W1) = VL IL cos (30 + ϕ)The reading of second wattmeter (W2) = VL IL cos (30 - ϕ)Total power in the circuit (P) = W1 + W2Total reactive power in the circuit Q = √3 (W_1-W_2 )Power FACTOR = cos ϕϕ = tan^(-1) ((√3 (W_1-W_2 ))/(W_1 + W_2 ))Calculation:GIVEN that, W1 = 1.7 kWW2 = 1.1 kWW1 – W2 = 0.6 kWW1 + W2 = 2.8 kWtanϕ = (√3 × 0.6)/2.8 = 0.371⇒ ϕ = 20.36° ⇒ Power factor = cos ϕ = 0.937 lagging

"> W1) = VL IL cos (30 + ϕ)The reading of second wattmeter (W2) = VL IL cos (30 - ϕ)Total power in the circuit (P) = W1 + W2Total reactive power in the circuit Q = √3 (W_1-W_2 )Power FACTOR = cos ϕϕ = tan^(-1) ((√3 (W_1-W_2 ))/(W_1 + W_2 ))Calculation:GIVEN that, W1 = 1.7 kWW2 = 1.1 kWW1 – W2 = 0.6 kWW1 + W2 = 2.8 kWtanϕ = (√3 × 0.6)/2.8 = 0.371⇒ ϕ = 20.36° ⇒ Power factor = cos ϕ = 0.937 lagging

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The two-wattmeter method is used to measure the input power of a three-phase induction motor. If the two wattmeter readings are 1,700 W and 1,100 W, determine the power factor of the motor.

General Knowledge General Awareness in General Knowledge 8 months ago

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Concept:In a two-wattmeter method, for lagging loadThe reading of first wattmeter (W1) = VL IL cos (30 + ϕ)The reading of second wattmeter (W2) = VL IL cos (30 - ϕ)Total power in the circuit (P) = W1 + W2Total reactive power in the circuit Q = √3 (W_1-W_2 )Power FACTOR = cos ϕϕ = tan^(-1) ((√3 (W_1-W_2 ))/(W_1 + W_2 ))Calculation:GIVEN that, W1 = 1.7 kWW2 = 1.1 kWW1 – W2 = 0.6 kWW1 + W2 = 2.8 kWtanϕ = (√3 × 0.6)/2.8 = 0.371⇒ ϕ = 20.36° ⇒ Power factor = cos ϕ = 0.937 lagging

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Posted on 30 Nov 2024, this text provides information on General Knowledge related to General Awareness in General Knowledge. Please note that while accuracy is prioritized, the data presented might not be entirely correct or up-to-date. This information is offered for general knowledge and informational purposes only, and should not be considered as a substitute for professional advice.

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