Mesh Analysis on AC Circuit

Mesh Analysis on AC Circuit


Since KVL is valid for phasors, we can analyze AC circuits by  MESH analysis

Example :

Determine current Io in the circuit of Fig. 10.7 using mesh analysis.

Solution:
Applying KVL to mesh 1, we obtain

 (8+j10−j2)I1 −(−j2)I2 −j10I3 = 0     (10.3.1)

For mesh 2, (4−j2−j2)I2 −(−j2)I1 −(−j2)I3 +20 90◦ = 0     (10.3.2) 

For mesh 3, I3 = 5.

 Substituting this in Eqs. (10.3.1) and (10.3.2), we get

 (8+j8)I1 +j2I2 = j50 (10.3.3) j2I1 +(4−j4)I2 =− j20−j10    (10.3.4)

Equations (10.3.3) and (10.3.4) can be put in matrix form as 

Example 2: