A balanced delta-connected load of (4+j3) Ω per phase is connected to a balanced three-phase 440V supply. The phase current is 10A. Find the total active power.
Answer & step-by-step solution
A is the intended option if the stated phase current of 10 A and the 440 V line voltage are used: P = √3 × 440 × 17.32 × 0.8 ≈ 10.56 kW. However, the supplied voltage, impedance and current are not mutually consistent.
Balanced Delta Load: Active Power and the Data Inconsistency
A is the intended option if the stated phase current of 10 A and the 440 V line voltage are used: P = √3 × 440 × 17.32 × 0.8 ≈ 10.56 kW. However, the supplied voltage, impedance and current are not mutually consistent.
Step 1: Calculate the power factor
The phase impedance is Z = 4 + j3 Ω. Its magnitude is |Z| = √(4² + 3²) = 5 Ω. Therefore cosφ = R/|Z| = 4/5 = 0.8, lagging because the reactance is inductive. The value 3/5 = 0.6 is sinφ, not the power factor.
Step 2: Use the current stated in the question
For a balanced delta connection, V_ph = V_L and I_L = √3 I_ph. Taking the given I_ph = 10 A gives I_L = 10√3 ≈ 17.32 A.
Total active power is P = √3 V_L I_L cosφ = √3 × 440 × 17.32 × 0.8 ≈ 10,560 W. Equivalently, P = 3V_phI_phcosφ = 3 × 440 × 10 × 0.8 = 10,560 W. This is the route represented by option A.
Data check: why this is only an intended answer
The same delta connection also requires I_ph = V_ph/|Z| = 440/5 = 88 A, not 10 A. If 440 V and 4 + j3 Ω are retained as the physical data, the active power is P = 3 × 440 × 88 × 0.8 = 92,928 W = 92.928 kW, which none of the options represents.
Conversely, a 10 A phase current through the stated 5 Ω impedance requires 50 V per phase and would give P = 3I_ph²R = 1,200 W. Thus all three given values cannot describe one load operating condition. The original paper or an official correction is needed to resolve the conflicting datum; the existing answer A is retained only as the intended option.
Useful Formula
Δ: V_ph = V_L; I_L = √3I_ph; cosφ = R/|Z|; P = 3V_phI_phcosφ = √3V_LI_Lcosφ. Given-current route: 10.56 kW. Voltage–impedance route: 92.928 kW.
Why the Other Options Are Not the Answer
- B — Uses 3 with line current 17.32 A and also substitutes sinφ = 0.6 for cosφ.
- C — Uses the correct line-quantity coefficient √3 but the wrong power factor 0.6.
- D — Uses the correct power factor 0.8 but incorrectly combines coefficient 3 with line current 17.32 A.
Remember for the Exam
Before calculating three-phase power, label every voltage and current as line or phase and check V_ph = I_ph|Z|.
Common mistake: Do not hide the inconsistent data by using |Z| only to find power factor. A correct-looking option does not make conflicting input values consistent.
Quick Trick
Use 3 with phase voltage and phase current; use √3 with line voltage and line current.
Common Mistake
Do not hide the inconsistent data by using |Z| only to find power factor. A correct-looking option does not make conflicting input values consistent.
प्रति फेज़ (4+j3) Ω का एक संतुलित डेल्टा-संयोजित लोड संतुलित तीन-फेज़ 440V आपूर्ति से जुड़ा है। फेज़ धारा 10A है। कुल सक्रिय शक्ति ज्ञात कीजिए।
उत्तर और चरणबद्ध हल
A — डेल्टा में I_L = √3 × 10 = 17.32 A और cosφ = 4/5 = 0.8। P = √3 × 440 × 17.32 × 0.8 ≈ 10.56 kW। ध्यान दें: दिए गए वोल्टेज, प्रतिबाधा और धारा आपस में पूरी तरह सुसंगत नहीं हैं।
विस्तृत हल (मूल भाषा में)
Balanced Delta Load: Active Power and the Data Inconsistency
A is the intended option if the stated phase current of 10 A and the 440 V line voltage are used: P = √3 × 440 × 17.32 × 0.8 ≈ 10.56 kW. However, the supplied voltage, impedance and current are not mutually consistent.
Step 1: Calculate the power factor
The phase impedance is Z = 4 + j3 Ω. Its magnitude is |Z| = √(4² + 3²) = 5 Ω. Therefore cosφ = R/|Z| = 4/5 = 0.8, lagging because the reactance is inductive. The value 3/5 = 0.6 is sinφ, not the power factor.
Step 2: Use the current stated in the question
For a balanced delta connection, V_ph = V_L and I_L = √3 I_ph. Taking the given I_ph = 10 A gives I_L = 10√3 ≈ 17.32 A.
Total active power is P = √3 V_L I_L cosφ = √3 × 440 × 17.32 × 0.8 ≈ 10,560 W. Equivalently, P = 3V_phI_phcosφ = 3 × 440 × 10 × 0.8 = 10,560 W. This is the route represented by option A.
Data check: why this is only an intended answer
The same delta connection also requires I_ph = V_ph/|Z| = 440/5 = 88 A, not 10 A. If 440 V and 4 + j3 Ω are retained as the physical data, the active power is P = 3 × 440 × 88 × 0.8 = 92,928 W = 92.928 kW, which none of the options represents.
Conversely, a 10 A phase current through the stated 5 Ω impedance requires 50 V per phase and would give P = 3I_ph²R = 1,200 W. Thus all three given values cannot describe one load operating condition. The original paper or an official correction is needed to resolve the conflicting datum; the existing answer A is retained only as the intended option.
Useful Formula
Δ: V_ph = V_L; I_L = √3I_ph; cosφ = R/|Z|; P = 3V_phI_phcosφ = √3V_LI_Lcosφ. Given-current route: 10.56 kW. Voltage–impedance route: 92.928 kW.
Why the Other Options Are Not the Answer
- B — Uses 3 with line current 17.32 A and also substitutes sinφ = 0.6 for cosφ.
- C — Uses the correct line-quantity coefficient √3 but the wrong power factor 0.6.
- D — Uses the correct power factor 0.8 but incorrectly combines coefficient 3 with line current 17.32 A.
Remember for the Exam
Before calculating three-phase power, label every voltage and current as line or phase and check V_ph = I_ph|Z|.
Common mistake: Do not hide the inconsistent data by using |Z| only to find power factor. A correct-looking option does not make conflicting input values consistent.
Quick Trick
Use 3 with phase voltage and phase current; use √3 with line voltage and line current.
Common Mistake
Do not hide the inconsistent data by using |Z| only to find power factor. A correct-looking option does not make conflicting input values consistent.
પ્રતિ ફેઝ (4+j3) Ω નો એક સંતુલિત ડેલ્ટા-જોડાણવાળો લોડ સંતુલિત ત્રણ-ફેઝ 440V પુરવઠા સાથે જોડાયેલ છે. ફેઝ પ્રવાહ 10A છે. કુલ સક્રિય પાવર શોધો.
જવાબ અને પગલાવાર ઉકેલ
A — ડેલ્ટામાં I_L = √3 × 10 = 17.32 A અને cosφ = 4/5 = 0.8. P = √3 × 440 × 17.32 × 0.8 ≈ 10.56 kW. નોંધ: આપેલ વોલ્ટેજ, અવબાધ અને પ્રવાહ પરસ્પર સંપૂર્ણ સુસંગત નથી.
વિગતવાર ઉકેલ (મૂળ ભાષામાં)
Balanced Delta Load: Active Power and the Data Inconsistency
A is the intended option if the stated phase current of 10 A and the 440 V line voltage are used: P = √3 × 440 × 17.32 × 0.8 ≈ 10.56 kW. However, the supplied voltage, impedance and current are not mutually consistent.
Step 1: Calculate the power factor
The phase impedance is Z = 4 + j3 Ω. Its magnitude is |Z| = √(4² + 3²) = 5 Ω. Therefore cosφ = R/|Z| = 4/5 = 0.8, lagging because the reactance is inductive. The value 3/5 = 0.6 is sinφ, not the power factor.
Step 2: Use the current stated in the question
For a balanced delta connection, V_ph = V_L and I_L = √3 I_ph. Taking the given I_ph = 10 A gives I_L = 10√3 ≈ 17.32 A.
Total active power is P = √3 V_L I_L cosφ = √3 × 440 × 17.32 × 0.8 ≈ 10,560 W. Equivalently, P = 3V_phI_phcosφ = 3 × 440 × 10 × 0.8 = 10,560 W. This is the route represented by option A.
Data check: why this is only an intended answer
The same delta connection also requires I_ph = V_ph/|Z| = 440/5 = 88 A, not 10 A. If 440 V and 4 + j3 Ω are retained as the physical data, the active power is P = 3 × 440 × 88 × 0.8 = 92,928 W = 92.928 kW, which none of the options represents.
Conversely, a 10 A phase current through the stated 5 Ω impedance requires 50 V per phase and would give P = 3I_ph²R = 1,200 W. Thus all three given values cannot describe one load operating condition. The original paper or an official correction is needed to resolve the conflicting datum; the existing answer A is retained only as the intended option.
Useful Formula
Δ: V_ph = V_L; I_L = √3I_ph; cosφ = R/|Z|; P = 3V_phI_phcosφ = √3V_LI_Lcosφ. Given-current route: 10.56 kW. Voltage–impedance route: 92.928 kW.
Why the Other Options Are Not the Answer
- B — Uses 3 with line current 17.32 A and also substitutes sinφ = 0.6 for cosφ.
- C — Uses the correct line-quantity coefficient √3 but the wrong power factor 0.6.
- D — Uses the correct power factor 0.8 but incorrectly combines coefficient 3 with line current 17.32 A.
Remember for the Exam
Before calculating three-phase power, label every voltage and current as line or phase and check V_ph = I_ph|Z|.
Common mistake: Do not hide the inconsistent data by using |Z| only to find power factor. A correct-looking option does not make conflicting input values consistent.
Quick Trick
Use 3 with phase voltage and phase current; use √3 with line voltage and line current.
Common Mistake
Do not hide the inconsistent data by using |Z| only to find power factor. A correct-looking option does not make conflicting input values consistent.
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