---
title: "Delta load 20∠60° power and wattmeter readings"
subject: "Basic Electrical Engineering"
chapter: 8
key: "semester_1/basic_electrical_engineering:delta-load-20-60-440v-two-wattmeter"
source: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/delta-load-20-60-440v-two-wattmeter
site: BCT Engineering
---

# Delta load 20∠60° power and wattmeter readings

Basic Electrical Engineering · Chapter 8: Three-Phase Circuit Analysis

- Asked in: 2071 Chaitra, 6 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/delta-load-20-60-440v-two-wattmeter
- Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-8

## Question

Each phase of a 3-phase, delta-connected load consists of an impedance Z = 20∠60° ohm. The line voltage is 440 V at 50 Hz. Compute the power consumed by each phase impedance and the total power. What will be the readings of the two wattmeters connected?

## Short answer

Each phase impedance 20∠60° Ω draws 4840 W; total power 14.52 kW. Wattmeter readings: W1 = 14.52 kW, W2 = 0 W (cos φ = 0.5).

## Answer

Each phase impedance Z = 20∠60° Ω, delta connected, V_L = 440 V, 50 Hz.

#### Phase current
In delta V_ph = V_L = 440 V.

I_ph = 440/20 = 22 A, lagging by 60° (cos φ = 0.5)

#### Power per phase
P_ph = V_ph I_ph cos φ = 440 × 22 × 0.5 = **4840 W** (= I²R = 22² × 10)

#### Total power
P = 3 × 4840 = **14 520 W = 14.52 kW**

#### Wattmeter readings
Line current I_L = √3 I_ph = 1.732 × 22 = 38.1 A, φ = 60°.

- W₁ = V_L I_L cos(30° − φ) = 440 × 38.1 × cos(−30°) = 440 × 38.1 × 0.866 = **14 520 W**
- W₂ = V_L I_L cos(30° + φ) = 440 × 38.1 × cos 90° = **0 W**

Sum = 14.52 kW, equal to the total power. At power factor 0.5 one wattmeter reads zero and the other reads the whole power.

## Check yourself

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

1. What is the phase current of the delta load?
   A. 22 A
   B. 38.1 A
   C. 12.7 A
   D. 8.8 A
   Answer: A. I_ph = 440/20 = 22 A. (Key point: Phase current)

2. What is the total power?
   A. 4840 W
   B. 14.52 kW
   C. 25.1 kW
   D. 8.4 kW
   Answer: B. Each phase 440 x 22 x 0.5 = 4840 W, times three = 14.52 kW. (Key point: Total power)

3. With φ = 60°, what do the two wattmeters read?
   A. 7.26 kW each
   B. 14.52 kW and 0
   C. 14.52 kW and 14.52 kW
   D. 0 and 0
   Answer: B. W2 = V_L I_L cos(30° + 60°) = 0 and W1 = total power. (Key point: Wattmeter readings)

## Related in Basic Electrical Engineering

- [Delta connected coils line current and power](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Adelta-coils-7ohm-0-03h-400v) — Basic Electrical Engineering, chapter 8
- [Delta load phase currents, power, line currents](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Adelta-load-100v-6-8-4-3-ohm) — Basic Electrical Engineering, chapter 8
- [Delta load (15+j20) phase and line currents](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Adelta-load-15j20-220v-phasors) — Basic Electrical Engineering, chapter 8
- [Delta load line currents and power, ABC sequence](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Adelta-unbalanced-52-45-380v-abc) — Basic Electrical Engineering, chapter 8
- [Delta capacitor per phase for 50 kW motor](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Acapacitor-delta-415v-50kw-0-72) — Basic Electrical Engineering, chapter 8
- [Two wattmeter method proof](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Atwo-wattmeter-method-proof) — Basic Electrical Engineering, chapter 8, asked 3×

## Related in other subjects

- [Two-wattmeter method star load](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_and_electronics_engineering%3Atwo-wattmeter-star-load) — Basic Electrical and Electronics Engineering, chapter 1
- [150 mile, 60 Hz line: ABCD and sending end](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3A150-mile-60hz-line-abcd) — Power System Analysis I, chapter 5
- [280 km, 60 Hz line: nominal-π vs long line](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3A280km-60hz-nominal-pi-long) — Power System Analysis I, chapter 5, asked 2×
- [Line with 50+j50 load, first maximum, power](https://www.bctengineer.com/question.md?key=semester_3%2Felectromagnetism%3Aline-50-j50-load-first-maximum) — Electromagnetism, chapter 5