Your current will be 30/R Amps. Where R is the resistance in Ohms.
Real-world batteries do not have zero internal resistance. When one connects a load (resistance) to a battery, current begins to flow and the open-circuit potential is divided between the battery's internal resistance and the resistance of the load. Thus, one will measure a lower voltage at the battery terminals when a load is connected, compared to no-load conditions.
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.
I have no idea
Internal resistance is approximately equal to 94.667
The current in a circuit, expressed in milliamperes, is1,000 x (battery or power supply voltage)/(resistance connected between the power supply terminals)If you increase the voltage of the battery or power supply, the current in the circuitincreases proportionally, at least until something in the circuit gets hot, melts, fuses,and opens the circuit.
The resistance of the load is what causes an electric current to flow in a circuit.
The resistance of the load is what causes an electric current to flow in a circuit.
The current depends on what is connected to the battery's terminals. If nothing is connected to it, then there is no current, and the battery lasts quite a while. In general, the current is 1.5/resistance of the external circuit connected to the battery until that number gets too big, and then the voltage of the battery sags, because it can't deliver that much current.
A direct current is obtained from a connection to a battery.
Real-world batteries do not have zero internal resistance. When one connects a load (resistance) to a battery, current begins to flow and the open-circuit potential is divided between the battery's internal resistance and the resistance of the load. Thus, one will measure a lower voltage at the battery terminals when a load is connected, compared to no-load conditions.
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.
If a battery is "shorted", meaning that its terminals are connected together through a low resistance, high current flows in the connection and the battery becomes discharged very soon. It makes no difference whether any part of the battery is connected to ground.
The voltage (V) between its terminals. ~Gabby
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.
sure whatever?
-- 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'.)
That depends on the resistance connected. Use Ohm's Law: V=IR. Solving for current: I = V/R. If nothing is connected, there will be no current (infinite resistance).