# Electrical Engineering Questions and Solutions Set 11

Electrical Engineering Questions and Solutions Set 11
Question:-
Option (A)
1136 r.p.m
Option (B)
1236 r.p.m
Option (C)
1436 r.p.m
Option (D)
1500 r.p.m
Correct Option:
C
Question Solution:
Ish = (220/110) 2A
Ia1 = 5-2 = 3A
Ia2 = 52 -2 = 50A
Here Eb1 = 220 - 3 x 0.2 = 219.4 volt
Eb2 = 220 - 50 x 0.2 = 210.0 volt
So (N2/1500) = (210/219.4)
→ N2 1436 r.p.m
Question:-
Option (A)
12.5 A
Option (B)
16.9 A
Option (C)
20.5 A
Option (D)
21.7 A
Correct Option:
B
Question Solution:
turn - off time (tc) = ((2π -β)/ω)
=((360-210) π)/(180 x 2π x 50 ))
Tc = 8.33 m sec
Average output voltage (v0)
V0 = (√2.230/π )(cos40° - cos210°)
= 84.477 volt
I0 = (84.477/5) = 16.8954A
Question:-
Option (A)
75 A
Option (B)
120 A
Option (C)
150 A
Option (D)
106 A
Correct Option:
C
Question Solution:
V0 =380 + I0 x 0.5
((3√ 2 x 440) /π ) cos40° = 380 +0.5I0
I0 = 150.38 A
Question:-
Option (A)
16.5°
Option (B)
8.25°
Option (C)
33.0°
Option (D)
None
Correct Option:
A
Question Solution:
V = 1 p.u.
Xd = 0.7 p.u.
Xq = 0.4 p.u.
cos Æ = 0.8 => Æ = 36.9°, Ia = 1 p.u.
tan δ= (Ia Xq cos Æ) /( V+ Xq sin Æ)
=(1*0.4*0.8)/(1+0.4 x 0.6)
=0.258
d = 16.5°
Question:-
Option (A)
5.77 mF
Option (B)
11.54 mF
Option (C)
34.62 mF
Option (D)
69.24 mF
Correct Option:
B
Question Solution:
100 x 10-6 = 25 x c ln (2)
C = (100 x 10-6)/(25 x ln(2) ) = 5.77 μF
safety factor ‘2’ → C = 11.54 μF.
Question:-
Option (A)
(V/4fL)
Option (B)
(V/2fL)
Option (C)
(ÖV/4fL)
Option (D)
(V/4f ÖL)
Correct Option:
A