---
title: "Starting current; speed and % drop of shunt motor"
subject: "Electrical Machines"
chapter: 4
key: "semester_4/electrical_machines:dcm-high-starting-current-speed-drop"
source: https://www.bctengineer.com/q/semester_4/electrical_machines/dcm-high-starting-current-speed-drop
site: BCT Engineering
---

# Starting current; speed and % drop of shunt motor

Electrical Machines · Chapter 4: DC Motor

- Asked in: 2079 Jestha, 2+6 marks
- Video: [Numerical on speed control of DC shunt motors  #electricmotors #electricalengineering](https://www.youtube.com/watch?v=T7KmAY7FYA4) — EEE, SJBIT
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/electrical_machines/dcm-high-starting-current-speed-drop
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines/chapter-4

## Question

Why is the starting current very high in dc motor? A 4 pole DC shunt motor working on 250 V, takes a current of 2 A when running on 1000 rpm. What will be its speed and percentage speed drop if the motor takes 51 A at certain load. Given armature and field winding resistance are $0.2\ \Omega$ and $250\ \Omega$ respectively.

## Short answer

At starting, back emf is zero, limiting current only by R_a: I=V/R_a. For the shunt motor, field current is 1 A. No-load: Eb1=249.8 V, 1000 rpm; at 51 A line current, Eb2=240 V. Speed = 960.8 rpm, drop = 3.92%.

## Answer

#### Why the starting current is high

**Reason:** the armature current of a DC motor is

$$
I_a = \frac{V - E_b}{R_a}
$$

At the instant of starting the armature is at rest, so the back emf $E_b = \dfrac{\Phi ZNP}{60A} = 0$. The current is then limited only by the very small armature resistance:

$$
I_{a(start)} = \frac{V}{R_a}
$$

For example, a 220 V motor with $R_a = 0.5\ \Omega$ would draw $220/0.5 = 440$ A, about 15–20 times its full-load current (say 25 A).

#### Speed at 51 A

The field is across the supply, so $I_{sh} = 250/250 = 1$ A and the flux is constant. Then $N\propto E_b$.

**Condition 1 (1000 rpm, 2 A):**

$$
I_{a1} = 2 - 1 = 1\ \text{A}, \qquad E_{b1} = V - I_{a1}R_a = 250 - 1(0.2) = 249.8\ \text{V}
$$

**Condition 2 (51 A):**

$$
I_{a2} = 51 - 1 = 50\ \text{A}, \qquad E_{b2} = 250 - 50(0.2) = 240.0\ \text{V}
$$

$$
\frac{N_2}{N_1} = \frac{E_{b2}}{E_{b1}} \;\Rightarrow\; N_2 = 1000\times\frac{240.0}{249.8} = 960.8\ \text{rpm}
$$

#### Percentage speed drop

$$
\frac{N_1 - N_2}{N_1}\times 100 = \frac{1000 - 960.8}{1000}\times 100 = 3.92\%
$$

**Answer:** speed at 51 A $= 960.8\ \text{rpm}$; speed drop $= 3.92\%$.

## Check yourself

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

1. Why is the starting current of a DC motor very high?
   A. Flux is maximum at start
   B. Back emf is zero at standstill so only Ra limits the current
   C. Supply voltage is higher at start
   D. The field is open at start
   Answer: B. Ia = (V - Eb)/Ra with Eb = 0 gives V/Ra. (Key point: Why starting current is high)

2. For the 250 V motor with Rsh = 250 ohm taking 2 A, the armature current is:
   A. 2 A
   B. 0.2 A
   C. 1 A
   D. 3 A
   Answer: C. Ish = 250/250 = 1 A, so Ia = 2 - 1 = 1 A. (Key point: Field and armature current)

3. With flux constant, what is the speed at 51 A line current?
   A. 960.8 rpm
   B. 1000 rpm
   C. 1040 rpm
   D. 921.6 rpm
   Answer: A. Eb2 = 250 - 50(0.2) = 240 V; N2 = 1000 x 240/249.8 = 960.8 rpm. (Key point: Speed at 51 A)

4. The percentage speed drop (1000 to 960.8 rpm) is about:
   A. 0.4%
   B. 3.92%
   C. 39.2%
   D. 9.6%
   Answer: B. (1000 - 960.8)/1000 x 100 = 3.92%. (Key point: Percentage drop)

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