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Example 3.27 For the circuit shown in Fig. 3.58 calculate the current in
each branch and total current by the admittance method. Also calculate power
and power factor of the total circuit.
Figure 3.58
Solution:
I =VY = 230 × 0.0589 ∠ − 45° = 13.54 ∠ − 45°A
1
1
I =VY = 230 × 0.0559 ∠ − 64° = 12.85 ∠ − 64°A
2
2
I = I + I = 13.54 ∠ − 45° + 12.85 ∠ − 64°
2
1
or, I = 15.2 − j21 = 25.9 ∠ − 54°A
3.2.8 AC Series—Parallel Circuits
Consider the series–parallel circuit consisting of three branches A, B, and C
as shown. In Fig. 3.59.
