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B00A73GILA Used good or better, we ship best copy available! May have signs of use, may be ex library copy. Book Only. Expedited shipping is 2-6 business days after shipment, standard is 4-14 business days after shipment. Used items do not include access codes, cd's or other accessories, regardless of what is stated in item title. If you need to guarantee that these items are included, please purchase a brand new copy. Bookseller Inventory #

Title: **Alternating-Current Electricity and Its ...**

Publisher: **Ulan Press**

Book Condition: **Good**

Published by
Rarebooksclub.com, United States
(2013)

ISBN 10: 1230177590
ISBN 13: 9781230177595

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**Book Description **Rarebooksclub.com, United States, 2013. Paperback. Book Condition: New. 246 x 189 mm. Language: English . Brand New Book ***** Print on Demand *****. This historic book may have numerous typos and missing text. Purchasers can usually download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1915 edition. Excerpt: .would send through the several parallel paths of Fig. 191. Component of 0.25 amp. current in phase with voltage = 0.25 X cos 0 = 0.25 X1 = 0.25 Component of 6.125 amp. current in phase with voltage = 0.125 X cos 90 = 0.125 X 0 = 0.00 Component of 0.20 amp. current in phase with voltage = 0.20 X cos 37 = 0.20 X 0.80 = 0.16 Component of 0.10 amp. current in phase with voltage = 0.10 X cos 60 = 0.10 X 0.50 = 0.05 Total current in phase with voltage = 0.46 amp. Component of 0.25 amp. current lagging ao behind voltage, = 0.25 X sin 0 = 0.25 X 0 = 0.000 Component of 0.125 amp. current lagging oo behind voltage = 0.125 X sin 90 = 0.125 X 1 = 0.125 Component of 0.20 amp. current lagging oo behind voltage = 0.20 X sin 37 = 0.20 X 0.6 = 0.120 Component of 0.10 amp. current lagging go behind voltage = 0.10 X sin 60 = 0.10 X 0.866 = 0.087 Total current lagging 90 behind voltage = 0.332 Construct Fig. 193 drawing vector for 0.332 amp. 90 behind vector for 0.46 amp. and find resultant. / = V0.332 + 0146 = Vo.3225 = 0.568 amp. The current through the combination equals 0.57 ampere per volt. The power-factor of the combination can be computed as follows: 0.568 Power-factor = 81 per cent. By this method the resulting impedance, power-factor, etc., of all parallel combinations can be found. Prob. 60-6. An arc lamp containing 11 ohms resistance and 9 ohms reactance at 60 cycles is placed in parallel with a choke coil having 0.03 henry inductance and 6 ohms resistance across a 60-cyclc circuit. What is the impedance of the combination? Prob. 61-6. If the parallel combination of Prob. 50 were placed across a 25-cycle circuit, what would be the impedance of the combination? Prob. 62-6. What is the power-factor of the combination of Prob. 50?. Bookseller Inventory # APC9781230177595

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RareBooksClub

ISBN 10: 1230177590
ISBN 13: 9781230177595

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**Book Description **RareBooksClub. Paperback. Book Condition: New. This item is printed on demand. Paperback. 134 pages. Dimensions: 9.7in. x 7.4in. x 0.3in.This historic book may have numerous typos and missing text. Purchasers can usually download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1915 edition. Excerpt: . . . would send through the several parallel paths of Fig. 191. Component of 0. 25 amp. current in phase with voltage 0. 25 X cos 0 0. 25 X1 0. 25 Component of 6. 125 amp. current in phase with voltage 0. 125 X cos 90 0. 125 X 0 0. 00 Component of 0. 20 amp. current in phase with voltage 0. 20 X cos 37 0. 20 X 0. 80 0. 16 Component of 0. 10 amp. current in phase with voltage 0. 10 X cos 60 0. 10 X 0. 50 0. 05 Total current in phase with voltage 0. 46 amp. Component of 0. 25 amp. current lagging ao behind voltage, 0. 25 X sin 0 0. 25 X 0 0. 000 Component of 0. 125 amp. current lagging oo behind voltage 0. 125 X sin 90 0. 125 X 1 0. 125 Component of 0. 20 amp. current lagging oo behind voltage 0. 20 X sin 37 0. 20 X 0. 6 0. 120 Component of 0. 10 amp. current lagging go behind voltage 0. 10 X sin 60 0. 10 X 0. 866 0. 087 Total current lagging 90 behind voltage 0. 332 Construct Fig. 193 drawing vector for 0. 332 amp. 90 behind vector for 0. 46 amp. and find resultant. V0. 332 0146 Vo. 3225 0. 568 amp. The current through the combination equals 0. 57 ampere per volt. The power-factor of the combination can be computed as follows: 0. 568 Power-factor 81 per cent. By this method the resulting impedance, power-factor, etc. , of all parallel combinations can be found. Prob. 60-6. An arc lamp containing 11 ohms resistance and 9 ohms reactance at 60 cycles is placed in parallel with a choke coil having 0. 03 henry inductance and 6 ohms resistance across a 60-cyclc circuit. What is the impedance of the combination Prob. 61-6. If the parallel combination of Prob. 50 were placed across a 25-cycle circuit, what would be the impedance of the combination Prob. 62-6. What is the power-factor of the combination of Prob. 50. . . This item ships from La Vergne,TN. Paperback. Bookseller Inventory # 9781230177595

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Published by
Rarebooksclub.com, United States
(2013)

ISBN 10: 1230177590
ISBN 13: 9781230177595

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Paperback
Quantity Available: 10

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**Book Description **Rarebooksclub.com, United States, 2013. Paperback. Book Condition: New. 246 x 189 mm. Language: English . Brand New Book ***** Print on Demand *****.This historic book may have numerous typos and missing text. Purchasers can usually download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1915 edition. Excerpt: .would send through the several parallel paths of Fig. 191. Component of 0.25 amp. current in phase with voltage = 0.25 X cos 0 = 0.25 X1 = 0.25 Component of 6.125 amp. current in phase with voltage = 0.125 X cos 90 = 0.125 X 0 = 0.00 Component of 0.20 amp. current in phase with voltage = 0.20 X cos 37 = 0.20 X 0.80 = 0.16 Component of 0.10 amp. current in phase with voltage = 0.10 X cos 60 = 0.10 X 0.50 = 0.05 Total current in phase with voltage = 0.46 amp. Component of 0.25 amp. current lagging ao behind voltage, = 0.25 X sin 0 = 0.25 X 0 = 0.000 Component of 0.125 amp. current lagging oo behind voltage = 0.125 X sin 90 = 0.125 X 1 = 0.125 Component of 0.20 amp. current lagging oo behind voltage = 0.20 X sin 37 = 0.20 X 0.6 = 0.120 Component of 0.10 amp. current lagging go behind voltage = 0.10 X sin 60 = 0.10 X 0.866 = 0.087 Total current lagging 90 behind voltage = 0.332 Construct Fig. 193 drawing vector for 0.332 amp. 90 behind vector for 0.46 amp. and find resultant. / = V0.332 + 0146 = Vo.3225 = 0.568 amp. The current through the combination equals 0.57 ampere per volt. The power-factor of the combination can be computed as follows: 0.568 Power-factor = 81 per cent. By this method the resulting impedance, power-factor, etc., of all parallel combinations can be found. Prob. 60-6. An arc lamp containing 11 ohms resistance and 9 ohms reactance at 60 cycles is placed in parallel with a choke coil having 0.03 henry inductance and 6 ohms resistance across a 60-cyclc circuit. What is the impedance of the combination? Prob. 61-6. If the parallel combination of Prob. 50 were placed across a 25-cycle circuit, what would be the impedance of the combination? Prob. 62-6. What is the power-factor of the combination of Prob. 50?. Bookseller Inventory # APC9781230177595

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Published by
Palala Press
(2016)

ISBN 10: 1357208553
ISBN 13: 9781357208554

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**Book Description **Palala Press, 2016. HRD. Book Condition: New. New Book.Shipped from US within 10 to 14 business days.THIS BOOK IS PRINTED ON DEMAND. Established seller since 2000. Bookseller Inventory # IP-9781357208554

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ISBN 10: 1357695926
ISBN 13: 9781357695927

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**Book Description **2016. HRD. Book Condition: New. New Book.Shipped from US within 10 to 14 business days.THIS BOOK IS PRINTED ON DEMAND. Established seller since 2000. Bookseller Inventory # IP-9781357695927

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Published by
Palala Press
(2016)

ISBN 10: 1357208553
ISBN 13: 9781357208554

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**Book Description **Palala Press, 2016. HRD. Book Condition: New. New Book. Delivered from our US warehouse in 10 to 14 business days. THIS BOOK IS PRINTED ON DEMAND.Established seller since 2000. Bookseller Inventory # IP-9781357208554

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Published by
Palala Press
(2016)

ISBN 10: 1357695926
ISBN 13: 9781357695927

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**Book Description **Palala Press, 2016. HRD. Book Condition: New. New Book. Delivered from our US warehouse in 10 to 14 business days. THIS BOOK IS PRINTED ON DEMAND.Established seller since 2000. Bookseller Inventory # IP-9781357695927

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Published by
Rarebooksclub.com, United States
(2012)

ISBN 10: 123655874X
ISBN 13: 9781236558749

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**Book Description **Rarebooksclub.com, United States, 2012. Paperback. Book Condition: New. 246 x 189 mm. Language: English . Brand New Book ***** Print on Demand *****. This historic book may have numerous typos and missing text. Purchasers can download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1914 edition. Excerpt: . 0.16 Component of 0.10 amp. current in phase with voltage = 0.10 X cos 60 = O.io X 0.50 = O05 Total current in phase with voltage = 0.46 amp. Component of 0.25 amp. current lagging 90 behind voltage = 0.25 X sin 0 = 0.25 X 0 = 0.000 Component of 0.125 amp. current lagging 90 behind voltage = 0.125 X sin 90 = 0.125 X 1 = 0.125 Component of 0.20 amp. current lagging 90 behind voltage = 0.20 X sin 37 = 0.20 X 0.6 = 0.120 Component of 0.10 amp. current lagging 90 behind voltage = 0.10 X sin 60 = 0.10 X 0.866 = 0.087 Total current lagging 90 behind voltage = 0.332 Construct Fig. 193 drawing vector for 0.332 amp. 90 behind vector for 0.46 amp. and find resultant. / = V0.3322 + 0.462 = /0.3225 = 0.568 amp. The current through the combination equals 0.57 ampere per volt The power-factor of the combination can be computed as follows: By this method the resulting impedance, power-factor, etc., of all parallel combinations can be found. Prob. 60-6. An arc lamp containing 11 ohms resistance and 9 ohms reactance at 60 cycles is placed in parallel with a choke coil having 0.03 henry inductance and 6 ohms resistance across a 60-cycle circuit. What is the impedance of the combination? Prob. 61-6. If the parallel combination of Prob. 50 were placed across a 25-cycle circuit, what would be the impedance of the combination? Prob. 62-6. What is the power-factor of the combination of Prob. 50? Prob. 63-6. What is the angle of phase difference between the current and voltage of the combination in Prob. 51? Prob. 64-6. An induction coil of 20 ohms impedance and 75 per cent power-factor is placed in parallel with a choke coil of 18 ohms impedance and 3 per cent power-factor. What is the impedance and power-factor of the combination? 69. Mutual Inductance. It has been. Bookseller Inventory # APC9781236558749

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Published by
RareBooksClub

ISBN 10: 123655874X
ISBN 13: 9781236558749

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**Book Description **RareBooksClub. Paperback. Book Condition: New. This item is printed on demand. Paperback. 132 pages. Dimensions: 9.7in. x 7.4in. x 0.3in.This historic book may have numerous typos and missing text. Purchasers can download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1914 edition. Excerpt: . . . 0. 16 Component of 0. 10 amp. current in phase with voltage 0. 10 X cos 60 O. io X 0. 50 O05 Total current in phase with voltage 0. 46 amp. Component of 0. 25 amp. current lagging 90 behind voltage 0. 25 X sin 0 0. 25 X 0 0. 000 Component of 0. 125 amp. current lagging 90 behind voltage 0. 125 X sin 90 0. 125 X 1 0. 125 Component of 0. 20 amp. current lagging 90 behind voltage 0. 20 X sin 37 0. 20 X 0. 6 0. 120 Component of 0. 10 amp. current lagging 90 behind voltage 0. 10 X sin 60 0. 10 X 0. 866 0. 087 Total current lagging 90 behind voltage 0. 332 Construct Fig. 193 drawing vector for 0. 332 amp. 90 behind vector for 0. 46 amp. and find resultant. V0. 3322 0. 462 0. 3225 0. 568 amp. The current through the combination equals 0. 57 ampere per volt The power-factor of the combination can be computed as follows: By this method the resulting impedance, power-factor, etc. , of all parallel combinations can be found. Prob. 60-6. An arc lamp containing 11 ohms resistance and 9 ohms reactance at 60 cycles is placed in parallel with a choke coil having 0. 03 henry inductance and 6 ohms resistance across a 60-cycle circuit. What is the impedance of the combination Prob. 61-6. If the parallel combination of Prob. 50 were placed across a 25-cycle circuit, what would be the impedance of the combination Prob. 62-6. What is the power-factor of the combination of Prob. 50 Prob. 63-6. What is the angle of phase difference between the current and voltage of the combination in Prob. 51 Prob. 64-6. An induction coil of 20 ohms impedance and 75 per cent power-factor is placed in parallel with a choke coil of 18 ohms impedance and 3 per cent power-factor. What is the impedance and power-factor of the combination 69. Mutual Inductance. It has been. . . This item ships from La Vergne,TN. Paperback. Bookseller Inventory # 9781236558749

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Published by
Arkose Press
(2015)

ISBN 10: 1346150109
ISBN 13: 9781346150109

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**Book Description **Arkose Press, 2015. HRD. Book Condition: New. New Book. Delivered from our US warehouse in 10 to 14 business days. THIS BOOK IS PRINTED ON DEMAND.Established seller since 2000. Bookseller Inventory # IP-9781346150109

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