Available Power?

How much can I disclose?

____________________ | _____ |_____

120 vac ) || ( 8 vac | | 60 HZ ) || ( R 1 R 2 6 A ) || ( 14 Ohms 21 Ohms 50 Turns ) || ( 7 Turns |__________ | |___________________|

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, 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.

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An Inquiring Mind
Randy Gross
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http://www.gomedia.ca/~aaawelder/
Chance favors the prepared mind whose hands do the work!