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(6a3 - 19a2 + 15a)/(2a - 3) = a(6a2 - 19a + 15)/(2a - 3) = a(2a - 3)(3a -5)/(2a -3) = a(3a - 5)
-4 + 2A = 12 - 15A + A-4 + 2A = 12 - 16A2A = 12 - 16A + 42A = 16 - 16AA = 8 - 8A
No
8a4+27a2-15a. contact at saqibahmad81@yahoo.com
No B/c ur unit Circuitry design for 1A u can put in 2A MAX 15A may cause burn ur unit
No, underpowering will not "power" the device, it will not run. You must get an adapter that is 6V and (2A or higher) will be ok as well.
The voltage across a DC device that draws 2A and consumes 12Wh/h is 12/2 or six volts.
Yes, there will be no problem with this adapter. The 1 amp device will only be drawing half of what the adapter can produce.
No... the source does not have the required capacity. The device (sink) would ask for more current which the source will not be able to provide. Do not use this source-sink pair.
Unfortunately no, if the device calls for 2000ma you will need a 2A (amp) power supply to adequately power it.
The adapter's voltage must match that of the device, and its current-rating must exceed that of the device. So the answer is yes.
Probably not. The other way around would be fine: a device that only needs 750mA will work fine on a 2A power supply, but one that needs 2A will NOT work on only 750mA, which is less than half the current required.