This companion volume to Electrical Engineering License Review presents the main book's end-of-chapter problems with detailed step-by-step solutions. A sample exam, also with step-by-step solutions, is included. 100% problems and solutions.x 208 25^0Ad amps 10x746 Im = -p = 35.3Z45Ad amps = 25 + 725 V3x 208x0.83x0. 707 Thus ImR=25ZOAd /my=25Z-90Ad /, = 25 + 25 + 725 = 50 + J25 ... wire size: 125 % /m NEW + /, = 1.25 x 30 + 25 = 62.5 amps From copper wire tables (NEC, Article 310, Table 310-16) for 70 amps ... From over current protection table at 30 amps (no 125% factor is needed) +25 =55 amps we obtain a fuse size of 120 amps.

Title | : | Electrical Engineering Problems and Solutions |

Author | : | Lincoln D. Jones |

Publisher | : | Dearborn Trade Publishing - 2003-09-01 |

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