No current flows through the battery.
There is a current through the external circuit. I = E/R = 9/10 = 0.9 amperes.
That will depend on the internal resistance of the battery. I = E / R Where I is the current, E is the open circuit battery voltage, and R is the internal resistance of the battery.
The battery is used by the meter to apply a voltage across the resistance being measured and determine the current which determines the resistance. If there is no battery it can't supply the voltage and can't measure current, so no reading.
If you put a current of 1 amp through a resistor, the voltage across it is equal to the resistance in ohms. This can also be done with lower currents and then the result must be multiplied up in the right ratio. An ohmmeter, or the ohm scale on a multimeter, uses a battery to supply the current.
in a parallel circuit resistance decreases increasing the current.
Voltage across a resistance = (resistance) x (current through the resistance) =4 x 1.4 = 5.6If the ' 1.4 ' is Amperes of current, then the required voltage is 5.6 volts.
That will depend on the internal resistance of the battery. I = E / R Where I is the current, E is the open circuit battery voltage, and R is the internal resistance of the battery.
The battery is used by the meter to apply a voltage across the resistance being measured and determine the current which determines the resistance. If there is no battery it can't supply the voltage and can't measure current, so no reading.
In a d.c. circuit, voltage drop is the product of resistance and current through that resistance.
4.8 ohms
Resistance = (voltage across the circuit) divided by (current through the circuit) =12 / (3 x 10-3) = 4 KΩ
Volt across a resistor = resistance x current through the resistor.
Battery maximum current is limited by the internal resistance of the battery. As the current is increased towards this maximum, you will notice the output voltage appear to shink towards zero. What this means is the voltage the battery is capable of supplying is being dropped almost completely across the internal resistance, so no real power is available to use.This internal resistance is dependent on the chemical and physical makeup of the battery.
-- The resistance of the wire is proportional to its length. -- When the length is reduced by 1/2 , the resistance is also reduced by 1/2 . -- Reducing the resistance across the battery by 1/2 causes the current to double. -- The new current is 100 mA. (Assumes zero internal resistance in the battery, and that the 4.5 volts doesn't 'sag'.)
"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 .
The amount of current that will pass through a resistance is dependant upon the voltage applied across the resistance. Voltage devided by resistance equals current. This is Ohm's Law.
If you put a current of 1 amp through a resistor, the voltage across it is equal to the resistance in ohms. This can also be done with lower currents and then the result must be multiplied up in the right ratio. An ohmmeter, or the ohm scale on a multimeter, uses a battery to supply the current.
3 ohms. 9 volts across a 3 ohm resistor becomes 9/3 or 3 amps.