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ECET 2026 EEE

Day 6 ECET 2026 Evening – Basic Electrical Engineering: Star/Delta & Energy Concepts

Concept Notes

1. Star (Y) & Delta (Δ) Connections

In 3-phase systems, loads can be connected either in Star or Delta.

Star (Y) Connection

  • Three windings’ similar ends joined at a neutral point.
  • Phase voltage:  V_{ph} = Line voltage / √3
  • Phase current:  I_{ph} = I_{line}

Formulas:
 V_{ph} = \frac{V_{line}}{\sqrt{3}}

 I_{line} = I_{ph}

Delta (Δ) Connection

  • Windings connected end-to-end to form a closed loop.
  • Line voltage = Phase voltage
  • Line current = √3 × Phase current

Formulas:
 V_{ph} = V_{line}

 I_{line} = \sqrt{3} I_{ph}

👉 Power in both connections:

 P = \sqrt{3} V_{line} I_{line} \cos \phi

Example:
If a balanced star system has  V_{line} = 400V, I_{line} = 10A, \cos \phi = 0.8 :
 P = \sqrt{3} \times 400 \times 10 \times 0.8 = 5542W .


2. Energy Concepts in Electrical Systems

Electric Energy

Work done by an electric circuit over time:
 W = V \cdot I \cdot t (Joules)

Electrical Power

  • Instantaneous Power:  P = V \cdot I
  • AC Power (with power factor):  P = V \cdot I \cos \phi

Energy Consumption in kWh

 E = P \cdot t (in Wh or kWh)

1 kWh = Energy consumed by 1 kW load in 1 hour = 1000 × 3600 =  3.6 \times 10^6 J

Example:
A 100W bulb runs for 10 hours → Energy =  100 \times 10 = 1000Wh = 1kWh .


⚙️ Formulas

  • Star:  V_{ph} = \frac{V_{line}}{\sqrt{3}}, ; I_{line} = I_{ph}
  • Delta:  V_{ph} = V_{line}, ; I_{line} = \sqrt{3} I_{ph}
  • 3-phase Power:  P = \sqrt{3} V_{line} I_{line} \cos \phi
  • Energy:  W = V \cdot I \cdot t
  • 1 kWh =  3.6 \times 10^6 J

🔟 10 MCQs

Q1. In a star connection, line voltage is 400V. Phase voltage is:
a) 400V
b) 230V
c) 115V
d) 690V

Q2. In a delta system, line current = 10A. Phase current is:
a) 10A
b) 5.77A
c) 17.32A
d) 20A

Q3. For a balanced 3-phase star load with  V_{line} = 400V, I_{line} = 5A, \cos \phi = 1 , find power.
a) 2000W
b) 3000W
c) 3464W
d) 4000W

Q4. In star connection, if line voltage = 415V, then phase voltage is approximately:
a) 415V
b) 240V
c) 720V
d) 120V

Q5. In delta connection, if  V_{line} = 400V , then  V_{ph} = ?
a) 400V
b) 230V
c) 690V
d) 115V

Q6. Energy consumed by a 2kW heater running for 3 hours is:
a) 2kWh
b) 3kWh
c) 5kWh
d) 6kWh

Q7. A 60W fan runs for 5 hours daily for 30 days. Energy used = ?
a) 9kWh
b) 12kWh
c) 15kWh
d) 18kWh

Q8. A load takes 10A at 230V, unity power factor for 2 hours. Energy consumed = ?
a) 2.3kWh
b) 4.6kWh
c) 1.15kWh
d) 5kWh

Q9. In a 3-phase delta, line current is related to phase current by:
a)  I_{line} = I_{ph}
b)  I_{line} = \sqrt{3} I_{ph}
c)  I_{ph} = \sqrt{3} I_{line}
d)  I_{line} = 2 I_{ph}

Q10. 1 unit of electricity (1 kWh) is equal to:
a)  1000 J
b)  3600 J
c)  1000 \times 3600 J
d)  3.6 \times 10^6 J


✅ Answer Key

Q.NoAnswer
1b
2b
3c
4b
5a
6d
7a
8a
9b
10d

🧠 Explanations

  • Q1:  V_{ph} = \frac{400}{\sqrt{3}} = 230V → (b).
  • Q2:  I_{ph} = \frac{I_{line}}{\sqrt{3}} = \frac{10}{1.732} = 5.77A → (b).
  • Q3:  P = \sqrt{3} V_{line} I_{line} = 1.732 \times 400 \times 5 = 3464W → (c).
  • Q4:  V_{ph} = \frac{415}{\sqrt{3}} ≈ 240V → (b).
  • Q5: In delta,  V_{ph} = V_{line} = 400V → (a).
  • Q6:  E = P \cdot t = 2 \times 3 = 6kWh → (d).
  • Q7:  E = 0.06 \times 5 \times 30 = 9kWh → (a).
  • Q8:  P = V \cdot I = 230 \times 10 = 2300W = 2.3kW . For 2h →  4.6kWh → (a).
  • Q9: By definition,  I_{line} = \sqrt{3} I_{ph} → (b).
  • Q10:  1kWh = 1000 \times 3600 J = 3.6 \times 10^6 J → (d).

🎯 Motivation / Why Practice Matters

Star/Delta and energy concepts are repeatedly tested in ECET exams because they link directly to power systems, machines, and utilisation topics. Mistakes often happen in line-to-phase conversions and unit conversions (kWh to Joules). If you master these basics now, later problems in transformers, AC machines, and power systems will feel easier.


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