apparent-powercurrent measurementpower-factor-correctionthree phase
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.
You are using \$I_{LINE}\$ in both.
There is no dot, so the current that flows in phase must flow in transmission line. So in a star or wye, \$I_{LINE}\ =\ I_{PHASE}\$ (and \$V_{LINE}\ =\ √ (3)\ V_{PHASE}\$). You show this in your answer.
In a delta, \$I_{LINE}\ =\ √ (3)\ I_{PHASE}\$ (and \$V_{LINE}\ =\ V_{PHASE}\$). A component of two phase currents make up the line current. There is a dot.
$$P_T = √ (3)\ V_{LINE}\ I_{LINE}\ cos\ θ $$ $$P_T = 3\ V_{PHASE}\ I_{PHASE}\ cos\ θ $$
So in your first answer, \$V_{L-L}\$ which is line voltage \$V_{LINE}\ =\ 415V\$, which means a \$V_{PHASE}\ =\ 415V / √ (3)\ =\ 239.6V\$. $$ I_{PHASE}\ =\ V_{PHASE} / Z\ = 239.6V / 10Ω = 24.0A$$ $$ P\ = I_{PHASE}^2\ R = (24.0A)^2\ x\ 8Ω\ =\ 4.59kW$$ $$ P_T\ = 3\ P = 3\ x\ 4.59kW\ =\ 13.8kW$$
Best Answer
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
$$ I = \frac{S_{total}}{3V_{phase}} $$