Answer to Question #199869 in Electrical Engineering for A jaif

Question

The circuit shown in Fig.5 is applied with a 240 V, 50 Hz supply. Find thevalues of R and  C so thatVb=3Va and Vb and Va are in phase quadrature

Answer 1

$∠ COA = φ = 53.13°$

$∠ BOE = 90° − 53.13° = 36.87°$

$∠ DOA = 34.7°$ Angle between V and I 

The angle between Vs and Vb = 18.43° 

$XL = 314 × 0.0255 = 8 ohms Zb = 6 + j 8 = 10 ∠ 53.13° ohms Vb = 10 I = 3 Va, hence Va = 3.33 I$

In the phasor diagram, I have been taken as a reference. Vb is in the first quadrant. Hence Va must be in the fourth quadrant since Za consists of R and Xc. The 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 an 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 = $∠ DOA = tan−1 (6/866) = 34.7°$