---
title: "LT feeder options: loss and voltage drop, 50 kW"
subject: "Transmission and Distribution System Design"
chapter: 7
key: "semester_8/transmission_and_distribution_system_design:lt-feeder-two-options-50kw"
source: https://www.bctengineer.com/q/semester_8/transmission_and_distribution_system_design/lt-feeder-two-options-50kw
site: BCT Engineering
---

# LT feeder options: loss and voltage drop, 50 kW

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

- Asked in: 2072 Asoj, 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-feeder-two-options-50kw
- Whole chapter: https://www.bctengineer.com/md/semester_8/transmission_and_distribution_system_design/chapter-7

## Question

A particular load center has a peak demand of 50 kW at a power factor of 0.8 lagging. For the following two cases of 3 phase LT feeder configuration at distribution transformer, compute the % peak power loss and maximum percentage voltage drop in LT. Option I: 4 feeders each of length 2 km. Option II: 2 feeders each of length 4 km. The resistance and reactance per phase of LT conductors are 1.1 Ω/km and 0.25 Ω/km. Assume uniformly distributed load along the feeders. The distribution transformer is rated 11/0.4 kV.

## Short answer

For uniformly distributed load, per-feeder current and length determine loss (∝ I²l) and drop (∝ Il). Option I (4 feeders × 2 km): 8.95% peak power loss, 10.06% max voltage drop. Option II (2 feeders × 4 km): 35.81% loss, 40.23% drop.

## 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: $P = 50$ kW, p.f. 0.8 ($\sin\phi = 0.6$), $V_L = 400$ V (secondary of 11/0.4 kV), $r = 1.1\ \Omega$/km, $x = 0.25\ \Omega$/km.
$r\cos\phi + x\sin\phi = 1.1(0.8) + 0.25(0.6) = 1.03\ \Omega$/km.

#### Option I: 4 feeders × 2 km
$$
\begin{aligned}
P_{feeder} &= 50/4 = 12.5\ \text{kW},\quad I = \frac{12\,500}{\sqrt3\times400\times0.8} = 22.55\ \text{A}\\
P_L &= 4\times(22.55^2\times1.1\times2) = 4\times1118.98 = 4475.9\ \text{W}\\
\%P_L &= \frac{4475.9}{50\,000}\times100 = 8.95\ \%\\
V_{D,ph} &= \frac12\times22.55\times1.03\times2 = 23.23\ \text{V}\\
\%V_D &= \frac{23.23}{230.94}\times100 = 10.06\ \%
\end{aligned}
$$

#### Option II: 2 feeders × 4 km
$$
\begin{aligned}
P_{feeder} &= 25\ \text{kW},\quad I = \frac{25\,000}{\sqrt3\times400\times0.8} = 45.11\ \text{A}\\
P_L &= 2\times(45.11^2\times1.1\times4) = 2\times8951.8 = 17\,903.6\ \text{W}\\
\%P_L &= \frac{17\,903.6}{50\,000}\times100 = 35.81\ \%\\
V_{D,ph} &= \frac12\times45.11\times1.03\times4 = 92.92\ \text{V}\\
\%V_D &= \frac{92.92}{230.94}\times100 = 40.23\ \%
\end{aligned}
$$

#### Comparison
| Option | I per feeder | % peak loss | % max drop |
|---|---|---|---|
| I: 4 × 2 km | 22.55 A | 8.95 % | 10.06 % |
| II: 2 × 4 km | 45.11 A | 35.81 % | 40.23 % |

Going from option I to option II doubles both the current and the length of each feeder. The loss per feeder ($\propto I^2 l$) rises 8 times, and with half as many feeders the total loss rises 4 times; the drop ($\propto I l$) rises 4 times. Option I is clearly better, but even it exceeds the usual LT drop limit (about 5–8 %), suggesting a bigger conductor or more feeders/transformers.

**Answer: Option I: 8.95 % loss, 10.06 % drop; Option II: 35.81 % loss, 40.23 % drop.**

## Check yourself

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

1. For a 3-phase feeder with uniformly distributed load and sending-end current I, what is the total power loss?
   A. 3 I^2 r l
   B. I^2 r l
   C. I^2 r l / 3
   D. I^2 r l / 2
   Answer: B. Loss integrates I^2 along the feeder giving 1/3 of the lumped value; times 3 phases gives I^2 r l. (Key point: UDL loss formula)

2. For a UDL feeder, the per-phase voltage drop is given by:
   A. I(r cosφ + x sinφ) l
   B. (1/3) I(r cosφ + x sinφ) l
   C. (1/2) I(r cosφ + x sinφ) l
   D. sqrt3 I r l
   Answer: C. Current falls linearly to zero, so the drop equals full current over half the length. (Key point: UDL voltage drop)

3. In Option I, how is the current per feeder found?
   A. Total 50 kW divided by 4, then I = P/(sqrt3 V cosφ)
   B. 50 kW with I = P/(3 V cosφ) for one feeder
   C. Total 50 kW divided by 2, then I = P/(V cosφ)
   D. 50 kW divided by 4, then I = P/(V sinφ)
   Answer: A. The load is shared equally among feeders and each carries a 3-phase current at 400 V line voltage and p.f. 0.8. (Key point: Current per feeder)

4. Going from 4 x 2 km to 2 x 4 km feeders, how do the % loss and % voltage drop change?
   A. Both double
   B. Loss doubles, drop stays the same
   C. Both become 4 times
   D. Both stay the same
   Answer: C. Current and length both double: loss ∝ I^2 l gives 4 times, drop ∝ I l gives 4 times. (Key point: Effect of fewer, longer feeders)

## 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×
- [Peak loss and voltage drop for LT feeder options](https://www.bctengineer.com/question.md?key=semester_8%2Ftransmission_and_distribution_system_design%3Alt-feeder-configuration-options) — Transmission and Distribution System Design, chapter 7, asked 2×

## Related in Transmission and Distribution System Design

- [Distribution loss and voltage drop, four LT feeders](https://www.bctengineer.com/question.md?key=semester_8%2Ftransmission_and_distribution_system_design%3Aload-center-40kw-four-lt-feeders) — Transmission and Distribution System Design, chapter 7
- [LT loss and voltage drop, four 2 km feeders](https://www.bctengineer.com/question.md?key=semester_8%2Ftransmission_and_distribution_system_design%3Alt-feeders-40kw-2km-380v) — Transmission and Distribution System Design, chapter 7, asked 2×
- [Peak demand of load center from load pattern](https://www.bctengineer.com/question.md?key=semester_8%2Ftransmission_and_distribution_system_design%3Apeak-demand-load-center-load-pattern) — Transmission and Distribution System Design, chapter 6, asked 2×

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