Question

For the circuit shown in Fig. , find the values of R and C so that V_b = 3V_a, and V_b and V_a are in phase quadrature. Find also the phase relationships between V_s and V_b, and V_b and I.

Answer 1

∠ COA = φ = 53.13° 

∠ BOE = 90° − 53.13° = 36.87° 

∠ DOA = 34.7° Angle between V and I 

Angle between V_s and V_b = 18.43° 

X_L = 314 × 0.0255 = 8 ohms 

Z_b = 6 + j 8 = 10 ∠ 53.13° ohms 

V_b = 10 I = 3 V_a, and hence V_a = 3.33 I 

In phasor diagram, I has been taken as reference. V_b is in first quadrant. Hence V_a must be in the fourth quadrant, since Z_a consists of R and X_c. Angle between V_a and I is then 36.87°. Since Z_a and Z_b are in series, V is represented by the phasor OD which is at angle of 34.7°. 

| V |= √10 V_a = 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 = V_b = 3 V_a 

Hence, V_a = 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 X_c = 2 ohms. For X_c = 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°

Comments

Can someone confirm if this soln is correct?

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Yes this soln is correct