Loop Analysis of Resistive Circuit in the Context of DC Voltages & Currents | Basic Electrical Technology - Electrical Engineering (EE) PDF Download

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.

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.

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.

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.) 

Applying KVL around mesh (loop)-2:(similarly in mesh-2, I₂ is local current and 1₁ & I₃ are known as foreign currents) 

Applying KVL around mesh (loop)-3: 

** In general, we can write for i^th mesh ( for i = 1, 2, .....N)

 simply means to take the algebraic sum of all voltage sources around the i^th mesh. 

R_ii → means the total self resistance around the ith mesh. 
R_ii → means the mutual resistance between the and j^th meshes.  

Note: Generally, R_ij = R_ji (true only for linear bilateral circuits)   

I_i → 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-: Find the current through 'ab-branch' (I_ab) and voltage (V_cg) 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) 

Loop - 2: (Appling KVL)
Let us assume the voltage across the current source is  and its top end is assigned with a positive sign. 

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

I_bg = I1 − I 2 = 1.304 A

- 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.

Another way: −v₁+v_bg + v_bc = 0 ⇒ v₁ = v_cg =6.27 volt.

Example-L-4.2 For the circuit shown Figure 4.3 (a) find V_x using the mesh current method. 

Solution:  One can easily convert the extreme right current source (6 A) into a voltage source. Note that the current source magnitude is 6 A and its internal resistance is 6 Ω . The given circuit is redrawn and shown in Figure 4.3 (c) 

Using the value of I₂ = 2A in equation (4.7), we get V_x = 48volt.

Example-L-4.3 Find V_R for the circuit shown in figure 4.4 using ‘mesh current method. Calculate the power absorbed or delivered by the sources and all the elements. 

Solution: Assume the voltage across the current source is ‘ v ’ and the bottom end of current source is marked as positive sign.  

For loop No. 1: (KVL equation)         

v − (I₁− I₂) ×100 − I₁ ×100 = 0 ⇒ v − 200 I₁ + 100 I₂ = 0    

It may be noted that from the figure that the current flowing through the 10 resistor (in the middle branch) is10 Ω. More specifically, one can write the following expression 

I₁ − I₂ = 10 ×10⁻³  

For loop No. 2: (KVL equation) 
−20− (I₂ − I₁) ×100 − v − I₂ ×100 = 0 ⇒ v + 200 I₂ − 100 I₁ = − 20   

Solving equations (4.8)-(4.10), one can obtained the loop currents as I₁ =-0.095 = - 95mA (-ve sign indicates that the assigned loop current direction is not correct or in other words loop current (I₁) direction is anticlockwise.) and I₂ =-0.105 = -105 mA (note, loop current (I₂) direction is anticlockwise). Now the voltage across the 100 Ω resistor (extreme right branch) is given by v_R = I₂ x100 = -0.105x100 =-10.5volt. .This indicates that the resistor terminal (b) adjacent to the voltage source is more positive than the other end of the resistor terminal

(a). From equation (4.8)  and this implies that the ‘top’ end of the current source is more positive than the bottom ‘end’. v =− 8.5volt
 Power delivered by the voltage source = 20 × 0.105 = 2.1W (note that the current is leaving the positive terminal of the voltage source). On the other hand, the power received or absorbed by the current source = 8.5 × 0.01 = 0.085W (since current entering to the positive terminal (top terminal) of the current source). Power absorbed by the all resistance is given 

= (0.105)² ×100+ (0.095)² ×100+ (10 ×10⁻³)² ×100 = 2.015W .

Further one can note that the power delivered (P_d = 2.1 W) = power absorbed (P_ab = 0.085 + 2.015 = 2.1W) = 2.1W