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Explain two wattmeter method for measurement of 3 - phase power.
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  • In Balanced 3-phase circuit, power can be measured by using 2-watt meters.
  • Let's consider, 2-wattmeter method to measure the power in a star connected balanced 3-phase load as shown in figure below-

     

  • As we know that for a stare connected balanced 3-phase load-
  1. Magnitude of Line Voltage is equal to$\sqrt3$ times of magnitude of phase voltage. Means,$|V_L| =\sqrt3|V_{R}|=\sqrt3|V_{Y}|=\sqrt3|V_{B}|\dots(1) $
  2. Line current s equal to phase current. Means,$|I_L| = |I_{ph}| = |I_R| = |I_Y|=|I_B|\dots(2)$

From above connection diagram, 

(a) Current through wattmeter W1, $I_1 =I_R$

(b) Voltage across wattmeter W1, $V_1 = V_{RB} $

(c) Current through wattmeter W2, $I_2 =I_Y$

(d) Voltage across wattmeter W2, $V_2 = V_{YB} $

 

  • As we know that the power measured by dynamometer type wattmeter in given by,

$P= VI\cos\phi .....(3)$

Where, V is the voltage across the wattmeter

I is the current through wattmeter

$\phi$is the angle between V & I

  • Hence from above circuit, Power measured by wattmeterW1,

​$W_1= |V_1||I_1|\cos (angle\ between \ V_{1}\ and\ I_1) =| V_{RB}||I_R|\cos(angle\ between \ V_{RB}\ and\ I_R)....(4)$

  • Similarly, Power measured by wattmeterW2,$W_2=|V_2||I_2|\cos (angle\ between \ V_{2}\ and\ I_2) =|V_{YB}||I_Y|\cos(angle\ between \ V_{YB}\ and\ I_Y)....(5)$
  • Lets, consider that load is working lagging factor of$\cos\phi$. Then the phasor diagram for above circuit is shown in figure below-

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