Its the nominal output of a electric charger required to run a certain device. The charger is designed to provide 9volts up to a current draw of .6 of an amp. The device it powers will need 9volts and a maximum current draw of .6 of an amp to charge/run it properly. Think of a water tank in a house being fed from the water main. The voltage is the head of water in the tank. Turn on the tap and the current is the rate of water flow. The water can only be supplied at a certain rate. Exceed the rate and the head of water (voltage) drops and needs to be replenished.
this is an expression with a variable that is "v" 9v+74
the solution to the equation -53 plus 1 -26 is -78.
First of all you need to observe maximum current at various possible supply condition (i. e. 220VAC to 260VAC). Then check the inrush current for the equipment. When you start an high power device then it draws 4 times more current than usual. Your fuse should also take care of inrush current. For example: if maximum possible current is 500mA then your inrush current may go upto 2A. Thus if you put a fuse of 600mA then everytime you turn on the device, it may burn the fuse. So the suitable fuse for such condition is 2A
Do you mean ''What does the AUM Mantra mean?''
No, but sometimes "average" means "mean" - when it doesn't mean median, geometric mean, or something else entirely.
You can use any 9v adapter as long as the output amperage is rated higher than the amperage rating of your appliance so yes a 600ma adapter can be used to power a 500ma or 400ma appliance
The capacity of the 800 mA adapter is 200 mA larger than the 600 mA adapter.
I think you mean to ask if one can use a 9v 600mA adapter to power a 9v 300mA appliance. Yes, you can do that. A 9v 600mA adapter will deliver 9v at up to 600mA. A mA is one milli amp, or one thousandth of an amp. 300mA is 300 thousandth of an amp, 300/1000 or 0.3 amps. 600mA is 600 thousandth of an amp, 600/1000 or 0.6 amps, and is twice the current of 300mA.
no .never
Using a 12v 600ma source is allowing 12 volts and 600 amps of power. If the device requires a 12v 500ma source, the larger source is acceptable. Please note that it will only be drawing on 500ma of the available 600ma.
this is an expression with a variable that is "v" 9v+74
No. The adaptor will overheat.
3v2 + 9v = 3v (v + 3)
3v2 + 9v = 3v(v + 3)
I = p/v = 100w/9v = 11.11ai = p*v = 100w*9v = 900ai = (p/2)*v = 50w*9v = 450ai = p*2*v = 100w*(9v + 9v) = 8100a
yes this will work fine
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