Find the net capacitance of the combination in which ten capacitors of 10 μF are connected in parallel

SSC JE Electrical 2019 with solution SET-1
Find the net capacitance of the combination in which ten capacitors of 10 μF are connected in parallel

1 μF

50 μF

100 μF

0.1 μF

1 μF

50 μF

100 μF

0.1 μF

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SSC JE Electrical 2019 with solution SET-1
In the full-wave rectifier circuit shown in the figure, the diodes that conduct when the positive half-wave of AC signal applied are

D3, D4

D1, D2

D4, D1

D2, D3

D3, D4

D1, D2

D4, D1

D2, D3

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SSC JE Electrical 2019 with solution SET-1
If the copper loss of a transformer at half full load is 400 W, then the copper loss corresponding to full load is

800 W

1600 W

400 W

1200 W

800 W

1600 W

400 W

1200 W

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SSC JE Electrical 2019 with solution SET-1
What is the reason behind using a centrifugal switch in a single-phase induction motor?

To out in the capacitor during running conditions

To cut-off the starting winding at an appropriate instant

To protect the motor from overloading

To improve the starting performance of the motor

To out in the capacitor during running conditions

To cut-off the starting winding at an appropriate instant

To protect the motor from overloading

To improve the starting performance of the motor

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SSC JE Electrical 2019 with solution SET-1
Z1 and Z2 are connected in series to form a load. A wattmeter’s current coil is connected in series with the load, whereas its pressure coil is connected across Z2. The wattmeter reads

Power consumed by Z1 and Z2

Power consumed by Z2

Zero always

Power consumed by Z1

Power consumed by Z1 and Z2

Power consumed by Z2

Zero always

Power consumed by Z1

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SSC JE Electrical 2019 with solution SET-1
The laws involved in the expression e = −dφ/dt are

Coulomb’s and Faraday’s laws

Faraday’s and Ampere’s laws

Faraday’s and Lenz’s laws

Coulomb’s and Ampere’s laws

Coulomb’s and Faraday’s laws

Faraday’s and Ampere’s laws

Faraday’s and Lenz’s laws

Coulomb’s and Ampere’s laws

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SSC JE Electrical 2019 with solution SET-1

SSC JE Electrical 2019 with solution SET-1 Find the net capacitance of the combination in which ten capacitors of 10 μF are connected in parallel

SSC JE Electrical 2019 with solution SET-1 In the full-wave rectifier circuit shown in the figure, the diodes that conduct when the positive half-wave of AC signal applied are

SSC JE Electrical 2019 with solution SET-1 If the copper loss of a transformer at half full load is 400 W, then the copper loss corresponding to full load is

SSC JE Electrical 2019 with solution SET-1 What is the reason behind using a centrifugal switch in a single-phase induction motor?

SSC JE Electrical 2019 with solution SET-1 Z1 and Z2 are connected in series to form a load. A wattmeter’s current coil is connected in series with the load, whereas its pressure coil is connected across Z2. The wattmeter reads

SSC JE Electrical 2019 with solution SET-1 The laws involved in the expression e = −dφ/dt are

SSC JE Electrical 2019 with solution SET-1
Find the net capacitance of the combination in which ten capacitors of 10 μF are connected in parallel

SSC JE Electrical 2019 with solution SET-1
In the full-wave rectifier circuit shown in the figure, the diodes that conduct when the positive half-wave of AC signal applied are

SSC JE Electrical 2019 with solution SET-1
If the copper loss of a transformer at half full load is 400 W, then the copper loss corresponding to full load is

SSC JE Electrical 2019 with solution SET-1
What is the reason behind using a centrifugal switch in a single-phase induction motor?

SSC JE Electrical 2019 with solution SET-1
Z1 and Z2 are connected in series to form a load. A wattmeter’s current coil is connected in series with the load, whereas its pressure coil is connected across Z2. The wattmeter reads

SSC JE Electrical 2019 with solution SET-1
The laws involved in the expression e = −dφ/dt are