The power supplied by each phase of the power system is approximately 41.67 W.
In a balanced delta-connected resistive load, each resistor in the delta is equal, and the load is symmetrically distributed. When connected to a three-phase power system, the power supplied to each phase can be determined as follows:
Let P_total be the total power dissipated by the load, which is 375 W.
In a balanced three-phase system, the total power (P_total) is given by:
P_total = √3 * V_line * I_line * cos(θ)
where:
√3 is the square root of 3 (approximately 1.732),
V_line is the line voltage,
I_line is the line current, and
cos(θ) is the power factor.
In a delta-connected load, the line voltage (V_line) is the same as the phase voltage (V_phase).
So, for each phase:
P_phase = V_phase * I_line * cos(θ)
To find the power supplied by each phase, we need to determine the line current (I_line) and power factor (cos(θ)).
Since the load is resistive, the power factor (cos(θ)) is 1 (unity) for purely resistive loads.
Therefore, for each phase:
P_phase = V_phase * I_line * 1
P_phase = V_phase * I_line
Now, we need to determine the line current (I_line).
The total power (P_total) is given as 375 W. In a balanced three-phase system, the total power is equally distributed among the three phases. Therefore, the power supplied by each phase (P_phase) is one-third of the total power:
P_phase = P_total / 3
P_phase = 375 W / 3
P_phase = 125 W
Now, we have the power supplied by each phase (P_phase), and we know that P_phase = V_phase * I_line. Since the load is resistive, the power factor (cos(θ)) is 1, and we can assume the line voltage (V_phase) is constant for each phase.
Therefore, for each phase:
V_phase * I_line = 125 W
Since V_phase is the same for each phase, we can say:
V_phase * (I_line_phase1 + I_line_phase2 + I_line_phase3) = 125 W
For a balanced load, the line currents are equal in magnitude and 120 degrees apart in phase.
Let's assume the line current as I_line. Then:
I_line_phase1 + I_line_phase2 + I_line_phase3 = 3 * I_line
So, we can rewrite the equation as:
V_phase * 3 * I_line = 125 W
Now, we can solve for I_line:
I_line = 125 W / (3 * V_phase)
Finally, to find the power supplied by each phase (P_phase), we can substitute the value of I_line:
P_phase = V_phase * I_line
P_phase = V_phase * (125 W / (3 * V_phase))
P_phase = 125 W / 3
Therefore, the power supplied by each phase of the power system is approximately 41.67 W.