If an alternator is operating at unity power factor, then its terminal voltage is

SSC JE Electrical 2019 with solution SET-1
If an alternator is operating at unity power factor, then its terminal voltage is

Less than the induced EMF, with positive voltage regulation

Equal to the induced EMF, with zero voltage regulation

Less than the induced EMF, with negative regulation

Greater than the induced EMF, with negative voltage regulation

Less than the induced EMF, with positive voltage regulation

Equal to the induced EMF, with zero voltage regulation

Less than the induced EMF, with negative regulation

Greater than the induced EMF, with negative voltage regulation

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SSC JE Electrical 2019 with solution SET-1
The Norton’s current in the circuit shown below is

4A

2A

12A

120A

4A

2A

12A

120A

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SSC JE Electrical 2019 with solution SET-1
In the two wattmeter method, the readings of the two wattmeters are 500W, 500W respectively. The load power factor in a balanced 3-phase 3-wire circuit is:

1

0.9

0.5

0.8

1

0.9

0.5

0.8

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SSC JE Electrical 2019 with solution SET-1
Extension of moving iron ammeter range can be done by using

Shunt

Inductor

Multiplier

Capacitor

Shunt

Inductor

Multiplier

Capacitor

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SSC JE Electrical 2019 with solution SET-1
Which of the following losses is together called iron losses?

Hysteresis loss and copper loss

Hysteresis loss and frictional loss

Eddy current loss and frictional loss

Eddy current loss and hysteresis loss

Hysteresis loss and copper loss

Hysteresis loss and frictional loss

Eddy current loss and frictional loss

Eddy current loss and hysteresis loss

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SSC JE Electrical 2019 with solution SET-1
Reflected Light/Incident Light =

Maintenance factor

Reflection factor

Waste light factor

Absorption factor

Maintenance factor

Reflection factor

Waste light factor

Absorption factor

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SSC JE Electrical 2019 with solution SET-1

SSC JE Electrical 2019 with solution SET-1 If an alternator is operating at unity power factor, then its terminal voltage is

SSC JE Electrical 2019 with solution SET-1 The Norton’s current in the circuit shown below is

SSC JE Electrical 2019 with solution SET-1 In the two wattmeter method, the readings of the two wattmeters are 500W, 500W respectively. The load power factor in a balanced 3-phase 3-wire circuit is:

SSC JE Electrical 2019 with solution SET-1 Extension of moving iron ammeter range can be done by using

SSC JE Electrical 2019 with solution SET-1 Which of the following losses is together called iron losses?

SSC JE Electrical 2019 with solution SET-1 Reflected Light/Incident Light =

SSC JE Electrical 2019 with solution SET-1
If an alternator is operating at unity power factor, then its terminal voltage is

SSC JE Electrical 2019 with solution SET-1
The Norton’s current in the circuit shown below is

SSC JE Electrical 2019 with solution SET-1
In the two wattmeter method, the readings of the two wattmeters are 500W, 500W respectively. The load power factor in a balanced 3-phase 3-wire circuit is:

SSC JE Electrical 2019 with solution SET-1
Extension of moving iron ammeter range can be done by using

SSC JE Electrical 2019 with solution SET-1
Which of the following losses is together called iron losses?

SSC JE Electrical 2019 with solution SET-1
Reflected Light/Incident Light =