Page 21 - NETWORK ANALYSIS

15
Mesh Analysis

Converting mesh equations into matrix form;

(20 + j20) (15 + j20) i1 22
[ ] [ ] = [ ]
(15 + j20) (25 + j10) i2 10

To find ∆:

(20 + j20) (15 + j20)
[ ]
(15 + j20) (25 + j10)

∆ = (20 + j20) (25 + j10) – (15 + j20) (15 + j20) = 475 + j100

To find ∆ i1

22 (15 + j20)
[ ]
10 (25 + j10)

∆ i1 = (22) (25 + j10) – (15 + j20) (10) = 400 + j20

∆ i1 (400 + j20)
i1 = = = (0.815 – j0.129) A
∆ (475 + j100)

To find ∆ i2

(20 + j20) 22
[ ]
(15 + j20) 10

∆ i2 = (20 + j20) (10) – (22) (15 + j20) = – 130 – j240

∆ i2 (– 130 – j240)
i2 = = = (– 0.364 – j0.429) A
∆ (475 + j100)

i5Ω = i1 = (0.815 – j0.129) A

i10Ω = i2 = (– 0.364 – j0.429) A

We know that, i1 and i2 in the same direction;

i15Ω = i1 + i2 = (0.815 – j0.129) + (– 0.364 – j0.429) = (0.451 – j0.558) A