SSC JE Electrical 2019 with solution SET-1

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

1 Ω

2 Ω

10 Ω

5 Ω

1 Ω

2 Ω

10 Ω

5 Ω

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SSC JE Electrical 2019 with solution SET-1
With the current direction marked in the circuit shown, the net voltage applied is

V1

−(V2 − V1)

(V2 − V1)

V2

V1

−(V2 − V1)

(V2 − V1)

V2

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SSC JE Electrical 2019 with solution SET-1
4F2D is a/an number

Octal

Binary

Hexadecimal

Decimal

Octal

Binary

Hexadecimal

Decimal

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SSC JE Electrical 2019 with solution SET-1
In a single-phase capacitor start-and-run motor, the minimum number of capacitors to be used in it is:

2

1

4

3

2

1

4

3

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SSC JE Electrical 2019 with solution SET-1
The Thevenin’s resistance as seen through the terminals A and B is:

6 Ω

4 Ω

5 Ω

8 Ω

6 Ω

4 Ω

5 Ω

8 Ω

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

2.4 V

1.2 V

12 V

24 V

2.4 V

1.2 V

12 V

24 V

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