---
title: "LT loss and voltage drop, four 2 km feeders"
subject: "Transmission and Distribution System Design"
chapter: 7
key: "semester_8/transmission_and_distribution_system_design:lt-feeders-40kw-2km-380v"
source: https://www.bctengineer.com/q/semester_8/transmission_and_distribution_system_design/lt-feeders-40kw-2km-380v
site: BCT Engineering
---

# LT loss and voltage drop, four 2 km feeders

Transmission and Distribution System Design · Chapter 7: Distribution System Design

- Asked in: 2072 Magh, 10 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_8/transmission_and_distribution_system_design/lt-feeders-40kw-2km-380v
- Whole chapter: https://www.bctengineer.com/md/semester_8/transmission_and_distribution_system_design/chapter-7

## Question

A particular load center with uniformly distributed load of peak demand 40 kW has a power factor of 0.8. The distribution transformer has four 3φ LT feeders at 380 V of equal length of 2 km each. If the per phase resistance and reactance of LT conductors are 1.1 Ω/km and 0.25 Ω/km, determine: i) LT percentage peak power loss ii) Maximum percentage voltage drop in LT

## Short answer

Per-feeder current = 18.99 A. Total peak power loss = 3.17 kW, i.e. 7.94% of 40 kW. Maximum voltage drop = 19.56 V per phase, which is 8.92% of the phase voltage.

## Answer

#### Formulas (uniformly distributed load, 3-phase)
For each feeder of length $l$ carrying sending-end current $I$:
$$
\begin{aligned}
P_L &= 3\cdot\frac13 I^2 r l = I^2 r l\\
V_{D,ph} &= \frac12 I(r\cos\phi + x\sin\phi)\,l\\
\%V_D &= \frac{V_{D,ph}}{V_L/\sqrt3}\times100
\end{aligned}
$$
The peak load is shared equally between the feeders.

Given: $P = 40$ kW, p.f. 0.8 ($\sin\phi = 0.6$), 4 feeders, $V_L = 380$ V, $l = 2$ km, $r = 1.1$, $x = 0.25\ \Omega$/km.

#### Current per feeder
$$
P_{feeder} = \frac{40}{4} = 10\ \text{kW},\quad I = \frac{10\,000}{\sqrt3\times380\times0.8} = 18.99\ \text{A}
$$

#### i) Percentage peak power loss
$$
\begin{aligned}
P_{L,feeder} &= I^2 r l = 18.99^2\times1.1\times2 = 793.5\ \text{W}\\
P_{L,total} &= 4\times793.5 = 3174.1\ \text{W}\\
\%P_L &= \frac{3174.1}{40\,000}\times100 = 7.94\ \%
\end{aligned}
$$

#### ii) Maximum percentage voltage drop
The maximum drop occurs at the far end of each feeder.
$$
\begin{aligned}
r\cos\phi + x\sin\phi &= 1.1(0.8)+0.25(0.6) = 1.03\ \Omega/\text{km}\\
V_{D,ph} &= \frac12\times18.99\times1.03\times2 = 19.56\ \text{V}\\
V_{ph} &= 380/\sqrt3 = 219.39\ \text{V}\\
\%V_D &= \frac{19.56}{219.39}\times100 = 8.92\ \%
\end{aligned}
$$
(Line value of drop $= \sqrt3\times19.56 = 33.88$ V.)

| Quantity | Value |
|---|---|
| Current per feeder | 18.99 A |
| Total peak loss | 3.17 kW |
| % peak loss | 7.94 % |
| % max voltage drop | 8.92 % |

The drop is above the common LT limit of about 5–8 %, so a larger conductor or a shorter LT reach would be needed in practice.

**Answer: i) 7.94 %  ii) 8.92 %.**

## Check yourself

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

1. With 40 kW shared by 4 feeders at 380 V and p.f. 0.8, the current per feeder is about:
   A. 18.99 A
   B. 37.98 A
   C. 32.9 A
   D. 10.96 A
   Answer: A. I = 10000/(sqrt3 x 380 x 0.8) = 18.99 A. (Key point: Current per feeder)

2. How is the % peak power loss obtained?
   A. I^2 r l for one feeder divided by 40 kW
   B. 4 x (I^2 r l) divided by 40 kW
   C. 3 x 4 x I^2 r l divided by 40 kW
   D. 4 x (I^2 x l) divided by 40 kW
   Answer: B. Per-feeder UDL loss is I^2 r l; multiply by 4 feeders and divide by total load, giving 7.94 %. (Key point: Peak power loss)

3. Where does the maximum voltage drop occur on each feeder?
   A. At the transformer end
   B. At the middle
   C. At the far end
   D. Equally everywhere
   Answer: C. The drop accumulates along the feeder, so it is largest at the far end. (Key point: Where max drop occurs)

4. The per-phase drop of 19.56 V is expressed as % of which voltage?
   A. 380 V
   B. 380/sqrt3 = 219.4 V
   C. 380 x sqrt3 V
   D. 11 kV
   Answer: B. Per-phase drop is compared with phase voltage, giving 8.92 %. (Key point: Per-unit base for % drop)

## Close variants set in other exams

- [LT loss and voltage drop, 60 kW four feeders](https://www.bctengineer.com/question.md?key=semester_8%2Ftransmission_and_distribution_system_design%3Alt-feeders-60kw-1-5km-380v) — Transmission and Distribution System Design, chapter 7, asked 3×

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