The power consumption of three phase circuit, which is connected with \$ \Delta\$ type, is \$3000W\$,now if we modified it with \$ Y\$ type,what is the power consumption of this circuit?

The answer is 1000W,because when we have the same impedance for each three wire,we can have \$P_Y=\frac{1}{3}P_{\Delta} \$

But now i am confused that in what condition,we can have the same power of \$ \Delta\$ type and \$ Y\$ type ,I mean,\$P_{\Delta}=P_{Y}\$,obviously if the impedance are the same between the \$ Y\$ type connection and \$ P_{\Delta}\$ type connection,the \$P_{\Delta }\neq P_{Y}\$

So in what condition,we can have \$P_{\Delta}=P_{Y}\$,just like the formula below:

\$P_{\Delta}=3\times \frac{V_L}{\sqrt{3}}\times I_Lcos\theta=\sqrt{3}V_LI_Lcos\theta\$

\$P_{Y}=3\times V_L\times \frac{I_L}{\sqrt{3}}cos\theta=\sqrt{3}V_LI_Lcos\theta\$

\$V_L\$ means line voltage,\$I_L\$ means line current

\$V_P\$ means phase voltage,\$I_P\$ means phase current

In the \$ Y\$ type,\$\sqrt{3}V_P=V_L\$,\$I_L=I_P\$

In the \$ \Delta\$ type,\$V_P=V_L\$,\$I_L=\sqrt{3}I_P\$

  • \$\begingroup\$ What does \$ V_L \$ mean in your equations? Hit the edit link. \$\endgroup\$ – Transistor May 12 '20 at 6:18
  • \$\begingroup\$ @Transistor i have modified the question \$\endgroup\$ – shineele May 12 '20 at 7:16
  • \$\begingroup\$ What is "i" tag? \$\endgroup\$ – Shadow May 12 '20 at 7:27
  • \$\begingroup\$ "i"??which part do you mention? \$\endgroup\$ – shineele May 12 '20 at 7:49
  • \$\begingroup\$ You tagged the question with "i". What does that mean? You can use the "current" tag. \$\endgroup\$ – Shadow May 12 '20 at 9:32

So in what condition,we can have \$P_Δ=P_Y\$

  • In a star connected load, the phase current is the line current.

  • In a delta connected load, the line current is \$\sqrt3\$ times greater than that for a star load using the same limb impedances.

So if you want a star load to match a delta load in terms of power (VA), each limb of the star has to have an impedance that is \$\sqrt3\$ times lower than each limb of the delta load. This will then make the phase current (also line current) \$\sqrt3\$ times bigger and therefore matches the delta load line current.


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