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
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 AÂ + I 2 + I 3 = I 4 ... (3)
Solving the three equations we obtain our final answers:
I 2 = 1.5 A
I 3 = - 1 A
I 4 = 1.5 A
Negative sign with current I 3 indicates that actual current flow is opposite to our assumed direction of the current.
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 AÂ + I 2 + I 3 = I 4 ... (3)
Solving the three equations we obtain our final answers:
I 2 = 1.5 A
I 3 = - 1 A
I 4 = 1.5 A
Negative sign with current I 3 indicates that actual current flow is opposite to our assumed direction of the current.
No comments:
Post a Comment