Problems on Matrix Method of Analysis

PROBLEMS ON MATRIX METHOD OF ANALYSIS

Mesh Method / Loop Method

Problem 1.14

Find the current through 2Ω resistor in network of figure 1.43 by mesh method. Also find the power supplied by the batteries.

Solution :

Apply KVL to both loops,

10I1 + 2 (I1 – I2) = 20

12 I1 -2 I1 = 20            …. .. (1)

2 (I2 − I1) + 30 I2 = −40.

−2 I1 + 32 I2  = −40     ...... (2)

Using the above 2 equations, form the matrix equation,

Here we have to solve the mesh currents I1 and I2. By applying Cramer's rule, We can obtain mesh currents.

Current through 2Ω resistor = I1 – I2 = 2.627 A

Power supplied by 20V source = 20 × 1.47 = 29.4 W

Power supplied by 40V source = 40 × -1.157 = -46.28 W.

= 46.28 W (positive value considered)

Problem 1.15

Find the current I which flows through 10 Ω resistor in the circuit shown in figure 1.45.

Solution :

Apply KVL, to obtain mesh equations.

In loop:

+ 30I1 + 30 (I1 – I2) = 90

60 I1 - 30 I2 = 90 …. (1)

In loop 2:

+ 10 I2 + 30 (I2 − I1) + 30 (I2 − I3) = 0

10 I2 + 30 I2 - 30 I1 + 30 I2 - 30 I3 = 0

-30 I1 + 70 I2 - 30 I3 = 0

30 I1 - 70 I2 + 30 I3 = 0 ….(2)

In loop 3 :

30I3 + 30 (I3 – I2) + 30 I3 = 0

- 30 I2 + 90 I3 = 0

30 I2 - 90 I3 = 0   .....(3)

Using above equations form the matrix equations

By applying Crammer's rule, we can

obtain mesh current I2

= 60 [(-70 × −90) - (30 × 30)] - (-30) [(30 × −90) – 0].

= 60 [(6300 -900)] + 30 [-2700]

= 243000.

Therefore, current through 10 Ω resistor is 1A.

Problem 1.16

Find the current through the 10 Ω resistor in network of figure 1.46 by mesh method.

Solution :

Apply kVL in loop 1.

5I1 + 5 (I1 – I3) + 6 (I1 − I2) = 5

5 I1 + 5 I1 - 5 I3 + 6 I1 − 6 I2 = 5

16 I1 - 6 I2 - 5 I3 = 5        ... (1)

In loop 2:

6 (I2 – I3) + 6 (I2 − I1) + 3 I2 = -3

6 I2 − 6 I3 + 6 I2 − 6 I1 + 3 I2 = -3

-6 I1 + 15 I2 - 6 I3 = -3      ……(2)

In loop 3:

5 (I3 − I1) + 6 (I3 − I2) + 10 I3 = 0

5 I3 - 5 I1 + 6 I3 - 6 I2 + 10 I3 = 0

-5 I1 - 6 I2 + 21 I3 = 0             ……(3)

Mesh matrix :

= 16 [(15 × 21) − (−6 × −6)] (-6 × -6)] - (-6) [(-6 × 21) − (−5×-6)] 

- 5 [(-6 × −6) − (−5 × 15)]

= 16 [315 - 36] + 6 [-126 - 30] 5 [36 + 75]

= 16 × 279 + 6(−156) 5 (111)

= 4464 - 936 – 555

= 2973

= 16 [0 - (18)] − (−6) [0 − (15)] + 5 [36 − (−75)]

= -288 - 90 + 555

= 177