By B.L. Teraja, A.K. Teraja
A textbook of electric expertise: quantity 2 ac and dc machines elaborately covers the subjects, ac and dc machines, that's part of electric expertise. electric know-how, as an issue, covers quite a few divisions of electric engineering like simple electric engineering, electronics, keep watch over structures, instrumentation and communique structures. The booklet discusses and explains a number of theories relating to ac and dc machines. The chapters include a number of routines, examples, and a number of illustrations that relief in figuring out the topic larger. This ebook is acceptable for college students, and has been written in basic language. numerous solved examples assist in simplifying the subject material, and the illustrations make it visually extra attractive to scholars. the themes coated within the e-book diversity from dc turbines and dc automobiles to alternators and synchronous automobiles. specified value has been laid on chapters discussing cars. the target attempt part has additionally been rigorously revised to help the readers greater. The twenty third revised version of a textbook of electric know-how: quantity 2 ac and dc machines was once released by way of s chand publishing in 2006. it's to be had in paperback. Key beneficial properties: this illustrated version provides advanced fabric in an built-in and rational series for b. E. , b. Tech. , and different comparable commencement classes.
Read or Download A Textbook of Electrical Technology in S.I Units, Vol. 2: AC and DC Machines PDF
Best electrical & electronics books
This lecture offers an advent to transmission line results within the time area. basics together with time of flight, impedance discontinuities, right termination schemes, nonlinear and reactive so much, and crosstalk are thought of. Required prerequisite wisdom is proscribed to traditional circuit conception.
Supplying a great transition from introductory to complex suggestions, Electromagnetics, moment variation builds a origin that enables electric engineers to expectantly continue with the improvement of complicated EM stories, study, and purposes. This moment variation of a favored textual content keeps to provide assurance that spans the full box, from electrostatics to the essential strategies of Maxwell’s equations.
- Electric power transformer engineering, 1st Edition
- Essential Formulae for Electronic and Electrical Engineers
Additional info for A Textbook of Electrical Technology in S.I Units, Vol. 2: AC and DC Machines
All resistances values are in ohm. 5 Ω] Fig. 33 Fig. 34 Fig. 35 9. In the circuit of Fig. 35, find the resistance between terminals A and B when switch is (a) open and (b) closed. Why are the two values equal ? [(a) 2 Ω (b) 2 Ω] 10. The total current drawn by a circuit consisting of three resistors connected in parallel is 12 A. The voltage drop across the first resistor is 12 V, the value of second resistor is 3 Ω and the power dissipation of the third resistor is 24 W. What are the resistances of the first and third Ω; 6Ω Ω] [2Ω resistors ?
IR = IR1 + IR2 + IR3 or R = R1 + R2 + R3 1 + 1 + 1 1 Also = G G2 G3 G 1 As seen from above, the main characteristics of a series circuit are : 1. same current flows through all parts of the circuit. 2. different resistors have their individual voltage drops. 3. voltage drops are additive. 4. applied voltage equals the sum of different voltage drops. 5. resistances are additive. 6. powers are additive. 15. Voltage Divider Rule Since in a series circuit, same current flows through each of the given resistors, voltage drop varies directly with its resistance.
Branch ABC has a resistance of (3 + 12) = 15 Ω and branch ABC has a resistance of (6 + 4) = 10 Ω. Obviously, each branch has 15 V applied across it. d. across points B and D = VB − VD = 12 − 6 = 6 V When S is closed, the circuit becomes as shown in Fig. 20 (b) where points B and D become electrically connected together. RAB = 3 || 6 = 2 Ω and RBC = 4 || 12 = 3 Ω RAC = 2 + 3 = 5 Ω ; I = 15/5 = 3 A Fig. 20 Current through arm AB = 3 × 6/9 = 2 A. The voltage drop over arm AB = 3 × 2 = 6 V. Hence, drop over arm BC = 15 − 6 = 9 V.