Answer / 7anoter4

If the change in load will be the same Pa1=Pa2 and the
rated frequency fo1=fo2

for the same time interval dt1=dt2

IF Pa=H*MVA/180/fo*d2R/dt2 [See IEEE-399 ch. 8.6.2 How
stability programs work]

and d2R/dt2=2*pi()*df/dt

then (df/dt)1=Pa/2/pi()*180*fo/MVA1/H1 and

(df/dt)2=Pa/2/pi()*180*fo/MVA2/H2

(df/dt)2/(df/dt)1=MVA1*H1/MVA2*H2

If MVA1=MVA2 then (df/dt)2=(df/dt)1*H1/H2

df2=12*30/42=8.57 Hz

i am using auto transformer for starting of induction motor to provide reduse voltages...but the principle of a tansformer according to definition"which transform energy to on circuit to another circuit with constant power and constant frequency". so if i am using reduse voltage then current obsely hige because power is same.but we use auto transfarmer for starting of induction motor to give low current.but here according to definition we are giving hige current.

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