In this example students learn how they can easily set the polarity of the elements of the circuit and how to define the loop direction . There is also solution with matrix inversion method. Which help the students to understand the implementation of matrix in real life.
From Loop ABCD
-20 + 4 I1 + 2 I2 = 0
4 I1 + 2 I2 = 20 ------- (i)
From Loop DCEF
-2 I2 + 3 I3 = 0 ------ (ii)
According to KCL
I1 = I2 + I3
I3 = 20/17
I3 = 1.176 A
Putting the value of I3 in Eq (ii)
-2 I2 + 3 (1.176) = 0
- 2 I2 = - 3.528
I2 = 1.764
We know that
I1 = I2 + I3
I1 = 1.764 + 1.176
I1 = 2.94 ≈ 3 A
For checking putting values of I2 and I3 in
Eq. (ii)
-2 (1.764)2 + 3(1.176) = 0
-3.528 + 3.528 = 0
-2 3 I2 = 0
3 4 I3 10
According to Matrix inversion Method
I2 = 1 Adj -2 3 0
I3 |A| 3 4 10
I2 = 1 4 -3 0
I3 (-2x4)-(3x3) -3 -2 10
I2 = 1 (4x0)+(-3x10)
I3 -17 (-3x0)+(-2x10)
I2 = 1 -30
I3 -17 -20
For I2 = -30/-17 = 1.76 A
For I3 = -20/-17 = 1.176 A
Same as our above solution.
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