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In this circuit with frequency of 50hz and line to line voltage of 400 V why is the current \$ I_A = S_t/3V_P\$ I don't understand the logic behind since there is one Delta connection, I understand why that formula works for a Star connection but don't for the total of this circuit. The Delta load also has 12,5kw.

Circuit

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  • \$\begingroup\$ Is \$S\$ your apparent power? I'm trying to understand your notation. \$\endgroup\$ – KingDuken Jun 27 '18 at 20:22
  • \$\begingroup\$ @KingDuken it is the total apparent power of the three loads, and Vp = 400/sqrt(3) =230 \$\endgroup\$ – Pedro Jun 27 '18 at 20:23
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If your network is strong which mean that your short-circuitpower is higher than the power demand, the voltage at the input will not drop under load, than the power of the loads add up. Therefore the total power at the input is

$$ S_{total} = S_{Z}+S_{c}+S_{equil} $$

The basic equation of apparent power for a generic device with no harmonics:

$$ S = 3 V_{phase} I $$

Therefore, in your case

$$ I = \frac{S_{total}}{3V_{phase}} $$

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