APPARENT power.It is represented by cos ϕIn AC circuits, real power (in kW) = VI cos ϕReactive power (in kVAR) = VI sin ϕApparent power (in KVA) = VI\(\cos \phi = \frac{{Real\;power\;\left( {in\;kW} \right)}}{{Apparent\;power\;\left( {in\;kVA} \right)}}\)Calculation:Given:Real power = 500 WI = 3 Amp, V = 220 Volt\(\cos \phi = \frac{500}{220\times3}=0.7575\)

"> APPARENT power.It is represented by cos ϕIn AC circuits, real power (in kW) = VI cos ϕReactive power (in kVAR) = VI sin ϕApparent power (in KVA) = VI\(\cos \phi = \frac{{Real\;power\;\left( {in\;kW} \right)}}{{Apparent\;power\;\left( {in\;kVA} \right)}}\)Calculation:Given:Real power = 500 WI = 3 Amp, V = 220 Volt\(\cos \phi = \frac{500}{220\times3}=0.7575\)

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An appliance is marked 500 W. At 220 V, it consumes 3 ampers of current. The power factor is approximately:

Power Engineering Gas And Vapor Power Cycles in Power Engineering 8 months ago

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Concept:In AC circuits, power factor can be defined as the ratio of real power to the APPARENT power.It is represented by cos ϕIn AC circuits, real power (in kW) = VI cos ϕReactive power (in kVAR) = VI sin ϕApparent power (in KVA) = VI\(\cos \phi = \frac{{Real\;power\;\left( {in\;kW} \right)}}{{Apparent\;power\;\left( {in\;kVA} \right)}}\)Calculation:Given:Real power = 500 WI = 3 Amp, V = 220 Volt\(\cos \phi = \frac{500}{220\times3}=0.7575\)

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Posted on 29 Oct 2024, this text provides information on Power Engineering related to Gas And Vapor Power Cycles in Power Engineering. 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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