Questions: Question 66 Draw a phasor diagram for a three-phase star connected four-wire system to determine the neutral conductor current for the listed phase currents. - A phase to neutral, 12 amperes purely resistive. - B phase to neutral, 20 amperes lagging phase voltage by pf 0.8 . - C phase to neutral, 20 amperes at unity power factor.

Solution Steps

• A phase: The current is purely resistive, so it is in phase with the voltage. Therefore, the current is \( I_A = 12 + j0 \, \text{A} \).
• B phase: The current lags the voltage by a power factor of 0.8. The angle \(\theta\) can be found using \(\cos(\theta) = 0.8\), which gives \(\theta = \cos^{-1}(0.8) \approx 36.87^\circ\). Therefore, the current is \( I_B = 20 \angle -36.87^\circ \, \text{A} \).
• C phase: The current is at unity power factor, so it is in phase with the voltage. Therefore, the current is \( I_C = 20 + j0 \, \text{A} \).

Convert \( I_B = 20 \angle -36.87^\circ \) to rectangular form: \[ I_B = 20 \times (\cos(-36.87^\circ) + j\sin(-36.87^\circ)) = 20 \times (0.8 - j0.6) = 16 - j12 \, \text{A} \]

The neutral conductor current \( I_N \) is the vector sum of the phase currents: \[ I_N = I_A + I_B + I_C = (12 + j0) + (16 - j12) + (20 + j0) \] \[ I_N = (12 + 16 + 20) + j(0 - 12 + 0) = 48 - j12 \, \text{A} \]

Final Answer