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Put my DC in a different spot this weekend with the pals, plugged into a newly installed 220v plug. After much fuss and ado, we finally got the thing mounted. Happily, I turned it on and boom, immediately tripped the breaker. Tried again…boom.

So I figured whatever it is, it will be an electrician thing, as the machine has worked in all plugs plugged in before this one.

So I proceeded to add the muffler and filter, and out of curiosity hit the green button and it worked fine. A few on and offs and it’s simply working again.

I know it’s supposed to draw 19.5 amp on a 20v circuit.

So, I don’t have a problem, but I’m curious as to how adding the resistance to the outflow affects the initial amperage draw at startup?

I'm thinking the startup current draw must be close to the breaker tripping point. Restriction/load is not really
present at startup...but is, in a small way. In your case, enough to trip the breaker.
In normal operation there's about a 15 to 20 percent current draw difference between fully open and
fully closed off. Fully open will draw more. The numbers you show seem to agree.

Ed, when an ac motor starts up it looks a lot like a short circuit until the inductive load ramps up so you a high surge current for a fraction of a second. We call this inrush current and it happens with inductive loads like motors and transformers. Most likely you need to change the breaker to a slow trip circuit breaker and you should be fine - assuming your wiring and such is correct. As a responsible engineer I have to advise you to consult an electrician.

From Grok:

Slow trip circuit breakers for 220V circuits are commonly referred to as "Type C" or "Type D" circuit breakers, depending on their tripping characteristics. These are designed for circuits with moderate to high inrush currents, such as motors or transformers, which require a delayed response to avoid nuisance tripping.
Type C: Trips at 5-10 times the rated current, suitable for general-purpose 220V circuits with moderate inrush loads.
Type D: Trips at 10-20 times the rated current, used for heavy inductive loads like large motors or equipment with high startup surges.
In some regions, they may also be called "slow blow" or "time-delay" circuit breakers.

That’s good to know right there.


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Turns out it’s grumpy like me this morning and I can’t figure out where to pour the coffee in, so I guess I’d better call the electrician.

I’m guessing it’s just a tired old breaker that’s been in service too long…

Thanks for the info!

Being a motor load with higher start up current we always built a motor circuit at 125% of name plate rating. So a motor rated at 19.5A at operating voltage would get a 30A breaker and be wired with #10 copper. Seems like that used to be in the code book but I haven't opened one up in over a decade.

See what your electrician tells you.

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Brian R, Wood Mechanic
N8ZED

Yep, #10 would be the minimum for a relatively short run IME.