---
title: "Delta load phase currents, power, line currents"
subject: "Basic Electrical Engineering"
chapter: 8
key: "semester_1/basic_electrical_engineering:delta-load-100v-6-8-4-3-ohm"
source: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/delta-load-100v-6-8-4-3-ohm
site: BCT Engineering
---

# Delta load phase currents, power, line currents

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

- Asked in: 2067 Asoj, 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-100v-6-8-4-3-ohm
- Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-8

## Question

For the delta connected load as below, determine (i) phase currents (ii) total power consumed (iii) line currents. [Figure: delta load supplied at 100 V; one arm has 6 Ω with an inductor; another arm 6 Ω and 8 Ω; third arm 8 Ω and 4 Ω with a 3 Ω capacitor; lines a, b and c.]

## Short answer

Phase currents: Iab=10 A, Ibc=20 A, Ica=10 A. Total power = sum of I²R in each branch = 3 kW. Line current magnitudes: |Ia|=18.56 A, |Ib|=12.39 A, |Ic|=29.81 A.

## Answer

From the figure: a delta load supplied from lines a, b and c at 100 V. The three arms are:

- Arm a–c: 8 Ω resistor in series with a 6 Ω inductive reactance, Z_ac = 8 + j6 = 10∠36.87° Ω
- Arm a–b: 6 Ω resistor in series with an 8 Ω inductive reactance, Z_ab = 6 + j8 = 10∠53.13° Ω
- Arm c–b: 4 Ω resistor in series with a 3 Ω capacitive reactance, Z_cb = 4 − j3 = 5∠−36.87° Ω

Assumptions: balanced 100 V line voltages with sequence a → b → c: V_ab = 100∠0°, V_bc = 100∠−120°, V_ca = 100∠120° V.

#### (i) Phase currents
- I_ab = 100/(6 + j8) = **10 A** ∠−53.13° = 6.00 − j8.00
- I_bc = 100∠−120°/(4 − j3) = **20 A** ∠−83.13° = 2.39 − j19.86
- I_ca = 100∠120°/(8 + j6) = **10 A** ∠83.13° = 1.20 + j9.93

#### (ii) Total power
- P_ab = 10² × 6 = 600 W
- P_bc = 20² × 4 = 1600 W
- P_ca = 10² × 8 = 800 W

**Total power = 3000 W = 3 kW** (reactive power per arm: 800 VAR, −1200 VAR and 600 VAR, so the net is 200 VAR lagging)

#### (iii) Line currents
- I_a = I_ab − I_ca = 4.80 − j17.93 = **18.56 A** (∠−75.0°)
- I_b = I_bc − I_ab = −3.61 − j11.86 = **12.39 A** (∠−106.9°)
- I_c = I_ca − I_bc = −1.19 + j29.79 = **29.81 A** (∠92.3°)

(Their sum is zero.) The line current magnitudes depend on the phase sequence assumed, whereas the phase currents and total power do not.

## Check yourself

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

1. In a delta load, what voltage is applied across each arm, and how is each phase current found?
   A. Phase voltage = line voltage / sqrt(3); I = V/Z
   B. Full line voltage (100 V) across each arm; I = V_arm / Z_arm
   C. Line voltage is shared equally among the three arms; I = (100/3)/Z
   D. Full line voltage across each arm; I = V x Z
   Answer: B. In a delta connection each arm is connected directly between two lines, so it sees the full 100 V line voltage and its current is that voltage divided by the arm impedance. (Key point: Phase current in a delta arm)

2. Why does the arm with 4 Ω resistance and 3 Ω capacitive reactance give the largest phase current (20 A)?
   A. Its impedance is 4 - j3 = 5 Ω, the smallest of the three arms, so I = 100/5 = 20 A
   B. Its impedance is 4 + j3 = 5 Ω, so the capacitor adds to the current
   C. Its impedance is 7 Ω, so the current is 100/7
   D. Its impedance is 10 Ω and the capacitor doubles the current
   Answer: A. Capacitive reactance is negative: |4 - j3| = 5 Ω, half of the 10 Ω of the other arms, so the current is 20 A against 10 A. (Key point: Impedance of the capacitive arm)

3. How is the total power of this unbalanced delta load obtained, and what is it?
   A. sqrt(3) x 100 x line current x cos(phi) = 3000 W using one arm
   B. Sum of I^2 x R for each arm: 600 + 1600 + 800 = 3000 W
   C. Sum of I^2 x Z for each arm: 1000 + 2000 + 1000 = 4000 W
   D. Sum of I^2 x X for each arm, including the capacitor: 200 W
   Answer: B. Only resistance consumes real power, so P = sum of I^2 R over the arms: 10^2x6 + 20^2x4 + 10^2x8 = 3000 W. The balanced-load formula cannot be used here. (Key point: Total power calculation)

4. How are the line currents found for this unbalanced delta load, and what is the value of I_a?
   A. I_a = sqrt(3) x I_ab = 17.3 A
   B. I_a = I_ab + I_ca, found by phasor addition
   C. I_a = I_ab - I_ca (phasor difference) = 18.56 A
   D. I_a = I_ab - I_bc (phasor difference) = 18.56 A
   Answer: C. By KCL at node a, the line current is the phasor difference of the two phase currents meeting there: I_a = I_ab - I_ca = 4.80 - j17.93, magnitude 18.56 A. The sqrt(3) rule works only for balanced loads. (Key point: Line currents from phase currents)

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