\[E_{2} \propto\phi\propto V \Rightarrow
T \propto sV^{2}\]

\(T\) at any \(N_r\) is proportional to \(V^2\)

If \(V\) decreases by 10%, then
\(T\) decreases by 20%

Changes in \(V\) not only
affect\(T_{st}\) but torque under
running conditions also

If \(V\) decreases, then \(T\) also decreases

Hence, for maintaining same \(T\), \(s\)
increases i.e. \(N_r\) falls

Let \(V \rightarrow V^{'}\),
\(s \rightarrow s^{'}\) and \(T \rightarrow T^{'}\)\[\dfrac{T}{T'}=\dfrac{sV^{2}}{s^{'}V^{'2}}\]

Hardly any important changes in \(f\) take place on a large distribution
system except during a major disturbance

However, large frequency changes often take place on isolated,
low-power systems in which electrical energy is generated by means of
diesel engine or gas turbines

Example of such systems are: emergency supply in hospital and the
electrical system on a ship etc.

The major effect of change in \(f\) is on motor speed

If \(f\) drops by 10% then \(N_r\) also drops by 10%

A 50-Hz motor connected to 60-Hz supply then motor runs \((60-50)\times 100/50 = 20\%\) faster than
normal

A 50-Hz motor operates well on a 60-Hz line provided its terminal
voltage is raised to \(60/50=120\%\) of
the name plate rating

In that case, the new breakdown torque becomes equal to the
original breakdown torque and the starting torque is only slightly
reduced

However, power factor, efficiency, and temperature rise remain
satisfactory

Similarly, a 60-Hz motor can operate satisfactorily on 50-Hz
supply provided its terminal voltage is reduced to \(5/6=80\%\) of its name-plate
rating

Full-load Torque and Maximum Torque

Let \(s_f\) be the slip
corresponding to the full-load torque \[\begin{aligned}
T_{f} &
\propto\dfrac{s_{f}R_{2}}{R_{2}^{2}+\left(s_{f}X_{2}\right)^{2}}~\mbox{and}~T_{max}\propto\dfrac{1}{2X_{2}}\\
\therefore\dfrac{T_{f}}{T_{max}} &
=\dfrac{2s_{f}R_{2}X_{2}}{R_{2}^{2}+\left(s_{f}X_{2}\right)^{2}}\\
\Rightarrow & \dfrac{T_{f}}{T_{max}}=\dfrac{2s_{f}\cdot
R_{2}/X_{2}}{\left(R_{2}/X_{2}\right)^{2}+s_{f}^{2}}=\dfrac{2s_{m}s_{f}}{s_{m}^{2}+s_{f}^{2}}
\end{aligned}\]

In general, \[\dfrac{\mbox{operating
torque at any slip s}}{\mbox{maximum torque}} =
\dfrac{2s_ms}{s_m^2+s_f^2}\]