# Answer to Question #199869 in Electrical Engineering for A jaif

Question #199869

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

1
2021-05-31T04:51:03-0400 "\u2220 COA = \u03c6 = 53.13\u00b0"

"\u2220 BOE = 90\u00b0 \u2212 53.13\u00b0 = 36.87\u00b0"

"\u2220 DOA = 34.7\u00b0" Angle between V and I

The angle between Vs and Vb = 18.43°

"XL = 314 \u00d7 0.0255 = 8 ohms Zb = 6 + j 8 = 10 \u2220 53.13\u00b0 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\u00b0. | V |= \u221a10 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\u00b0, OB \u2212RI = OE \u00d7 cos 36.87\u00b0 = 3.33 \u00d7 0.8 \u00d7 I = 2.66 I."

Hence, "R = 2.66 And BE = OE sin 36.87\u00b0 = 3.33 \u00d7 0.6 \u00d7 I = 2 I"

Hence "Xc = 2 ohms. For Xc = 2 ohms, C = 1\/(314 \u00d7 2) = 1592 \u03bcF"

Horizontal component of "OD = OB + OA = 8.66 I"

Vertical component of "OD = AC \u2212 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 = "\u2220 DOA = tan\u22121 (6\/866) = 34.7\u00b0"

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