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Example A 250-W, 230-V, 50-Hz capacitor-start motor has the following constants for the main and auxiliary windings: Main winding, Zm = (4.5 + j 3.7) ohm. Auxiliary winding Za = (9.5+ j 3.5) ohm. Determine the value of the starting capacitor that will place the main and auxiliary winding currents in quadrature at starting. Solution. Let XC be the reactance of the capacitor connected in the auxiliary winding. Then Za = 9.5 +j 3.5 - j XC= (9.5 + jX) where X is the net reactance. Now, Zm = 4.5 +j 3.7= 5.82 39.4° ohm Obviously, Im lags behind V by 39.4° Since, time phase angle between Im and Ia has to be 90°, la must lead V by (90° – 39.4°) = 50.6°. For auxiliary winding, tan pa = X/R or tan 50.6° = X/9.5 or X = 9.5. 1.217 = 11.56 2 (capacitive) .: XC = 11.56 + 3.5 = 15.06 Q :: 15.06 = 1/314 C; C = 211 µF.

Example A 250-W, 230-V, 50-Hz capacitor-start motor has the following constants for the main and auxiliary windings: Main winding, Zm = (4.5 + j 3.7) ohm. Auxiliary winding Za = (9.5+ j 3.5) ohm. Determine the value of the starting capacitor that will place the main and auxiliary winding currents in quadrature at starting. Solution. Let XC be the reactance of the capacitor connected in the auxiliary winding. Then Za = 9.5 +j 3.5 - j XC= (9.5 + jX) where X is the net reactance. Now, Zm = 4.5 +j 3.7= 5.82 39.4° ohm Obviously, Im lags behind V by 39.4° Since, time phase angle between Im and Ia has to be 90°, la must lead V by (90° – 39.4°) = 50.6°. For auxiliary winding, tan pa = X/R or tan 50.6° = X/9.5 or X = 9.5. 1.217 = 11.56 2 (capacitive) .: XC = 11.56 + 3.5 = 15.06 Q :: 15.06 = 1/314 C; C = 211 µF.

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...
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single phase machines
Example A 250-W, 230-V, 50-Hz capacitor-start motor has the following constants for the main and auxiliary windings: Main winding, Zm = (4.5 + j 3.7) ohm. Auxiliary winding Za = (9.5+ j 3.5) ohm. Determine the value of the starting capacitor that will place the main and auxiliary winding currents in quadrature at starting. Solution. Let XC be the reactance of the capacitor connected in the auxiliary winding. Then Za = 9.5 +j 3.5 - j XC= (9.5 + jX) where X is the net reactance. Now, Zm = 4.5 +j 3.7= 5.82 Z 39.4° ohm Obviously, Im lags behind V by 39.4° Since, time phase angle between Im and Ia has to be 90°, la must lead V by (90° – 39.4°) = 50.6°. For auxiliary winding, tan qa = X/R or tan 50.6° = X/9.5 or X = 9.5. 1.217 = 11.56 2 (capacitive) .: XC = 11.56 + 3.5 = 15.06 Q : 15.06 = 1/314 C; C = 211 µF. %3D
Transcribed Image Text:Example A 250-W, 230-V, 50-Hz capacitor-start motor has the following constants for the main and auxiliary windings: Main winding, Zm = (4.5 + j 3.7) ohm. Auxiliary winding Za = (9.5+ j 3.5) ohm. Determine the value of the starting capacitor that will place the main and auxiliary winding currents in quadrature at starting. Solution. Let XC be the reactance of the capacitor connected in the auxiliary winding. Then Za = 9.5 +j 3.5 - j XC= (9.5 + jX) where X is the net reactance. Now, Zm = 4.5 +j 3.7= 5.82 Z 39.4° ohm Obviously, Im lags behind V by 39.4° Since, time phase angle between Im and Ia has to be 90°, la must lead V by (90° – 39.4°) = 50.6°. For auxiliary winding, tan qa = X/R or tan 50.6° = X/9.5 or X = 9.5. 1.217 = 11.56 2 (capacitive) .: XC = 11.56 + 3.5 = 15.06 Q : 15.06 = 1/314 C; C = 211 µF. %3D
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