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8.6kw ÷ 240V = 35.8A. ����Ui`-G�5�p]�zYFߵ_?O���{��E��:J����~�����k�'��!-w�m2��SP����ʆ�G �Ktoʯ Table 310.15(B)(16) shows that a #6 THWN-2 copper conductor is rated for 75 amps, therefore: 75 x 0.65 = 48.75A. Transformer Transformer - U1 = 230 V, N1 = 300, N2 = 1,200, I1 = 4 A. Per Table 310.15(B)(16), #1/0 would be sufficient for 150A, that was increased to #350 kcmil. (���f:�;�%�%���/���r�|{?#L p!|w� in. h�b```���\�� cc`a�x-���$1�d.��X������V(W2Z\��Ҳe���?yI� �]&����l� �Y� �o��400L]�5)G6+h��eGv��f@\M�'4E���_N3[&:���E�VS���iKB��6(�v. in. Table 250.122 gives the EGC size for a 400a feed to be a #3 cu wire. Problem-solving skills. You must be very … H�E�A�0EO0w�KܔvJ.1h�q� S��Ą��6$�'���͗�
6�:9�/ State exams cover such areas as power and current formulas, branch circuit load calculations, voltage drop, raceway fill and sizing and more. The basic algebra students learn in high school is only the beginning, a necessary foundation for almost any further development in either mathematics or electrical engineering. Take a free Electrician Math Practice Test to see what kind of math questions are on actual electrician license exams. To determine ratio: 350,000 ÷ 105,600 = 3.314. from the previous question, equals 8.25 cu. H��W��(���� Electrical engineering - math word problems Electrical engineering is an engineering discipline that generally deals with the study and application of electricity, electronics, and electromagnetism. Here it refers you to the appropriate Tables in Ch. Take a free Electrician Math Practice Test to see what kind of math questions are on actual electrician license exams. For this environment, Table 352.44 shows a length change of 3.04"/100'. �?Ԩ��l*tҨ���¬��PEJ����6���2��xb�>D`)�h��ܼ����� ��1%���^��F���c�-�-�s��w���L�F"�u������n�Z]CdQD�ƼG Most states require an electrician to pass an exam to receive a journeyman or master electrician license. Using Ch.9 Table 5, you find the following wire area numbers: #500 kcmil THWN = 0.7073", 314.28(A)(1) requires 8x the largest raceway minimum: 8 x 3" = 24". Electrical Engineering. To determine how many receptacles are permitted to be connected to a circuit, first find the VA of the circuit (volts x amperes) and divide the result by 180 VA. in. Table 250.122 shows a #6 required; 250.122(B) states that if the ungrounded conductors are increased in size, the ECG must be increased proportionately. required, subtracting the available 21 cu. 9, you will need to use Tables 1, 4, and 5 for the following numbers: #14 = 0.0097 circ. State exams cover such areas as power and current formulas, branch circuit load calculations, voltage drop, raceway fill and sizing and more. With 24.75 cu. The calculation for 80 continuous feet of show window is as follows: 250.122(B) requires the EGC to be increased proportionately to the ungrounded conductors. To solve this problem, first find the temperature adder, 40ºF, from Table 310.15(B)(3)(c) due to the proximity of the conduit to the rooftop exposure to sunlight then, add this value to the outdoor ambient temperature: 352.44 states that an expansion coupling is needed if the conduit will change more than 1/4" in length in a straight run. Other adjustment factors could apply, but are here unstated. ��T��a�{�`V���"��?&�ָ�0�*�j
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K�r�|�c$���� 51:U<5���n(�h:�� O؟B�����pٰ"��Ez*v�n�ʊ[Ă���*�:v�)6���. 215.2(A)(1)(a,b) directs the load to be calculated as follows: 35A @ 100%, 85A @ 125%. For help with math calculations on electrician license exams, the following two books written by Ray Holder (Master Electrician and Certified Electrical Trade Instructor) focus on math and show you, in a step-by-step way, how to solve electrician math problems: To solve this problem apply the single-phase current formula: I = P ÷ E. Incorrect answer. in. Then apply the single-phase voltage-drop formula as follows: Table 310.15 (B)(2)(a) shows that a correction factor multiplier of 0.65 applies to this environment. Table 314.16(B) shows an allowance of 2.25 cu. Circular mil area of #6 is 26,240, so applying the ratio = 26,240 x 3.314 = 86,959 circ. Below is a list of math used in electrical engineering: Algebra. 1. 310.15(B)(3)(a) requires an ampacity adjustment for more than 3 current carrying conductors, and since you do not know what the loads are, you must count the neutral. in./ #12. One receptacle is required for every 12 linear ft. or "major fraction thereof" (6 ft. or more). leaves 3.75 cu. Ohm’s law and the electrical formulas related to it are the foundation of all electrical circuits. Chapter 1 Introduction 1.1Themes1 From its beginnings in the late nineteenth century, electrical engineering has blossomed from focusing on electrical circuits for power, telegraphy and telephony to focusing on a much broader range of disciplines. Inf. 314.28(A)(2) states that the distance from the raceway entry to the opposite side of the box shall not be less than 6 times the size of the largest conduit. 220.55 applies Table 220.55 column B, which shows a demand factor of 32% to be applied to the total appliance combined wattage. note 2 shows the need to maintain the voltage drop to less than 3%. 6 x 1.5" = 9". Please choose another answer. Even though the breaker and wiring will be rated at 40 amperes, the minimum ampacity needed is 36A. Copyright © 2009-2020 Tests.com LLC - All Rights Reserved. So, 11 conductors multiplied by 2.25 equals 24.75. ���B]=L1�������X�D�v��g# Number of problems found: 38. Thus. ��A�ekmT���I��"V����|c�3�4��H*�5bmwZ��>A��&��^ �c�c6����}�~{8�UM���`xf�$k�,r�����ts��b�)�K�\kJ4�B�Q���$�����Ʀ���kI@��6�2�H,53'��\xn*z
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