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.
The supply won't have to work as hard. It is perfectly acceptable, for example, to use a 1A, 12v supply to supply a 12v, .5A load. The current rating indicates the ability of the supply to dissipate heat caused by the current flowing. If the load current is above the power supply current rating, the power supply will overheat.
The current, if connected to a voltage source that can supply the needed current to (R1+R2) R3, will be unchanged. If the source cannot supply the needed current, the terminal voltage will decrease, which will change the current flowing through R1 and R2.
It depends on how the capacitor is connected and whether the supply voltage is a.c. or d.c. Assuming you are talking about a power-factor improvement capacitor (connected in parallel with an inductive load, supplied with a.c.), then the supply current will reduce.
3 amps
4 resistors were connected in parallel it yields 5A of current from 220V supply.
The supply won't have to work as hard. It is perfectly acceptable, for example, to use a 1A, 12v supply to supply a 12v, .5A load. The current rating indicates the ability of the supply to dissipate heat caused by the current flowing. If the load current is above the power supply current rating, the power supply will overheat.
From a soldering gun supply store
Generally, yes. If your devise will only draw 80mA, it will do it whether it is connected to a supply that is capable of supplying 300mA or 800mA. The amperage rating on the power supply is the highest current that it is rated for. It will easily and safely provide less current. The load that is connected to the power supply will determine the actual amount of current.
The current, if connected to a voltage source that can supply the needed current to (R1+R2) R3, will be unchanged. If the source cannot supply the needed current, the terminal voltage will decrease, which will change the current flowing through R1 and R2.
opposes changes in current
It depends on how the capacitor is connected and whether the supply voltage is a.c. or d.c. Assuming you are talking about a power-factor improvement capacitor (connected in parallel with an inductive load, supplied with a.c.), then the supply current will reduce.
3 amps
for reducing the leakage current.
You can measure the current and power of a 'power supply', using an ammeter and a wattmeter. With the power supply connected to its load, the ammeter must be connected in series with the power supply's input. The wattmeter's current coil must also be connected in series with the power supply's input, and its voltage coil must be connected in parallel with the supply, taking the instrument's polarity markings into account.
No, not at all. It will have current when it is charging only
1.7amp
Two batteries can supply more current than one if they are connected in parallel.