# Numerical - Power rating of electric motor, given r.p.m & torque - Basic Mechanical Engineering

Q.  Determine power rating of an electric motor if it runs at 1440 r.p.m and line shaft transmits torque of 75 Nm.  Assume Reduction ratio = 1.6
- Published on 01 Sep 15

a. 10.36 kW
b. 11.3 kW
c. 7.068 kW
d. 9.12 kW

#### Discussion

• Rohit Bagh   -Posted on 05 Jun 21
Mechanical engineering
• Uttamkumar Jagtap   -Posted on 19 Aug 17
Very nice
and Very help full sir
ITI FITTER SEM-IV- FINAL EXAM- JULY-2016
QUESTION NO. 19
The tendency of force to rotate an object about an axis is called as torque or moment.
Given: speed (n) : 1440 r.p.m, torque = 75 Nm, reduction ratio (I) = 1.6
Formula: P = 2 p N T / 60
Solution:
i = n / N
1.6 = 1440 / N
N = 900 r.p.m
P = (2 p N T)/ 60
P = (2 x 3.14 x 900 x 75 ) /60
P = 423900 /60
= 7065 W
= 7.065 kW
Power rating of an electric motor is 7.065 kW
• jaypalsinh jadeja (asst. professor-amec dhari)   -Posted on 30 May 17
nice explanation.....
• Sravanthi   -Posted on 06 Nov 15
The tendency of force to rotate an object about an axis is called as torque or moment.

Given: speed (n) : 1440 r.p.m, torque = 75 Nm, reduction ratio (I) = 1.6

Formula: P = 2 π N T / 60

Solution:

i = n / N

1.6 = 1440 / N

N = 900 r.p.m

P = (2 π x 1440 x 75) / 60

= 7068 W

= 7.068 kW

Power rating of an electric motor is 7.068 kW

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