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

# Distribution loss and voltage drop, four LT feeders

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

- Asked in: 2078 Chaitra, 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/load-center-40kw-four-lt-feeders
- 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 40 kW at pf of 0.85 lagging. The distribution transformer having 200 watt iron loss and 480 watt copper loss is used to supply four 3 phase LT feeders at 380 V of equal length of 1.5 km each. If LT feeder conductor has resistance and reactance of 0.9 ohm per km and 0.25 determine (i) % Distribution loss (ii) % voltage drop in LT line.

## Short answer

Each feeder carries 10 kW, current 17.88 A, R=1.35 Ω, X=0.375 Ω. Line loss per feeder = I²R = 431 W, total LT loss 1.73 kW; with transformer loss 0.68 kW, total 2.41 kW, %loss = 6.01%. Voltage drop = (√3×17.88×(1.35×0.85+0.375×0.5268))/2 = 20.8 V, 5.48%.

## Answer

**Assumptions:**

- The 40 kW peak is shared equally by the four feeders (10 kW each) and the load is **uniformly distributed** along each feeder.
- The reactance is 0.25 Ω/km per phase.
- The transformer iron loss (200 W) and copper loss (480 W) are the losses at this peak load (the transformer rating is not given).
- % distribution loss is taken as peak power loss as a percentage of peak demand.

#### Step 1: Current in each feeder

$$
I = \frac{P}{\sqrt{3} V \cos\phi} = \frac{10{,}000}{\sqrt{3} \times 380 \times 0.85} = 17.875 \text{ A}
$$

#### Step 2: Feeder impedance

$$
\begin{aligned}
R &= 0.9 \times 1.5 = 1.35\ \Omega, \quad X = 0.25 \times 1.5 = 0.375\ \Omega \\
\sin\phi &= \sqrt{1 - 0.85^2} = 0.5268
\end{aligned}
$$

#### Step 3: LT line loss

For uniformly distributed load, loss = 1/3 of the end-loaded value:

$$
\begin{aligned}
P_{loss,feeder} &= \frac{3 I^2 R}{3} = I^2 R = 17.875^2 \times 1.35 = 431.33 \text{ W} \\
P_{loss,LT} &= 4 \times 431.33 = 1725.31 \text{ W}
\end{aligned}
$$

#### Step 4: Total distribution loss

$$
\begin{aligned}
P_{loss} &= 1725.31 + 200 + 480 = 2405.31 \text{ W} \\
\%\text{ loss} &= \frac{2405.31}{40{,}000} \times 100 = 6.01\%
\end{aligned}
$$

(LT line alone: 4.31%; transformer: 1.70%. On the basis of input power, $2405.31/42{,}405.31 = 5.67\%$.)

#### Step 5: Voltage drop in LT line

For uniformly distributed load, drop = 1/2 of the end-loaded value:

$$
\begin{aligned}
\Delta V_L &= \frac{\sqrt{3}\, I (R\cos\phi + X\sin\phi)}{2} \\
&= \frac{\sqrt{3} \times 17.875 \times (1.35 \times 0.85 + 0.375 \times 0.5268)}{2} \\
&= \frac{\sqrt{3} \times 17.875 \times 1.3450}{2} = 20.82 \text{ V} \\
\%\Delta V &= \frac{20.82}{380} \times 100 = 5.48\%
\end{aligned}
$$

**Answer:**

- (i) % distribution loss ≈ **6.01%** of peak demand (2.41 kW)
- (ii) % voltage drop in each LT feeder ≈ **5.48%** (20.82 V)

The voltage drop is within the usual ±10% LT limit (and close to a 5% limit), and losses are acceptable.

## Check yourself

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

1. Current in each feeder (10 kW, 380 V, 0.85) is:
   A. 17.9 A
   B. 30.9 A
   C. 10 A
   D. 71.5 A
   Answer: A. I = 10000/(sqrt3 x 380 x 0.85) = 17.875 A. (Key point: Feeder current)

2. For a uniformly distributed load, the 3-phase loss per feeder is:
   A. 3 I^2 R
   B. I^2 R
   C. I^2 R/2
   D. I^2 R/9
   Answer: B. 1/3 of the end-loaded 3 I^2 R, so I^2 R = 431.3 W. (Key point: UDL loss)

3. Total distribution loss percentage (LT lines + transformer) is:
   A. 4.31%
   B. 1.70%
   C. 6.01%
   D. 5.48%
   Answer: C. 1725.3 + 200 + 480 = 2405.3 W; 6.01% of 40 kW. (Key point: Total loss)

4. The % voltage drop in the LT line is:
   A. 10.96%
   B. 5.48%
   C. 2.74%
   D. 1.3%
   Answer: B. Drop = sqrt3 I (R cos + X sin)/2 = 20.82 V = 5.48% of 380 V. (Key point: Voltage drop)

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