Page 28 - NETWORK ANALYSIS

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Mesh Analysis

Note:

The currents i1 and i2 are in the opposite direction.

The source 20<45° V and i1 are in the same direction

Loop i2

By applying Kirchhoff Voltage Law (KVL);

(6 – j4+ j2 – j8) i2 – (–j8) i1 = – j20 V

(j8) i1 + (6 – j10) i2 = – j20 V ----------------------------- (2)

Note:

The source j20 V and i2 are in the opposite direction

From equation (1) and (2), converting mesh equations into matrix form;

(12 + j2) j8 i1 20 < 45° V
[ ] [ ] = [ ]
j8 (6 – j10) i2 – j20 V

To find ∆:

(12 + j2) j8
[ ]
j8 (6 – j10)

∆ = (12 + j2) (6 – j10) – (j8) (j8)

= (92 − j108) – (– 64)

= 156 −