Page 13 - NETWORK ANALYSIS
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7
Mesh Analysis
∆ = (7 + j4) (5) – (−3) (−3) = +
To find ∆i1:
20 −3
[ ]
0 5
∆i1 = (20) (5) – (−3) (0) =
∆ i1 100
i1 = = = (2.416 – j1.859) A
∆ 26 + j20
The value of mesh current i1 = (2.416 – j1.859) Amps
Example 3
Calculate the value of mesh current I1 and I2 using mesh analysis.
Figure 1.3
Solution
Loop I1
By applying Kirchhoff Voltage Law (KVL);
o
(-j4 + j5) I1 – (j5) I2= +20∠45
o
(j1) I1 – (j5) I2 = 20∠45 --------------------------------- (1)
