∠ COA = φ = 53.13°
∠ BOE = 90° − 53.13° = 36.87°
∠ DOA = 34.7° Angle between V and I
Angle between Vs and Vb = 18.43°
XL = 314 × 0.0255 = 8 ohms
Zb = 6 + j 8 = 10 ∠ 53.13° ohms
Vb = 10 I = 3 Va, and hence Va = 3.33 I
In phasor diagram, I has been taken as reference. Vb is in first quadrant. Hence Va must be in the fourth quadrant, since Za consists of R and Xc. Angle between Va and I is then 36.87°. Since Za and Zb are in series, V is represented by the phasor OD which is at angle of 34.7°.
| V |= √10 Va = 10.53 I
Thus, the circuit has a total effective impedance of 10.53 ohms.
In the phasor diagram, OA = 6 I , AC = 8 I, OC = 10 I = Vb = 3 Va
Hence, Va = OE = 3.33 I,
Since BOE = 36.87°, OB −RI = OE × cos 36.87° = 3.33 × 0.8 × I = 2.66 I.
Hence, R = 2.66 And BE = OE sin 36.87° = 3.33 × 0.6 × I = 2 I
Hence Xc = 2 ohms. For Xc = 2 ohms, C = 1/(314 × 2) = 1592 μF
Horizontal component of OD = OB + OA = 8.66 I
Vertical component of OD = AC − BE = 6 I
OD = 10.54 I = Vs
Hence, the total impedance = 10.54 ohms = 8.66 + j 6 ohms
Angle between Vs and I = ∠ DOA = tan−1 (6/866) = 34.7°
