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I'm looking to modify one of my pre-amps to use a 9v AC adapter instead of 6 D-cell batteries, for financial, environmental, and convenience reasons.

According to the wiring diagram, it draws 450mW at 9V, which makes it seem like it draws 20 amps. I=V/R, 20=9/.45 That just doesn't seem correct though, since that means the batteries would have to be replaced every 2.5 hours (48 amp hours for six D batteries).

I know I'm doing something wrong here. I haven't touched an electric circuit in years. Any help figuring out what size 9v adapter I need is appreciated.

As a side note, I don't want to make any modifications to the input voltage. I just want to connect this to the existing leads in place of the batteries, if possible, so that I can undo it at any time.

Upon reflection it is possible that it is set up as two banks of three batteries, leading to an input voltage of 4.5, but I'm pretty confident the whole thing is in series. I'm not in front of it right now to check.

.05 amps

P=V*I

Sent from my Nexus 4 using Tapatalk 4

I=p/v , r is resistance not power

Edit: too slow

AllanMar wrote:I=p/v , r is resistance not power

Edit: too slow

That was a dumb mistake on my part. Thanks Allan & X. So any 9v, 0.05amp adapter should do it?

I may have one if you want it.

Sandy

Oh? That would be great!