01․ In an alternator, voltage drops occurs in

Under construction.

02․ The frequency of voltage generated by an alternator having 8 poles and rotating at 250 rpm is

Under construction.

03․ Unbalance 3 - ÃŽÂ¦ stator currents cause

Under construction.

04․ The maximum possible speed at which an alternator can be driven to generate 50 Hz and 4000 V is

Under construction.

05․ Two mechanically coupled alternators deliver power at 50 Hz and 60 Hz respectively. The highest speed of the alternators is

Under construction.

06․ A synchorous generator connected to an infinite bus is over excited. Considering only the reactive power from the point view of the system the machine acts as

Since overexcited synchorous generator conncted to an infinite bus. Here no single generator can influence either voltage or frequency.

07․ Two mechanically coupled alternators deliver power at 50 Hz and 60 Hz respectively. Highest speed of alternator is

[math] rac{F_1}{F_2} = rac{P_1}{P_2} = rac{50}{60}
ewline \Rightarrow 6P_1 = 5P_2 [/math]
For maximum speed, no. of poles should be minimum. If P

_{1}= 10, P_{2}= 10 as P_{1}, P_{2}should be even. [math] N_S = rac{50 imes 120}{10} = 600 \; rpm [/math]08․ A part an alternator winding consist of eight coils in series, each having an emf of 20 V rms induced in it. Coils are placed in successive slots and between each slot and the next, there is an electrical phase displacement of 30°. Find the emf of eight coils in series

β = 30° m = 8.
Arithmatic sum of voltage induced in coils = 8 X 20 = 160 V
Vector sum = K

_{d}X airmetic sum = 0.4182 X 160 = 66.91 V.09․ A 3-φ alternator has negligible stator resistance. A short circuit test is conducted on this alternator. At a particular speed, a field current of I

_{f1}is required to drive the rated armature current. If the speed of the alternator is reduced to half, the field current required to maintain rated armature currentIf speed is reduced to half, then induced emf will be reduced to half or in other words to maintain rated open circuit voltage, the field current required is two times the previous value
But impedence varies with field current and so unless we know that the value of Z

_{s}for various field currents it is impossible to maintain rated current at short circuited condition with the half reduced speed.10․ A single phase, 2 KV alternator has an armature resistance of 0.8 ohms and 4.94 ohms respectively. The voltage regulation of the alternator at 100 a load 0.8 loading power factor is

% Voltage regulation = (E

_{O}- V_{FL}) / V_{FL}X 100 where E_{O}= no load voltage. VFL= rated or full load voltage.<<<8283848586>>>