case of a transformer, frequency of the induced e.m.f. in the secondary
is same as the voltage applied to primary. Now in case of induction
motor at start N = 0 and slip s = 1. Under this condition as long as s =
1, the frequency of induced e.m.f. in rotor is same as the voltage
applied to the stator. But as motor gathers speed, induction motor has
some slip corresponding to speed N. In such case, the frequency of
induced e.m.f. in rotor is no longer same as that of stator voltage.
Slip affects the frequency of rotor induced e.m.f. Due to this some
other rotor parameters also get affected. Let us study the effect of
slip on the following rotor parameters.
At start when N = 0, s = 1 and stationary rotor has maximum relative
motion with respect to R.M.F. Hence maximum e.m.f. gets induced in the
rotor at start. The frequency of this induced e.m.f. at start is same as
that of supply frequency.
As motor actually rotates with speed N, the relative speed of rotor
with respect R.M.F. decreases and becomes equal to slip speed of Ns – N. The induced e.m.f. in rotor depends on rate of cutting flux i.e. relative speed Ns
– N. Hence in running condition magnitude of induced e.m.f.
decreases so as to its frequency. The rotor is wound for same number of
poles as that of stator i.e. P. If fr is the frequency of rotor induced e.m.f. in running condition at slip speed Ns – N then there exists a fixed relation between (Ns – N), fr and P similar to equation (1). So we can write for rotor in running condition,
(Ns – N)/Ns = (120 fr / P)/(120 f / P) but (Ns – N)/Ns = slip s
At start we have s = 1 hence rotor frequency is same as supply
frequency. As slip of the induction motor is in the range 0.01 to 0.05,
rotor frequency is very small in the running condition.
A 4 pole, 3 phase, 50 Hz induction motor runs at a speed of 1470 r.p.m.
speed. Find the frequency of the induced e.m.f in the rotor under this