# JNTU Question Papers-EEE-Electrical Machines-I-Set-III-Nov-2008

**JNTU II B.Tech I Semester Regular Examinations, November 2008**

**ELECTRICAL MACHINES-I**

**(Electrical & Electronic Engineering)**

**SET-3**

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1. (a) Prove that energy and coenergy in a linear magnetic system are given by identical expressions.

(b) A 10 KW, 1500rpm d.c shunt generator has a time constant Lf/Rf of 0.5 sec for its field winding. Under normal operating conditions, the I2f rf loss in the filed winding is 600 watts. Compute the energy stored in the magnetic field produced by the field winding, under normal operating conditions.

2. What are the ordinary types of armature winding for dc machine? Explain the essential difference between them and give relative merits and the applications of the two types windings.

3. (a) How demagnetizing and cross magnetizing ampere turns per pole are calculated in a DC machine?

(b) Determine per pole, the number of

i. across magnetizing ampere turns

ii. demagnetising ampere turns

iii. series field turns to balance the back ampere turns in the case of a dc generator having the following data

6000 conductors, total current 100A, 6 pole wave wound, angle of lead is

100 , leakage coefficient = 1.3

4. (a) List the conditions for building up of a dc shunt generator.

(b) A d.c. shunt generator is supplying load connected to a bus – bar voltage of 220 V. It has an armature resistance of 0.025 and field resistance of 110 . Calculate the value of load current and load power when it generates an emf of 230 V. Neglect the effect of armature reaction. Draw circuit diagram.

5. (a) What are the reasons for failure of voltage build up in a self excited D.C generator

(b) A shunt generator is to be converted into a level compounded generator by the addition of a series field winding. From a test on the machine with shunt excitation only, it to give 400V on no-load and 4.8A to give the same voltage when the machine is supplying its full load of 200A. the shunt winding has 1200 turns/pole. find the noos series turns required per pole.

6. (a) Draw and explain the dc Series motor characterstics.

(b) The magnetization characteristic of a 4-pole dc series motor may be taken as proportional to current over a part of the working range; on this basis the flux per pole is 4.5 mwb/A. The load requires a gross torque proportional to the square of the speed equal to 30 Nm at 1000 rev/min. The armature is wave-wound and has 492 active conductors. Determine the speed at which5rthe motor will run and the current it will draw when connected to a 220 V supply, the total resistance of the motor being 2.0 ohm.

7. A 250V dc series motor has armature and series field resistance of 0.25 and 0.15ohms

respectively.

(a) Calculate the current for developing a torque of 80Nm at 1200 rpm.

(b) Calculate the percentage reduction in flux when the motor runs at 1800 rpm

at half the current obtained in part (a).

8. A 50 Kw, 440 V shunt generator having an armature circuit resistance including inter-pole winding of 0.15 ohm at normal working temperature was run as a shunt motor on no-load at rated voltage and speed. The total current drawn by the motor was 5 A including shunt field current of 1.5 A. Calculate the efficiency of the shunt generator at 3/4th full-load.