![]() Limitation of this method is that it cannot be applied on unbalanced load. 0 Comments If the voltage and present to the motor are 400V and 8. Measurement of Three Phase Power by One Wattmeter Method 0 Comments If the voltage and present to the motor are 400V and 8.6A respectively, determine the energy factor of the program Power G 5000W, line voltage V T 400 V, line current, I L 8.A and energy, G 3 Sixth is v L I M cos Therefore power factor cos G 3 V T I M 5000 3 (400) (8.Two wattmeters are usually. The reading of wattmeter one can be written asīut, hence expression for total power will reduce to. When delta connected load, the diagram is shown in below Mathematically we can writeīut we have, hence putting the value of. Thus the total power of the circuit is sum of the reading of both the wattmeters. Similarly the reading of wattmeter two is the product of phase current and the voltage difference (V 2-V 3). We will see later that we need Type I for analysis lters and Type II for synthesis lters in a perfect reconstruction lter bank. When the load is star connected load, the diagram is shown in below-įor star connected load clearly the reading of wattmeter one is product of phase current and voltage difference (V 2-V 3). In the Type II, or reverse polyphase decomposition, the powers of zprogress in the opposite direction: H(z) NX1 l0 z(Nl1)R l(z N).
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