---
title: "DC shunt motor hoist: raising and rheostatic lowering"
subject: "Utilization of Electrical Energy"
chapter: 3
key: "semester_7/utilization_of_electrical_energy:dc-shunt-hoist-rheostatic-braking-400kg"
source: https://www.bctengineer.com/q/semester_7/utilization_of_electrical_energy/dc-shunt-hoist-rheostatic-braking-400kg
site: BCT Engineering
---

# DC shunt motor hoist: raising and rheostatic lowering

Utilization of Electrical Energy · Chapter 3: Control of Electric Drive

- Asked in: 2080 Bhadra, 8 marks
- Video: [Breaking of dc motor](https://www.youtube.com/watch?v=0S9HiKro8Gc) — Parmod Saharan
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/utilization_of_electrical_energy/dc-shunt-hoist-rheostatic-braking-400kg
- Whole chapter: https://www.bctengineer.com/md/semester_7/utilization_of_electrical_energy/chapter-3

## Question

A 220V, DC shunt motor having an efficiency of 85% drives a hoist having an efficiency of 80%. Calculate the current drawn from the supply to raise a load of 400 kg at 2.5 m/s. What resistance must be added in the armature circuit in order to lower the load at 2.5 m/s using rheostatic braking, assuming that the efficiency of the motor and the load will remain the same?

## Short answer

Lifting: load power 9810 W, motor input 14426.5 W, supply current 65.6 A. Lowering at 2.5 m/s with rheostatic braking: generated electrical power 6670.8 W, braking resistance 7.26 Ω.

## Answer

#### Raising the load
Power needed to lift the load:
$$
P_{load} = mgv = 400 \times 9.81 \times 2.5 = 9810\ \text{W}
$$
Motor output (hoist efficiency 80 %) and input (motor efficiency 85 %):
$$
\begin{aligned} P_{out} &= \frac{9810}{0.80} = 12262.5\ \text{W} \\ P_{in} &= \frac{12262.5}{0.85} = 14426.5\ \text{W} \\ I &= \frac{P_{in}}{V} = \frac{14426.5}{220} = 65.57\ \text{A} \end{aligned}
$$

#### Lowering at 2.5 m/s by rheostatic braking
When lowering, the falling load drives the motor, which runs as a generator disconnected from the supply with a resistor across its armature (field stays excited). Power now flows the other way, so each efficiency reduces it:
$$
\begin{aligned} \text{Shaft power to machine} &= 9810 \times 0.80 = 7848\ \text{W} \\ \text{Electrical power generated} &= 7848 \times 0.85 = 6670.8\ \text{W} \end{aligned}
$$
The speed is the same as when raising and the field is unchanged, so the generated emf is taken as about the supply voltage, $E \approx 220$ V (armature resistance not given; its drop is neglected). All this power is dissipated in the braking resistance:
$$
\begin{aligned} I_b &= \frac{6670.8}{220} = 30.32\ \text{A} \\ R &= \frac{E^2}{P} = \frac{220^2}{6670.8} = 7.26\ \Omega \end{aligned}
$$

**Answer: Supply current while raising = 65.6 A; braking resistance for lowering at 2.5 m/s ≈ 7.26 Ω.**

(If the armature resistance $R_a$ were known, the external resistance to add would be $7.26 - R_a$.)

## Check yourself

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

1. What power is needed to lift 400 kg at 2.5 m/s?
   A. 1000 W
   B. 9810 W
   C. 12262 W
   D. 24525 W
   Answer: B. P = mgv = 400 x 9.81 x 2.5 = 9810 W. (Key point: Load power)

2. How are the efficiencies used when raising the load?
   A. Multiply the load power by both efficiencies
   B. Divide the load power by both efficiencies to get input power
   C. Divide by the motor efficiency only
   D. Add the efficiencies
   Answer: B. Input = 9810/(0.80 x 0.85) = 14426 W, so I = 65.6 A. (Key point: Supply current when raising)

3. When lowering, how do the efficiencies act?
   A. They are divided out as before
   B. They multiply and reduce the electrical power: 9810 x 0.8 x 0.85 = 6670.8 W
   C. They are ignored
   D. They are added
   Answer: B. Power flows from load to machine, so each loss reduces the available electrical power. (Key point: Lowering efficiencies)

4. How is the braking resistance obtained?
   A. R = E^2/P with E about 220 V and P = 6670.8 W
   B. R = V/I raise
   C. R = P/E^2
   D. R = 220/65.6
   Answer: A. All generated power is dissipated in R: 220^2/6670.8 = 7.26 Ω. (Key point: Braking resistance)

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