magikal,
Next, what are you asking? Are you inquiring about driving the transformer higher than its design frequency? The voltage will then decrease due to parasitic effects as enumerated by the responders above. Otherwise within its design frequency, the voltage will not change much.
Think of it this way. Suppose dI/dt is 10 amps/1 sec for a value of 10 . A doubling of the frequency will decrease the time to 1/2 sec so that dI/dt = 10/(1/2) = 20 . Doubling of the frequency will also double the inductive reactance of the transformer. That in turn will halve the current so that dI/dt = 5/(1/2) = 10 , which is what we had before. So it balances out until the frequency becomes so high that secondary effects take over.
Ratch
You have not defined the question very well. A solenoid is not used to couple with another solenoid in order to transform a voltage/current. It is used to move a plunger in order do some physical work. A transformer has no moving parts, and forms an inductive storage system that transforms voltage/current.In a transformer with 2 solenoids of different turn number mounted on a magnetic core powered by an AC source, why does the secondary EMF decrease as frequency of the power source is increased? What is the theoretical cause of this?
Next, what are you asking? Are you inquiring about driving the transformer higher than its design frequency? The voltage will then decrease due to parasitic effects as enumerated by the responders above. Otherwise within its design frequency, the voltage will not change much.
Think of it this way. Suppose dI/dt is 10 amps/1 sec for a value of 10 . A doubling of the frequency will decrease the time to 1/2 sec so that dI/dt = 10/(1/2) = 20 . Doubling of the frequency will also double the inductive reactance of the transformer. That in turn will halve the current so that dI/dt = 5/(1/2) = 10 , which is what we had before. So it balances out until the frequency becomes so high that secondary effects take over.
Ratch