H
henry chow
- Jan 1, 1970
- 0
Last time I posted a question about power lost on the transmission line and
I thank everyone for answering my question. Right now I get a better
understanding now because the V (voltage) (used on the power calculation) is
measured across the single line rather than between the two high-voltage
lines which is over ten thousand volts.
Here is another "basic" question on a transformer. Suppose I have a
transformer with a primary windings of 100 turns and a secondary windings of
200 turns. If I apply an AC voltage of 100 volts and 2 A to the primary, I
will get an AC voltage of 200 volts and 1A from the secondary.
If I put a load of, say 50 ohms, across the secondary windings output, what
should the current be? According to Ohms' law, it should be I=V/R=200
volts/50 ohms=4 A. But the max current you can get from the secondary
windings is 1A. Why the contradiction? If the current across the load is
1A, should the voltage drop across the load be V=IR=1A * 50 ohms=50 volts,
which is not right because the secondary windings have a voltage of 200
volts? Can you explain the contradiction?
Again, suppose instead I apply a load of 400 ohms to the secondary windings,
I should get a current of 200 volts/400 ohms=0.5 A? Is that correct? Then
what happens to the 1A I am supposed to get from the secondary windings?
I thank everyone for answering my question. Right now I get a better
understanding now because the V (voltage) (used on the power calculation) is
measured across the single line rather than between the two high-voltage
lines which is over ten thousand volts.
Here is another "basic" question on a transformer. Suppose I have a
transformer with a primary windings of 100 turns and a secondary windings of
200 turns. If I apply an AC voltage of 100 volts and 2 A to the primary, I
will get an AC voltage of 200 volts and 1A from the secondary.
If I put a load of, say 50 ohms, across the secondary windings output, what
should the current be? According to Ohms' law, it should be I=V/R=200
volts/50 ohms=4 A. But the max current you can get from the secondary
windings is 1A. Why the contradiction? If the current across the load is
1A, should the voltage drop across the load be V=IR=1A * 50 ohms=50 volts,
which is not right because the secondary windings have a voltage of 200
volts? Can you explain the contradiction?
Again, suppose instead I apply a load of 400 ohms to the secondary windings,
I should get a current of 200 volts/400 ohms=0.5 A? Is that correct? Then
what happens to the 1A I am supposed to get from the secondary windings?