No because the adaptor must supply the required current for the device (or more). A device that needs 2 amps will not work if the adaptor can only supply 200 mA (0.2 amp).
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The adapter's output is 200 mA, the load demand is 2 amps which is the same as 2000 mA.
I don't suggest you try it. I thought the same thing and plugged in a 12v adapted into my DM6 Drum Module that required only 9v and it burned down. Same thing happened to my modem, and my 3D speakers. Anyway... I ended up wasting like $350 buying my module back and lets not even talk about the rest of the stuff I lost... Just go to eBay or Amazon and find the right AC/DC cable that suits your system. I hope I answered your question. ^^
It will operate the 1600 mA device but the adapter will run hotter that it was designed to do. There will also be a bit of a voltage drop due to trying to draw 1600 mA from a 1500 mA adapter.
No
Yes, the rating of the adapter is the maximum amount of current allowed to be drawn from it. The adapter you state has a maximum of 1000 mA or 1 amp. If the device draws 700 mA's you will have 300 mA's of spare capacity.
No. The device will try to draw 700 milliamps which will overload the 300 milliamp adapter most likely causing it to burn out and possibly catch fire.
No
1000-300 = 700
700 + 300 = 1,000
No. The 200mA adapter will only produce 1/8th the current of the 1600mA adapter, and probably won't even power what you intend to use. If it does, the extreme undercurrent will likely damage the equipment itself. A possible solution is a universal adapter, which may allow you to set the voltage and/or current for use with the intended device.
Yes, the 1000 mA has more that enough capacity to operate a 700 mA device. If the 700 ma adapter was original equipment with the device it was to charge, then the device is more than likely drawing about 500 to 600 mA. Look on the device's nameplate and you should see the mA draw.
As long as the current output is greater than power supply it replaces it will work fine. If the replacement is rated for less current (amperes) than the original you don't want to use it. Of course a DC supply has to be replaced with a DC power supply. AC with an AC supply. The polarity should be the same and the voltage should be the same and the pin and barrel should be the same diameter as well.
1700
700*3/7 = 300
700
The range is 400. (700 - 300 = 400)