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

Power factor
Load factor
Utilization factor
Form factor

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

120A
4A
2A
12A

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SSC JE Electrical 2019 with solution SET-1
 Let T be the net torque developed by the rotor runs at ω rad/s. What is the mechanical power developed?

Pmech = ωT
Pmech = (ωT)/60
Pmech = (2πωT)
Pmech = (2πωT)/60

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SSC JE Electrical 2019 with solution SET-1
 If dΦ is the luminous flux incident normally on an area dA, then Illuminance is given as

dΦ − dA
dΦ ⁄ dA
dA ⁄ dΦ
Φ × A

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