You have to put a DC current ammeter between one pole of the battery and the place that particular wire goes to on the saw.. You would have to first remove the handle part of the saw and possibly disconnect the wire to the switch then measure between that wire and switch terminal. - As it is DC, there will of course be NO outlet ( as someone else has incorrectly suggested )
Put a current meter in between the saw and the outlet.
You are confusing terms. Current is measured in amps or some fraction thereof. If you have a 24 volt battery (All batteries are DC) you need to connect a load to the battery and then measure the current that flows in the circuit. The current is measured in amps, milliamps or perhaps microamps for a very high resistance load.
You need a DC power supply in order to connect a 24v DC motor to household power.
It depends. The 2A current, did you measure that while the soldering iron was on? Or is it rated at 2A current consumption on the device itself (on the powercable or the stem of the soldering iron). Generaly speaking you can calculted the real power consumption by using P=V*I (thus 2A*24V = 48W). But do remeber that this power consumption is in the steady state, that is, after its switch on and all transient effects have died down. To be safe allow for 3A-4A switching currents that occurs at power on.
No its not. Its for 24v security cameras.
24 volts (joules per coulomb), alternating current
You don't. 2 12v panels together don't make 24v - doesn't work that way. If you need 24v you need to get 24v panels.
Yes. You can surely use the same one. What matters more is the current rating. If the current rating is more than that required and the voltage is same then it will do.
Of course it is bad, whether it`s humming or not.
Apply 24v and listen for click Ck coil for open with ohmmeter Clean contacts if possible
It is possible but it is too involved to explain here. You would need a printed circuit board and a schematic to build what you want.
2.7A on the primary25A on the secondaryThis is easily calculated usingP = VIAs long as we make sure we use RMS values for current and voltage.
Buy two strips and install them in series. Measure the current through one strip at 12 volts. Divide 12 by that current to get ohms. Install a resistor of that resistance in series with the strip. Calculate the required power by multiplying 12 by the current, and make sure the resistor can handle that amount of power.