---
title: "Series motor voltage to drive fan at 300 rpm"
subject: "Electrical Machines I"
chapter: 4
key: "semester_4/electrical_machines_i:series-motor-fan-400v-300rpm"
source: https://www.bctengineer.com/q/semester_4/electrical_machines_i/series-motor-fan-400v-300rpm
site: BCT Engineering
---

# Series motor voltage to drive fan at 300 rpm

Electrical Machines I · Chapter 4: DC Motor

- Asked in: 2071 Bhadra, 8 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/electrical_machines_i/series-motor-fan-400v-300rpm
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines_i/chapter-4

## Question

A series motor takes 20 A at 400 V to drive a fan at 200 RPM. Its armature resistance is 0.5 Ω and field resistance is also 0.5 Ω. If the torque required to drive the fan varies as the square of the speed, find the necessary applied voltage and current to drive the fan at 300 RPM.

## Short answer

Load torque varies as square of speed, so current is proportional to speed: I₂ = 30 A. Back emf at 300 rpm is 855 V, requiring applied voltage V₂ = 855 + 30×1 = 885 V.

## Answer

For a series motor the field current equals the armature current, so (below saturation) $\phi \propto I_a$. Then $T \propto \phi I_a \propto I_a^2$ and $N \propto E_b/\phi \propto E_b/I_a$.

**Given:** $V_1 = 400$ V, $I_1 = 20$ A, $N_1 = 200$ rpm, $R_a = 0.5\ \Omega$, $R_{se} = 0.5\ \Omega$, so the total series resistance is $R = R_a + R_{se} = 1\ \Omega$. $N_2 = 300$ rpm. Load torque $T \propto N^2$.

#### Step 1: New current

The motor torque must equal the fan torque:

$$
T \propto I^2 \quad \text{and} \quad T \propto N^2 \;\Rightarrow\; I \propto N
$$

$$
\begin{aligned}
I_2 &= I_1 \times \frac{N_2}{N_1} = 20 \times \frac{300}{200} = 30\ \text{A}
\end{aligned}
$$

#### Step 2: Back emf at 200 rpm

$$
\begin{aligned}
E_{b1} &= V_1 - I_1 (R_a + R_{se}) \\
&= 400 - 20 \times 1 = 380\ \text{V}
\end{aligned}
$$

#### Step 3: Back emf at 300 rpm

Since $N \propto E_b / \phi$ and $\phi \propto I$:

$$
\frac{N_2}{N_1} = \frac{E_{b2}}{E_{b1}} \times \frac{I_1}{I_2}
$$

$$
\begin{aligned}
E_{b2} &= E_{b1} \times \frac{N_2}{N_1} \times \frac{I_2}{I_1} \\
&= 380 \times \frac{300}{200} \times \frac{30}{20} \\
&= 380 \times 1.5 \times 1.5 = 855\ \text{V}
\end{aligned}
$$

#### Step 4: Required applied voltage

$$
\begin{aligned}
V_2 &= E_{b2} + I_2 (R_a + R_{se}) \\
&= 855 + 30 \times 1 = 885\ \text{V}
\end{aligned}
$$

| Quantity | 200 rpm | 300 rpm |
|---|---|---|
| Current | 20 A | 30 A |
| Back emf | 380 V | 855 V |
| Applied voltage | 400 V | 885 V |

Assumption: the magnetic circuit is unsaturated ($\phi \propto I$) and brush drop is neglected.

**Answer: To drive the fan at 300 rpm the motor needs an applied voltage of 885 V and takes a current of 30 A.**

## Check yourself

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

1. Fan torque varies as N^2 and series motor torque as I^2 (unsaturated). How are current and speed related?
   A. I is proportional to N^2
   B. I is proportional to N
   C. I is inversely proportional to N
   D. I is independent of N
   Answer: B. I^2 is proportional to N^2, so I is proportional to N; I2 = 20 x 300/200 = 30 A. (Key point: Current relation)

2. Why does torque of a series motor vary as I^2 below saturation?
   A. The field and armature carry the same current, so flux is proportional to Ia
   B. The field is across the supply
   C. Back emf is proportional to current
   D. Resistance is proportional to current
   Answer: A. Flux is proportional to Ia and T is proportional to flux x Ia, hence I^2. (Key point: Why T is proportional to I^2)

3. Eb at 300 rpm equals Eb1 x (N2/N1) x (I2/I1). Why is the current ratio included?
   A. Because flux rises with current, so emf for a given speed is larger
   B. Because armature resistance changes
   C. Because supply voltage changes
   D. Because torque is constant
   Answer: A. Eb is proportional to phi x N and phi is proportional to I, so Eb2 = 380 x 1.5 x 1.5 = 855 V. (Key point: Back emf at 300 rpm)

4. What applied voltage is needed at 300 rpm?
   A. 885 V
   B. 855 V
   C. 600 V
   D. 400 V
   Answer: A. V2 = Eb2 + I2(Ra + Rse) = 855 + 30 x 1 = 885 V. (Key point: Required voltage)

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