Page 15 - NETWORK ANALYSIS

9
Mesh Analysis

Example 4

Calculate the value of current I1 using Mesh Analysis.

Figure 1.4

Solution

Loop I1

By applying Kirchhoff Voltage Law (KVL);

(−j6 + 3) − 3 = +25∠0 --------------------------------- (1)
2
1

Loop I2

o
I2 = 5∠30 A --------------------------------- (2)

Insert equation (2) into (1);

(−j6 + 3) − 3 = +25∠0 --------------------------------- (1)
1
2
o

(−j6 + 3) − 3(5∠30 ) = +25∠0
1

(−j6 + 3) = 25∠0 + 3(5∠30 )

1
(−j6 + 3) = (25) + (12.990 + j7.5)
1
(−j6 + 3) = (37.990 + j7.5)
1
(37.990 + j7.5)
= = ( . + . )
1
(−j6 + 3)