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Electrical Engineering Questions and Solutions Set 8

Electrical Engineering Questions and Solutions Set 8
Question:-
Option (A)
3.0 sec
Option (B)
4.0 sec
Option (C)
5.0 sec
Option (D)
6.0 sec
Correct Option:
C
Question Solution:
H(S) = (K/1+tS)
For Step i/p Y(S) = (K/S(1+tS))=K[(1/S)-( t/1+tS)]
Þ y(t) = K(1- e-t/t) u(t)
At t = ¥, y(t) = 2 Þ K = 2
Now 1.1 = 2 (1 – e-4/t)(at t=4 y(t) is 1.1)
Þ t= 5.0 sec
Question:-
Option (A)
10mWb
Option (B)
30mWb
Option (C)
35mWb
Option (D)
40mWb
Correct Option:
B
Question Solution:
Since shunt current is negligible, there is no shunt Cu loss. The copper loss occurs in armature only.
I = Ia = 10,000/250 = 40 A; I2aRa= Arm. Cu loss or 402×Ra = 0.64×103; Ra = 0.4 ohm
IaRa drop = 0.4×40 = 16 V; Brush drop = 2×1 = 2 V
\ generated e.m.f. Eg = 250 + 16 + 1 = 267 V
Now,Eg=(j ZN/60)(P/A) volt
\ 267=(jx534x1000/60)(6/6)
\j=30x10-3 Wb=30mWb
Question:-
Option (A)
2.35
Option (B)
5.32
Option (C)
3.45
Option (D)
1.23
Correct Option:
D
Question Solution:
when T remains constant, S1V21 =S2V22
\ (S2/S1)=(V1/V2)2=(1/0.9)2=1.23
Again I2 a sV
\ (I'2/I2)=S2V2/S1V1)=1.2x0.9=1.107
Now, Cu losses are nearly proportional toI22
(Cu loss in the 2nd case)/(Cu loss in the 1st case )=I'2/I2=1.1072=1.23
Question:-
Option (A)
0.24 ohm
Option (B)
0.67 ohm
Option (C)
0.95 ohm
Option (D)
1.24 ohm
Correct Option:
A
Question Solution:
Tst/Tmax=2a/(1+a2); Since Tst=Tmax
\ 1=2a/(1+a2) or a=1 Now , a=(Ra+r)/X2
where r = external resistance per phase added to the rotor circuit
1= (0.06+r)/0.3
\ r=0.3-0.06=0.24 ohm
Question:-
Option (A)
10 kW
Option (B)
20kW
Option (C)
30kW
Option (D)
40kW
Correct Option:
D
Question Solution:
% R = % Cu loss = (Cu loss/VA )x100
Now % R = 0.02 × 100 = 2% \ 2 = (Cu loss/2,000)x100 \ Cu Loss=40kW
Question:-
Option (A)
0.012Wb
Option (B)
0.024Wb
Option (C)
0.048Wb
Option (D)
0.006Wb
Correct Option:
A
Question Solution:
E=(jZN/60)(P/A)volt
\240={(jx175x792)/60}x(4/2)
\ Working flux/pole, j=0.0121 Wb