2)R4=0

Va Vc=(R2+R4)I1 R4I2 R2I3=R11I1 R12I2 R13I3 (4.1) Applying KVL around mesh (loop)-2:(similarly in mesh-2, I2 is local current and

I1&I3are known as foreign currents) Vb (I2 I3)R3 (I2 I1)R4=0

Vb= R4I1+(R3+R4)I2 R3I3 = R21I1+R22I2 R23I3

Applying KVL around mesh (loop)-3:

Vc I3R1 (I3 I2)R3 (I3 I1)R2=0

** In general, we can write for ith mesh ( for i=1,2,.....N) (4.2) Vc= R2I1 R3I2+(R1+R2+R3)I3 = R31I1 R32I2+R33I3 (4.3)

→ simply means to take the algebraic sum of all voltage sources around the ith

mesh. ii

thRii → means the total self resistance around the i mesh. ∑V∑Vii= Ri1I1 Ri2I2........+RiiIi Ri,i+1Ii+1 ....RiNIN

thRij → means the mutual resistance between the and j meshes.

Note: Generally, Rij=Rji ( true only for linear bilateral circuits)

Ii → the unknown mesh currents for the network.

Summarize:

Step-I: Draw the circuit on a flat surface with no conductor crossovers.

Step-2: Label the mesh currents (Ii) carefully in a clockwise direction.

Step-3: Write the mesh equations by inspecting the circuit (No. of independent mesh (loop) equations=no. of branches (b) - no. of principle nodes (n) + 1).

Note:

To analysis, a resistive network containing voltage and current sources using ‘mesh’

equations method the following steps are essential to note:

If possible, convert current source to voltage source.

Otherwise, define the voltage across the current source and write the mesh equations as if these source voltages were known. Augment the set of equations

with one equation for each current source expressing a known mesh current or

difference between two mesh currents.

Mesh analysis is valid only for circuits that can be drawn in a two-dimensional

plane in such a way that no element crosses over another.

Example-L-4.1: Find the current through 'ab-branch' (Iab) and voltage (Vcg) across the current source using Mesh-current method.

Solution: Assume voltage across the current source is v1 (‘c’ is higher potential than ‘g’ (ground potential and assumed as zero potential) and note I2 = -2A (since assigned current direction (I2) is opposite to the source current)

Loop - 1: (Appling KVL)

Va (I1 I3)R2 (I1 I2)R4=0 3=3I1 2I2 I3

3I1 I3= 1 (4.4) Loop - 2: (Appling KVL)

Let us assume the voltage across the current source is v1 and its top end is assigned with a positive sign.

v1 (I2 I1)R4 (I2 I3)R3=0 v1= 2I1+6I2 4I3

2I1+12+4I3=v1 (note:I2= 2A ) (4.5) Loop - 3: (Appling KVL)

I3R1 (I3 I2)R3 (I3 I1)R2=0 I1 4I2+8I3=0

I1 8I3=8 (Note, I2= 2A) (4.6)

Solving equations (4.4) and (4.6), we get I1=

I3= 48= 0.6956A and 6925= 1.0869A, Iab=I1 I3=0.39A , Ibc=I2 I3= 0.913A and 23

Ibg=I1 I2=1.304A

- ve sign of current means that the current flows in reverse direction (in our case, the current flows through 4Ω resistor from ‘c’ to ‘b’ point). From equation (4.5), one can get v1 == 6.27 volt.

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