While measuring power of a three-phase balanced load by the two-wattmeter method, the readings are 100W and 250 W. The power factor of the load is Your answer
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The correct answer is - 0.8
As given in the question, the readings for the measured power by using the 2 two-wattmeter methods for a 3-phase balanced load is = 100 W and 250 W.
It is known that Power factor is denoted by cos Ф.
= cos {tan⁻¹ [√3 (ω₁ - ω₂) / (ω₁ + ω₂)]}
= cos {tan⁻¹ [√3 (100 - 250 / 100 + 250)]}
= cos {tan⁻¹ [√3 (-150) / (350)]}
By using the trigonometric table values, we can get the value of the above equation as -
= 0.8029
Therefore, the power factor of the load is - 0.8029
As given in the question, the readings for the measured power by using the 2 two-wattmeter methods for a 3-phase balanced load is = 100 W and 250 W.
It is known that Power factor is denoted by cos Ф.
= cos {tan⁻¹ [√3 (ω₁ - ω₂) / (ω₁ + ω₂)]}
= cos {tan⁻¹ [√3 (100 - 250 / 100 + 250)]}
= cos {tan⁻¹ [√3 (-150) / (350)]}
By using the trigonometric table values, we can get the value of the above equation as -
= 0.8029
Therefore, the power factor of the load is - 0.8029
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