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SSC JE Electrical 2019 with solution SET-2

SSC JE Electrical 2019 with solution SET-2
 The number of valence electrons of P and Si are respectively?

4 and 5
4 and 4
5 and 4
3 and 4

ANSWER EXPLANATION DOWNLOAD EXAMIANS APP

Phosphorus (P) has 15 electron i.e 2,8,5. Hence the number of electron in its outermost orbit is 5.
Silicon (Si) has 14 electron i.e 2,8,4. Hence the number of electrons in its outermost orbit is 4.

SSC JE Electrical 2019 with solution SET-2
In Boolean algebra (A. Ā) + A

1
0
Ā
A

ANSWER EXPLANATION DOWNLOAD EXAMIANS APP

According to Boolean algebra law
A. Ā = A
A + A = A
Therefore in the given question
(A. Ā) + A = A + A = A
 

SSC JE Electrical 2019 with solution SET-2
The function of compensation winding in AC series motor is to

Reduce the effects of armature reaction
Improve commutation
Increase field flux
Reduced the armature resistance

ANSWER DOWNLOAD EXAMIANS APP

SSC JE Electrical 2019 with solution SET-2
In electric traction, the specific energy consumption is measured in

Watt-hour/ ton-km
Watt-hour/RPM
Ton-km watt-hour
RPM/watt-hour

ANSWER DOWNLOAD EXAMIANS APP

SSC JE Electrical 2019 with solution SET-2
If the number of poles in an 11 KV transmission line is 80, then how many disc insulators are required?

80
160
6
240

ANSWER EXPLANATION DOWNLOAD EXAMIANS APP

No of disc required for 11 kV transmission line = 1 disc
Since the transmission line is 3-phase, therefore, the number of discs required for a single-pole will be 3.
Total no. of poles = 80
No of disc = 3 × 80 = 240
 

SSC JE Electrical 2019 with solution SET-2
In a 4 pole. 20 kW, 200 V wave wound DC shunt generator, the current in each parallel path will be

25 A
50 A
10 A
100 A

ANSWER EXPLANATION DOWNLOAD EXAMIANS APP

Given
Power (P) = 20 kW = 20 × 103 W
Voltage (V) = 200 V
P = VI = I = P/V
I = (20 × 103)/200
I = 100A
For wave wound No. of Parallel Path = 2
Current in Each parallel Path for wave wound
I = 100/2 = 50 A
 

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