2: Fill in the blanks: 1- Conversion (0.88µA) to (nA) is 2- The number of electrons pass through the conductor in 2 min and 30sec, if the current (4mA) is 3- The maximum current that can be passed through the coil at temperature rise of 55C° and limited cupper losses is to limited to (600 watts) is resistance of (0.2 N) at (15C°). (ao=0.0043/C) where this coil has

Electricity for Refrigeration, Heating, and Air Conditioning (MindTap Course List)
10th Edition
ISBN:9781337399128
Author:Russell E. Smith
Publisher:Russell E. Smith
Chapter12: Electronic Control Devices
Section: Chapter Questions
Problem 14RQ
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2 : Fill in the blanks:
1- Conversion (0.88µA) to (nA) is
2- The number of electrons pass through the conductor in 2 min and 30sec, if the current
(4mA) is
3- The maximum current that can be passed through the coil at temperature rise of 55C°
and limited cupper losses is to limited to (600 watts) is
resistance of (0.2 N) at (15C°). (ao=0.0043/Cº)
where this coil has
Transcribed Image Text:2 : Fill in the blanks: 1- Conversion (0.88µA) to (nA) is 2- The number of electrons pass through the conductor in 2 min and 30sec, if the current (4mA) is 3- The maximum current that can be passed through the coil at temperature rise of 55C° and limited cupper losses is to limited to (600 watts) is resistance of (0.2 N) at (15C°). (ao=0.0043/Cº) where this coil has
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