In our previous tutorial, we learned the application of Branch current analysis in a 2 loop circuit with an example.

In this example, a network having three loops is solved using branch current analysis.

In this example, a network having three loops is solved using branch current analysis.

Step 1: Let's first label the circuit:

Step 2: Apply the Kirchhoff's voltage law in the loop between I2 and I3.

5 V = 2 I

Step 3: Apply KVL in the right loop.

1 V = 2 I

Step 4: Apply Kirchhoff's current law at node a.

1 A + I

Solving the three equations we obtain our final answers:

I

I

I

Negative sign with current I

_{2}- 2 I_{3 }... (1)Step 3: Apply KVL in the right loop.

1 V = 2 I

_{3}+ 2 I_{4 }... (2)Step 4: Apply Kirchhoff's current law at node a.

1 A + I

_{2 }+ I_{3}= I_{4}... (3)Solving the three equations we obtain our final answers:

I

_{2 }= 1.5 AI

_{3 }= - 1 AI

_{4 }= 1.5 ANegative sign with current I

_{3 }indicates that actual current flow is opposite to our assumed direction of the current.
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