Reply to "Protecting a Lionel KW transformer."

Ingeniero No1OGR Forum Member

GGG, Dale -

So a good approach would be to divide the load(s) that the KW will supply, and select the proper breaker or fuse for each output.

Then, the sum of all the expected current draws times their individiual voltages should not exceed 140 (watts), correct?

For example:
Variable A not to exceed 14 volts: Large area building lights, 4-amp breaker or fuse = 56W
Variable B not to exceed 14 volts: Small area building lights, 2-amp breaker or fuse = 42W
Fixed 14v = Operating accessories, 2-amp breaker or fuse = 28W
Fixed 20v = Turnouts, 1-amp breaker or fuse = 20W

These add up to 146 watts, but since the operating accessories and turnouts are intermittent, even if they all were to draw the maximum current simultaneously, it would be for a very short interval.

Is the above about right, ball-parkish speaking?

Thx!

Alex

1/16/124:36 PM

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