Categories

Electrical Engineering Questions and Solutions Set 12

Electrical Engineering Questions and Solutions Set 12
Question:-
Option (A)
140°
Option (B)
250°
Option (C)
180°
Option (D)
360°
Correct Option:
A
Question Solution:

From above figure it is clear that current through inductor starts increasing from 110° to 180° due to positive value of voltage across it ( LΔI = vxΔt). Now after wt = 180°, inductor is subjected to negative voltage. Due to symmetry of positive & negative voltage across inductor, the current through inductor decreases at the same rate at each instant after wt = 180, as it was increasing during wt = 110° to wt = 180°. Hence the current through inductor will become zero at wt = 180 + 70 = 250°. Note that Thyristor can’t turnoff until current through it becomes zero & negative voltage appears across it. Hence Thyristor will turnoff at wt = 250°. So it conducts for q = 70 x 2= 140°.
Question:-
Option (A)
200 Hz
Option (B)
600 Hz
Option (C)
100 Hz
Option (D)
500 Hz
Correct Option:
D
Question Solution:
Line current through a particular phase will appear.

Its Fourier series expansion is
ån=1,3,5(4I0/np) cos(np/6) sin. N wt
Its clear that the term for .n = 3, becomes zero.
So lowest harmonic frequency component is corresponding to n = 5.
Hence f = 5x100 = 500Hz
Question:-
Option (A)
0.5
Option (B)
-4
Option (C)
-0.25
Option (D)
-10
Correct Option:
B
Question Solution:
The root locus exists on real axis at a point if numbers of open-loop poles and zeroes to the right of section is odd.
Root locus is present between 0 and -1, to the left of -6
Question:-
Option (A)
T > 0
Option (B)
T < 0
Option (C)
T > K
Option (D)
T > 1/K
Correct Option:
A
Question Solution:
N = - 2; P+ = 2
N = Z+ - P+ = - 2 => Z+ = 0
\ (-1/T) < 0 Þ T > 0
Hence (A)
Question:-
Option (A)
Both A and B
Option (B)
A only
Option (C)
B only
Option (D)
neither A nor B
Correct Option:
D
Question Solution:
V1 = 15I1 + 5I2
V2 = 5I1 + 25I2
Z11 ¹ Z22 Þ circuit is not symmetric
Z12 ¹ Z21 Þ circuit is bilateral
Question:-
Option (A)
Peaked and flat topped
Option (B)
Peaked and peaked
Option (C)
Flat topped and flat topped
Option (D)
Flat topped and peaked
Correct Option:
D
Question Solution:
Its fact that when input magnetizing current is sinusoid, saturation flux will be flat topped. When a function is flat topped its derivative is always peaked & emf is nothing but the derivative of flux.