Already have an account? Get multiple benefits of using own account!
Login in your account..!
Remember me
Don't have an account? Create your account in less than a minutes,
Forgot password? how can I recover my password now!
Enter right registered email to receive password!
The creation of a magnetic field in a ring may need to be accessed by means of an air gap. This is typically used to apply the magnetic field to a current carrying conductor to produce a force - e.g. in an electric motor.]
Consider therefore, a ring with an air gap present.
Note that since the flux lines are continuous, the flux density in the air gap and the ring will be the same.
However, the air gap represents a much larger reluctance to the flux than the high µ ring, so for the same magneto-motive force, the flux density in the ring will be reduced significantly compared to the case when no air gap was present.
theory and general purposes
The sinusoidal voltage source in the circuit shown in Fig. is developing an rms voltage of 2000 V. The 4 ? load in the circuit is absorbing four times as much average power as the
Control Bus The control bus comprises of various single lines that carry various control signals for synchronizing various devices and performing different task. The m
RE should be made large enough to swamp out rB/ B. how does making RE large saturate the transistor b
Q. Split-phase or resistance-split-phasemotor? Split-phase or resistance-split-phasemotors: Split-phasemotors have two statorwindings (amainwinding and an auxiliarywinding)with
Define HRQ? The hold demand output requests the access of the system bus. In non- cascaded 8257 systems, this is linked with HOLD pin of CPU. In cascade mode, this pin of a sla
what are the advantages and disadvantages of Thevenin theorm over Norton theorem
Q. (i) What are the different types of plots for frequency response in an RC coupled amplifier? (ii) What is a Bode Plot? What are it's uses? The different types of
Q. A 100-kVA, 2300:230-V, 60-Hz, single-phase transformer has the following parameters: R 1 = 0.30 , R 2 = 0.003 , RC 1 = 4.5k, X 1 = 0.65 , X 2 = 0.0065 , and Xm 1 = 1.
What current must flow if 0.45 coulombs is to be transferred in 5 ms? Quantity of electricity, Q = It, then:
Get guaranteed satisfaction & time on delivery in every assignment order you paid with us! We ensure premium quality solution document along with free turntin report!
whatsapp: +91-977-207-8620
Phone: +91-977-207-8620
Email: [email protected]
All rights reserved! Copyrights ©2019-2020 ExpertsMind IT Educational Pvt Ltd