Loop Analysis Technique

Problem-Solving Strategy for solving the Loop Analysis

STEP 1. Determine the number of independent loops in the circuit. Assign a loop current to

each independent loop. For an $N$-loop circuit, there are $N$-loop currents. As a result, $N$ linearly independent equations must be written to solve for the loop currents. If current sources are present in the circuit, either of two techniques can be employed. In the first case, one loop current is selected to pass through one of the current sources. The remaining loop currents are determined by open-circuiting the current sources in the circuit and using this modified circuit to select them. In the second case, a current is assigned to each mesh in the circuit.

STEP 2. Write a constraint equation for each current source—independent or dependent—in

the circuit in terms of the assigned loop current using KCL. Each constraint equation

represents one of the necessary linearly independent equations, and $N_I$ current

sources yield $N_I$ linearly independent equations. For each dependent current source,

express the controlling variable for that source in terms of the loop currents.

STEP 3. Use KVL to formulate the remaining $N−N_I$ linearly independent equations. Treat dependent voltage sources like independent voltage sources when formulating the

KVL equations. For each dependent voltage source, express the controlling variable

in terms of the loop currents.

PROBLEM:

Note: This post is based on Irwin's Basic Engineering Circuit Analysis 11th ed. 

If there are any errors in solution, please let me know in the comments

SOLUTION:

Applying KVL to each loop is as follows:

$12+1k(I_1-I_4)+1kI_1=0$

$-12+1k(I_2-I_3)+1k(I_2-I_5)=0$

$I_3=-6mA$

$2V_x+1k(I_4-I_1)+1k(I_4-I_5)=0$

$I_6-I_5=2I_x$

$1k(I_5-I_4)+1k(I_5-I_2)+1k(I_6-I_3)+1kI_6=0$

The control variables for the two dependent sources are

$V_x=1kI_6$

$I_x=I_4-I_5$

Combining these equations yields

$2kI_1-1kI_4=-12$

$2kI_2-1kI_5=6$

$-1kI_1+2kI_4-1kI_5+2kI_6=0$

$2I_4-I_5-I_6=0$

$-1kI_2-1kI_4+2kI_5+2kI_6=-6$

Thus, each loop current is

$I_1=-6.48mA$

$I_2=3.12mA$

$I_4=-0.96mA$

$I_5=0.24mA$

$I_6=-2.16mA$

Since $I_o=I_5-I_2$

$I_o=-2.88mA$

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