A circuit with a 2 ohm resistor and a 4 ohm resistor in series with a 12 volt battery will have 2 amps flowing through each resistor. The current is the same in each resistor because they are in series, and a series circuit has constant current throughout.
An ampere-foot is a unit used in calculating the fall of pressure in distributing electrical mains, equivalent to a current of one ampere flowing through one foot of conductor.
Use the symbols "I" for current, "E" for voltage, "R" for resistance.One of the forms in which Ohm's Law can be written is:I = E / R.Applying this equation to the question:I = 5 / 8 = 0.625 Ampere = 625 milliamperes
One ohm.
No, not unless it has a voltage or current regulator or series resistance to limit the current (built in somewhere).
Questions seem to be two! First is the direct current in a circuit with 12 V battery and 3 ohm resistor in it. The answer is 4 ampere if there is negligible internal resistance in the battery. By ohm's law V = R I So, I = V/R Second one is the alternating current with 120V given to 60 W bulb. The answer is 0.5 ampere. This 0.5 is the rms value of alternating current. Power P = V I So, I = P/V
3 Ampere
3 ampere
The unit of electrical current is Ampere, or 'Amps' for short.
The number of Ah (ampere-hours) or milli-ampere-hours tells you NOTHING about the voltage. It is the amount of current (amperes), multiplied by the time such a current can be extracted from the battery.If you multiply the number of Ah by the voltage, you will get the energy stored in the battery.
amperes or A.
The path where current flows through.
The unit for measuring current is the ampere, symbolized as A.
ampere is the unit in all the systems for electric current
I have no idea what a normal household battery is. Generally you would have to match the current and voltage of the car battery and then you have the issue of how long the battery can sustain the current, or ampere hours.
The battery life (assuming it is a primary cell) is determined by the Ampere-hour drawn from it. You cannot connect a 3.5V bulb directly to a 9V battery. The bulb will fuse.
An ampere-foot is a unit used in calculating the fall of pressure in distributing electrical mains, equivalent to a current of one ampere flowing through one foot of conductor.
If a battery sends a current of 10A through a circuit for one hour how many coulombs will flow through the circuit?