---
title: "Delta load (15+j20) phase and line currents"
subject: "Basic Electrical Engineering"
chapter: 8
key: "semester_1/basic_electrical_engineering:delta-load-15j20-220v-phasors"
source: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/delta-load-15j20-220v-phasors
site: BCT Engineering
---

# Delta load (15+j20) phase and line currents

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

- Asked in: 2070 Chaitra, 8 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-15j20-220v-phasors
- Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-8

## Question

A 220 V, 3-phase voltage is applied to a balanced delta connected 3-phase load of phase impedance (15+j20)Ω. Calculate: i) The phase voltages ii) The phasor current in each line iii) The power consumed per phase iv) Draw the phasor diagram v) What is the phasor sum of three line currents? Why does it have this value?

## Short answer

Phase voltage equals line voltage 220 V. Line currents: 15.24 A at −83.13°, 156.87°, 36.87°. Power per phase is 1161.6 W. Sum of three line currents is zero (balanced system with no neutral).

## Answer

Balanced delta load Z_ph = 15 + j20 = 25∠53.13° Ω, V_L = 220 V, RYB sequence assumed.

#### i) Phase voltages
In delta, the phase voltage equals the line voltage: **V_ph = 220 V**. V_RY = 220∠0°, V_YB = 220∠−120°, V_BR = 220∠120° V.

#### ii) Currents
Phase current: I_ph = 220/25 = 8.8 A, lagging its phase voltage by 53.13°.

- I_RY = 8.8∠−53.13° A, I_YB = 8.8∠−173.13° A, I_BR = 8.8∠66.87° A

Line current magnitude: I_L = √3 × 8.8 = 15.24 A. In delta, each line current lags the corresponding phase current by 30°.

- I_R = I_RY − I_BR = **15.24∠−83.13° A**
- I_Y = I_YB − I_RY = **15.24∠−203.13° = 15.24∠156.87° A**
- I_B = I_BR − I_YB = **15.24∠+36.87° A**

#### iii) Power per phase
P_ph = I_ph² R = 8.8² × 15 = **1161.6 W** (total 3484.8 W)

#### iv) Phasor diagram (V_RY as reference)

| Phasor | Angle |
|---|---|
| V_RY, V_YB, V_BR | 0°, −120°, +120° |
| I_RY, I_YB, I_BR (53.13° behind their voltages) | −53.13°, −173.13°, +66.87° |
| I_R, I_Y, I_B (30° behind the phase currents) | −83.13°, +156.87°, +36.87° |

```text
           V_BR (120°)      I_BR (66.9°)
               ↖            ↗
                 ↖        ↗      ↗ I_B (36.9°)
                   ↖    ↗    ↗
   I_Y (156.9°) ←──────  ●  ──────────────► V_RY (0°)
   I_YB (−173.1°)      ↙ │ ↘
                     ↙   │   ↘ I_RY (−53.1°)
                   ↙     ↓
         V_YB (−120°)    I_R (−83.1°)
```
The three line voltages are 120° apart, each phase current is 53.1° behind its voltage, and each line current is 30° behind its phase current.

#### v) Phasor sum of the line currents
I_R + I_Y + I_B = 0.

For a balanced load the three line currents have equal magnitude and are 120° apart, so they form a closed triangle. This is also required by Kirchhoff's current law, because a delta-connected three-wire system has no path (neutral) for any residual current.

## Check yourself

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

1. What is the phase voltage of a delta load on a 220 V line?
   A. 127 V
   B. 220 V
   C. 381 V
   D. 110 V
   Answer: B. In delta V_ph = V_L. (Key point: Phase voltage)

2. What is the line current for I_ph = 8.8 A?
   A. 8.8 A
   B. 15.24 A
   C. 5.08 A
   D. 26.4 A
   Answer: B. I_L = √3 x 8.8 = 15.24 A. (Key point: Line current magnitude)

3. In delta, how does each line current relate in angle to the phase current?
   A. Leads by 30°
   B. Lags by 30°
   C. In phase
   D. Lags by 120°
   Answer: B. I_R = I_RY - I_BR lags I_RY by 30°. (Key point: Phase relation)

4. Why is I_R + I_Y + I_B = 0?
   A. The line currents are equal in magnitude and 120° apart, and KCL holds with no neutral path
   B. The loads are resistive
   C. The voltage is zero
   D. The sum is not zero
   Answer: A. Balanced currents form a closed triangle, and a three-wire delta has no path for residual current. (Key point: Sum of line currents)

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