Page 19 - NETWORK ANALYSIS

13
Mesh Analysis

To find ∆:

(5 + j10) (– j10)
[ ]
(– j10) (– j5)

∆ = (5 + j10) (–j5) – (– j10) – j10 = 150 – j25

To find ∆ i1:

−10 < 0° (– j10)
[ ]
12 < 0° (– j5)

∆ i1 = (– 10 <0°) (–j5) – (– j10) (12 <0°) = j170

∆ i1 ( )
i1 = = = – 0.184 + j1.103 A
∆ ( – )

To find ∆ i2:

(5 + j10) −10 < 0°
[ ]
(– j10) 12 < 0°

∆ i2 = (5 + j10) (12 <0°) – (– 10 <0°) (– j10) = 60 + j20

∆ i2 (60 + j20)
i2 = = = 0.368 + j0.195 A
∆ ( – )

The current that goes through j10Ω;

ij10Ω = i1 – i2 = (– 0.184 + j1.103) – (0.368 + j0.195) = – 0.552 + j0.908 A