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Electrical Engineering Questions and Solutions Set 9

Electrical Engineering Questions and Solutions Set 9
Question:-
Option (A)
x y z
Option (B)
x Å y Å z
Option (C)
x y z
Option (D)
`x`y+z
Correct Option:
D
Question Solution:
f = `x`y + x `y + `xy + xyz

By K – Map f = `x + `y + z
Question:-
Option (A)
1120 rpm
Option (B)
1150 rpm
Option (C)
1170 rpm
Option (D)
1210 rpm
Correct Option:
D
Question Solution:

Assuming linear magnetization characteristics, EbÆ N ∝ IN
Eb1 = 220 - 0.7 x 20 = 206 V
At the new speed, current through series field = ((20 x 0.4)/0.6) = 13.3 A
Eb2 = 220 - [0.5 x 20 + 0.2 x 13.03] = 220 - 12.66 = 207.34 V
Eb1/Eb2) = (I1N1/I2N2)
N2 =(20 x 800 x 207.34 )/(13.3 x 206 ) = 1,210 rpm
Question:-
Option (A)
50 %
Option (B)
70%
Option (C)
90%
Option (D)
100%
Correct Option:
C
Question Solution:

Output torque = 10 x (75/100) = 7.5 kg-m
= 7.5 x 9.81 Nw-m
= 73.575 Nw-m.
Power output = 2p x (1400/60) x 73.575
= 10786.6 W
Power input = 240 x 50 = 12000 W
Therefore, the efficiency = (10786.6/12000) x 100 = 89.9%.
Question:-
Option (A)
5%
Option (B)
7.5%
Option (C)
25%
Option (D)
37%
Correct Option:
D
Question Solution:
N = (1-5) x ((120 x f)/p)
=(1-0.05) x ((120 x 50)/6)
= 950 rpm.
N5new=((120 x f)/p) = ((120 x 50)/4) = 1500
Slipnew=((1500-950)/1500)=0.3667
=36.67%
Question:-
Option (A)
180 MJ
Option (B)
120 MJ
Option (C)
220 MJ
Option (D)
200MJ
Correct Option:
A
Question Solution:
Stored energy is
GH = 20 x 9 = 180 MW. sec. = 180 MJ
Question:-
Option (A)
0.0025
Option (B)
0.00125
Option (C)
0.0050
Option (D)
Data is insufficient
Correct Option:
A
Question Solution:
Strain (e) = (dR/2R)
So, e =(0.6)/(2 x 120) = (1/400) = 0.0025