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What will be the total active power consumed by a 3-phase, delta-connected system, which is supplied with a line voltage of 230 V, when the value of the phase current is 15 A and the current lags the voltage by 30°?
10.25 kW
8.963 kW
14.63 kW
12.26 kW
8.963 kW
The total active power in a balanced 3-phase system is calculated using the phase voltage and phase current. Since it is a delta-connected system, the line voltage is equal to the phase voltage, and the power is determined by P=3×Vph×Iph×cos(ϕ).
IS 10810-7:1984
The total active power in a balanced 3-phase system is calculated using the phase voltage and phase current. Since it is a delta-connected system, the line voltage is equal to the phase voltage, and the power is determined by P=3×Vph×Iph×cos(ϕ).
P=3×Vph×Iph×cos(ϕ) — Formula for total active power in a 3-phase delta system
Vph=VL — Relation between phase voltage and line voltage in delta connection
In a 3-phase delta system, the line voltage VL is equivalent to the phase voltage Vph. The total active power is the sum of the power consumed by each of the three phases, defined by the product of phase voltage, phase current, and the power factor (cosϕ).
In delta connection, line voltage equals phase voltage (VL=Vph)
Phase current is given as 15 A
Power factor is cos(30°)=0.866
Total Power P=3×230×15×cos(30°)=8963.5 W or 8.963 kW
Higher power handling capability
Balanced loading on phases
Requires three-phase supply
Complex wiring compared to single-phase systems
Industrial motor drives
Power distribution grids
Given VL=230 V, Iph=15 A, ϕ=30°. Calculation: P=3×230×15×cos(30°)=10350×0.866025=8963.36 W≈8.963 kW.
Option A, C, and D are incorrect due to misapplication of the phase power sum or incorrect power factor values.
B is correct — The total active power calculation for a 3-phase delta system results in 8.963 kW.
Always verify if the given current is line current (IL) or phase current (Iph) before applying the power formula, as IL=3×Iph in delta systems.
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