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SSC JE Electrical 2019 with solution SET-1

SSC JE Electrical 2019 with solution SET-1
 The cheap and temporary system of the internal wiring is?

Casing-capping
Conduit wiring
CTS or TRS wiring
Cleat wiring

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SSC JE Electrical 2019 with solution SET-1
 Two coupled coils with L1 = L2 = 0.5H have a coupling coefficient of K = 0.75. The turn ratio N1 ⁄ N2 = ?

2
0.5
4
1

ANSWER EXPLANATION DOWNLOAD EXAMIANS APP

The self-inductance is given as
L = μN2A/I
L ∝ N2
where
N is the number of turns of the solenoid
A is the area of each turn of the coil
l is the length of the solenoid
and μ is the permeability constant
L1/L2 = N21/N22
0.5/0.5 = N21/N22
N1/N2 = 1
 

SSC JE Electrical 2019 with solution SET-1
 A coil is wound with 50 turns and a current 8A produces a flux of 200μWb. Calculate the inductance of the coil

2.5 mH
1.25 mH
0.25 mH
0.125 mH

ANSWER EXPLANATION DOWNLOAD EXAMIANS APP

The inductance of the coil is given by the relation
L = Nφ/I
Where
N = number of turns = 50
φ = flux = 200μWb
I = current = 8 A
L = 50 × 200 ×  10−6 ⁄ 8
L = 1.25 mH

SSC JE Electrical 2019 with solution SET-1
 The ratio of average energy demand to maximum demand during a specific period in

Form factor
Utilization factor
Load factor
Power factor

ANSWER EXPLANATION DOWNLOAD EXAMIANS APP

The ratio of average energy demand to maximum demand during a specific period is called the load factor.
Load Factor = Average Load/Maximum Demand
 

SSC JE Electrical 2019 with solution SET-1
Please check out below figure.

1 Ω
10 Ω
2 Ω
5 Ω

ANSWER EXPLANATION DOWNLOAD EXAMIANS APP

From the figure it can be concluded that the voltmeter reads 5 volts as shown in the figure below. Based on the voltmeter and ammeter readings in the measuring network, determine the value of the resistor R Here Current I = 1/2 A = 0.5 A Voltage V = 5 V R = V/I = 0.5/5 R = 10Ω


SSC JE Electrical 2019 with solution SET-1
 The voltmeter is shown in the circuit reads:

1.2 V
12 V
24 V
2.4 V

ANSWER EXPLANATION DOWNLOAD EXAMIANS APP

Total resistance in the given circuit
R = (250 + 250)MΩ = 500 MΩ
Current I = V/R = 24/(500 × 103)
Now the Voltage in the voltmeter
= \dfrac{{24}}{{500 \times {{10}^3}}} \times 250 \times {10^3}
V = 12 V
 

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