(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
The voltage across a DC device that draws 2A and consumes 12Wh/h is 12/2 or six volts.
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.
Yes, there will be no problem with this adapter. The 1 amp device will only be drawing half of what the adapter can produce.
Not for very long. If the device actually requires 2A in order to do its job, then the adapter ... which is only designed to deliver 0.5A or less ... will get warm, then overheat, and then fail. Possibly in as little as a few seconds.
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.