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Subject: Formula for Available Power?

Date: 08/22/03 at 12:50 AM
Posted by: Randy Gross
E-mail: rgwg99[AT55]acsplus.com (replace [AT55] with @ to reply)
Message Posted:

Circuit:

120 vac ___ ____________ .
6 A )||( ____|____ .
)||( 8 vac | | .
50 Turns )||( R1 R2 .
)||( 14 ohms 21 ohms
)||( |_________| .
___)||(____________|

From this schematic ( I hope it is readable ), I can determine R total as:

1 / 14 + 1 / 21 = 1 / .119 = 8.4 Ohms.

The current in this circuit is:

I = 8 / 8.4 = 952 mA (800 mA actual reading).

Question: From this data, how can I determine the available current in the secondary?
952 mA is the apparent current but, the actual available secondary current is much, much, higher, approximately 100 amps. Basically, what I am looking for is a calculation.

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