The question i'm having trouble with is:

Calculate the speed of the rotor of a 6 pole induction motor with a slip of 2.2%. The motor is connected to a 230V single-phase supply.

I am familiar with the table below and a certain formulae but i'm doing something incorrect and am confused on how to tackle this question.

Any formulaes or help would be appriciated as I am really stuck

enter image description here

  • \$\begingroup\$ "i'm doing something incorrect " Start with showing us what you do, even if you know it is incorrect. Homework question with no attempt at a solution will be closed. See also here \$\endgroup\$
    – Oldfart
    Nov 2, 2019 at 10:07
  • \$\begingroup\$ Which "certain formulae" have you tried? Do you know the supply frequency? \$\endgroup\$
    – user80875
    Nov 2, 2019 at 13:53
  • 1
    \$\begingroup\$ I'm voting to close this question as off-topic because this is a homework or study problem with no demonstration of attempt to solve? \$\endgroup\$
    – user80875
    Nov 2, 2019 at 13:54
  • \$\begingroup\$ If the voltage is 230V what is the likely supply frequency? Having got that go to your course documentation and look up the formula for slip.; \$\endgroup\$
    – RoyC
    Nov 2, 2019 at 14:40
  • \$\begingroup\$ the calculation im doing is 120-50hz/6 = 1000 \$\endgroup\$
    – JakeeYm
    Nov 3, 2019 at 0:40

1 Answer 1


This is pretty straight forward. The slip= (ws-wr)/ws, where ws= 1000 rpm at 50 Hz and 1200 at 60 Hz , wr is the speed of the motor ,

wr at 50 Hz = 1000(1-2.2%) = 978 rpm, wr at 60 Hz = 1200(1-2.2%) = 1173.6 rpm


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.