ELECTRICAL WIRING:RESIDENTAL-6 PLANS
18th Edition
ISBN: 9781305098329
Author: MULLIN
Publisher: CENGAGE L
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Chapter 1, Problem 22R
What do the letters “PPE” stand for?
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Chapter 1 Solutions
ELECTRICAL WIRING:RESIDENTAL-6 PLANS
Ch. 1 - Prob. 1RCh. 1 - Prob. 2RCh. 1 - Does the NEC provide minimum or maximum standards?...Ch. 1 - What do the letters UL signify?Ch. 1 - What section of the NEC states that all listed or...Ch. 1 - Prob. 6RCh. 1 - Prob. 7RCh. 1 - Does compliance with the NEC always result in an...Ch. 1 - Name two nationally recognized testing...Ch. 1 - Prob. 10R
Ch. 1 - Prob. 12RCh. 1 - A junction box on a piece of European equipment is...Ch. 1 - Prob. 14RCh. 1 - You will learn in Chapter 3 that residential...Ch. 1 - Prob. 16RCh. 1 - What can you do to reduce or eliminate the...Ch. 1 - What is the NEC definition of a qualified person?Ch. 1 - Are low-voltage systems totally safe? Explain.Ch. 1 - Prob. 21RCh. 1 - What do the letters PPE stand for?...Ch. 1 - Prob. 23RCh. 1 - Where might you obtain information about...Ch. 1 - Prob. 25R
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- Q10 The full-load copper loss on the H.V. side of 100KVA, 11000/317 V, single-phase transformer is 0.62 kw and on the L.V. side is 0.48 kW. i) Calculate R1, R2 in ohms ii) Find X1,X2,if the percentage equivalent reactance is 4%, and reactance is divided in same proportion as resistance. Ans, 27.30, 0.175), 0.00482 . (7.5)arrow_forwardFind the binary sequence, for the following Differential Manchester code.arrow_forwardQ2- What are the parameters and loss that can be determined during open-circuit test of singlephase transformer. Draw the circuit diagram of open-circuit test and explain how can you calculate the Parameters and loss.arrow_forward
- Q6- the open circuit and short circuit tests on a 10 KVA, 125/250 v, 50 Hz single phase transformer gave the following results: O.C. Test: 125 V,0.6 A, 50 W ( on L.V.) S.C. Test: 20 V, 40 A, 177.78 W (on H.V. side) Calculate: i) Copper losses on half load ii) Full load efficiency at 0.8 leading p.f. iii) Half load efficiency at 0.8 leading p.f. iv) Regulation at full load at 0.9 leading p.f. Ans: 44.445 W, 97.23%, 97.69%, -1.8015%arrow_forwardQ3-A two-winding transformer has a primary winding with 208 turns and a secondary winding with 6 turns. The primary winding is connected to a 4160V system. What is the secondary voltage at no load? What is the current in the primary winding with a 50-amp load connected to the secondary winding? What is the apparent power flowing in the primary and secondary circuits? Ans. 120 V, 1.44 A, 6000 VAarrow_forwardQ12- A three phase transformer 3300/400 V,has D/Y connected and working on 50Hz. The line current on the primary side is 12A and secondary has a balanced load at 0.8 lagging p.f. Determine the i) Secondary phase voltage ii) Line current iii) Output power Ans. (230.95 V, 99.11 A, 54.94 kW)arrow_forward
- Q1- A single phase transformer consumes 2 A on no load at p.f. 0.208 lagging. The turns ratio is 2/1 (step down). If the loads on the secondary is 25 A at a p.f. 0.866 lagging. Find the primary current and power factor.arrow_forwardQ7- A 5 KVA, 500/250 V,50 Hz, single phase transformer gave the following reading: O.C. Test: 250 V,2 A, 50 W (H.V. side open) S.C. Test: 25 V10 A, 60 W (L.V. side shorted) Determine: i) The efficiency on full load, 0.8 lagging p.f. ii) The voltage regulation on full load, 0.8 leading p.f. iii) Draw the equivalent circuit referred to primary and insert all the values it.arrow_forwardQ4- A single phase transformer has 350 primary and secondary 1050 turns. The primary is connected to 400 V,50 Hz a.c. supply. If the net cross sectional area of core is 50 cm2, calculate i) The maximum value of the flux density in the core. ii) The induced e.m.f in the secondary winding. Ans: 1.029 T, 1200Varrow_forward
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