---
title: "3-phase 500 kVA transformer primary current"
subject: "Electrical Machines"
chapter: 2
key: "semester_4/electrical_machines:tf-500kva-11kv-400v-primary-current"
source: https://www.bctengineer.com/q/semester_4/electrical_machines/tf-500kva-11kv-400v-primary-current
site: BCT Engineering
---

# 3-phase 500 kVA transformer primary current

Electrical Machines · Chapter 2: Transformer

- Asked in: 2066 Magh (old course), 8 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/electrical_machines/tf-500kva-11kv-400v-primary-current
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines/chapter-2

## Question

A 500 kVA, 50 Hz, 11 kV/400 V, 3-phase transformer has delta/star connection. Calculate the current drawn by the transformer from primary side when it delivers full load at rated terminal voltage at 0.8 lagging p.f. Assume ideal transformer operation.

## Short answer

For ideal transformer, primary line current I_L1 = S/(√3 V_L) = 500×10³/(√3×11000) = 26.24 A. Phase current in delta winding = 26.24/√3 = 15.15 A. Current drawn is 26.24 A at 0.8 lagging pf.

## Answer

For an ideal transformer, losses and voltage drops are zero, so input VA equals output VA. The primary is delta connected, so the line current is $\sqrt{3}$ times the phase current.

Output at full load: $S = 500$ kVA, power factor 0.8 lagging.

#### Primary line current

$$
S = \sqrt{3}\,V_L I_L \quad\Rightarrow\quad I_{L1} = \frac{500\times 10^3}{\sqrt{3}\times 11000} = 26.24\ \text{A}
$$

#### Primary phase current (delta)

$$
I_{ph1} = \frac{I_{L1}}{\sqrt{3}} = \frac{26.24}{\sqrt{3}} = 15.15\ \text{A}
$$

(Check: $I_{ph1} = \dfrac{500000}{3\times 11000} = 15.15$ A.)

#### Power drawn

$$
P = S\cos\phi = 500 \times 0.8 = 400\ \text{kW}, \qquad Q = 500\times 0.6 = 300\ \text{kVAR}
$$

For an ideal transformer the input power factor equals the load power factor (0.8 lagging), so the current lags the voltage by $\cos^{-1}0.8 = 36.87^\circ$.

**Answer:** the transformer draws a line current of $26.24\ \text{A}$ from the 11 kV supply (phase current $15.15\ \text{A}$ in each delta winding) at 0.8 lagging power factor.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. For an ideal transformer, the input VA equals
   A. Output VA, since losses and drops are zero
   B. Output VA plus core loss
   C. Half the output VA
   D. Output kW
   Answer: A. Ideal means no losses and no voltage drop, so input VA = output VA = 500 kVA. (Key point: Ideal transformer assumption)

2. The primary line current at full load is
   A. 500000/11000 = 45.45 A
   B. 500000/(sqrt(3) x 11000) = 26.24 A
   C. 500000/(3 x 11000) = 15.15 A
   D. 500000 x 0.8/(sqrt(3) x 11000) = 21 A
   Answer: B. S = sqrt(3) VL IL gives IL = 26.24 A; power factor does not change the VA-based current. (Key point: Primary line current)

3. In the delta-connected primary, the winding (phase) current is
   A. 26.24 x sqrt(3) = 45.45 A
   B. 26.24 A
   C. 26.24/sqrt(3) = 15.15 A
   D. 26.24/3 = 8.75 A
   Answer: C. In delta, IL = sqrt(3) Iph, so Iph = 26.24/sqrt(3) = 15.15 A. (Key point: Delta phase current)

4. For an ideal transformer with a 0.8 lagging load, the input power factor is
   A. 1.0
   B. 0.8 lagging, the same as the load
   C. 0.6 lagging
   D. 0.8 leading
   Answer: B. With no magnetising current or impedance drops, the primary current has the same phase angle (36.87° lagging) as the load. (Key point: Input power factor)

## Related in Electrical Machines

- [25 kVA transformer: load voltage and primary current](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines%3Atf-25kva-11kv-load-voltage) — Electrical Machines, chapter 2
- [Primary current and efficiency of 11 kV/380 V Δ/Y transformer](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Adelta-star-11kv-380v-40kw-efficiency) — Electrical Machines I, chapter 2, asked 3×
- [500 kVA transformer efficiency, full and half load](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines%3Atf-500kva-6600-400-efficiency) — Electrical Machines, chapter 2
- [500 kVA 3-phase transformer efficiency](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines%3Atf-500kva-33-11kv-efficiency) — Electrical Machines, chapter 2
- [Primary current and efficiency of 100 kVA Δ/Y transformer](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Adelta-star-100kva-11kv-100a-efficiency) — Electrical Machines I, chapter 2
- [Full-load efficiency of 500 kVA transformer](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3A500kva-90pc-half-load-full-load) — Electrical Machines I, chapter 2

## Related in other subjects

- [500 kVA delta/star transformer efficiency](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machine%3Athree-phase-500kva-33-11-efficiency) — Electrical Machine, chapter 2
- [Design: 500 kVA 66/11 kV power transformer](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machine_design%3Adesign-500kva-66-11kv) — Electric Machine Design, chapter 3, asked 6×
- [500 kVA transformer efficiency full/half load](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machine%3Atransformer-500kva-efficiency-full-half) — Electrical Machine, chapter 2
- [Delta/star 120 kVA transformer values](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machine%3Adelta-star-120kva-line-phase-values) — Electrical Machine, chapter 2