Final answer:
To find the total power drawn by the load, calculate the power dissipated by each impedance and sum them up. The total power drawn by the load is 9.41 kW.
Step-by-step explanation:
To find the total power drawn by the load, we need to calculate the power dissipated by each impedance and then sum them up. The power dissipated by an impedance can be found using the formula: P = I^2 * R, where I is the current and R is the resistance of the impedance. In this case, the line impedance is (1+j2), so the total impedance seen by each impedance is (6+j9) + (1+j2) = (7+j11).
We can calculate the current flowing through the load using Ohm's Law: I = V/Z, where V is the supply voltage and Z is the total impedance seen by each impedance. In this case, the supply voltage is 400V, so the current is I = 400V / (7+j11) = 35.5∠58.7° A.
Finally, we can calculate the total power drawn by the load using the formula: P_total = 3 * P_individual, where P_individual is the power dissipated by each impedance. Substituting the values, we get P_total = 3 * (35.5∠58.7°)^2 * 7 = 9.41 kW.