In a given R - L circuit, I = 10 A and power loss in circuit is 1 kW, then
A
Inductance of the circuit is √310πH
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B
Resistance of the circuit is 10√3Ω
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C
Power factor is 0.5
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D
Voltage is 60o ahead of current
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Solution
The correct options are A Inductance of the circuit is √310πH C Power factor is 0.5 D Voltage is 60o ahead of current P=i2R⇒1000=102×R⇒R=10Ω V=IZ⇒200=10√R2+X2L⇒XL=10√3 XL=ωL⇒L=XL2π50=√310π ϕ=60o∴Power factor =cosϕ=12