Your description of the system is either contradictory or incomplete, therefor no answer can be given.
24 volts
Short
Two 6-ohm resistors in parallel have a net effective resistance of 3 ohms.With 3 ohms connected across a 12-volt supply, the current is 12/3 = 4 amperes.
A 4000-Ω resistor is connected across 220 V will have a current flow of 0.055 A.Ohm's law: Voltage equals Current times Resistance
It depends on the construction of the motor. If the field winding is connected in series with the armaturewinding, the motor is like a universal motor and will probably operate normally. If the field winding is connected in parallel with the armature, the motor will not run, but will draw current, make noise and get hot.
The current flowing through a bulb is equal to the (voltage across the bulb) divided by the (bulb resistance), and can be expressed in Amperes. The rate at which the bulb dissipates energy is equal to (voltage across the bulb) times (current through the bulb), and can be expressed in watts.
Power is current times voltage, so a current of 0.5 amperes and a voltage of 220v across a bulb will yield a power of 110 watts.
Two 6-ohm resistors in parallel have a net effective resistance of 3 ohms.With 3 ohms connected across a 12-volt supply, the current is 12/3 = 4 amperes.
"Ohm" and its multiples is not a unit of current."Ampere" and its multiples is.The current through a 2.2-megohm resistance is(the voltage across the resistance)/(2,200,000) amperes .
A 4000-Ω resistor is connected across 220 V will have a current flow of 0.055 A.Ohm's law: Voltage equals Current times Resistance
It depends on the construction of the motor. If the field winding is connected in series with the armaturewinding, the motor is like a universal motor and will probably operate normally. If the field winding is connected in parallel with the armature, the motor will not run, but will draw current, make noise and get hot.
opposes changes in current
Yes
Potential Source connected across a Closed Circuit Path.
That will depend on the sum of the load resistance and the internal resistance of the battery (this is true for all power sources, not just 6 volt batteries). Small compact batteries tend to have higher internal resistance and therefore are more limited in the current they can deliver to a given load than larger batteries.
sure whatever?
The current flowing through a bulb is equal to the (voltage across the bulb) divided by the (bulb resistance), and can be expressed in Amperes. The rate at which the bulb dissipates energy is equal to (voltage across the bulb) times (current through the bulb), and can be expressed in watts.
Power is current times voltage, so a current of 0.5 amperes and a voltage of 220v across a bulb will yield a power of 110 watts.
No current flows through the battery. There is a current through the external circuit. I = E/R = 9/10 = 0.9 amperes.