If both terminals had the same potential there would be no current flow.
Yes. If the termals are connected by wires, charges will flow from one end to the other.
The low slip test is used to determine the D and Q axis impedance of salient pole machines. The machine terminals are short circuited, each phase voltage and current are measured. Since the terminals are short circuited, machine voltage must be reduced to prevent excessive output currents that would damage the machine.
current will increase
I would suspect the battery has been discharged, is defective, or you have loose or corroded battery terminals. Remove and clean the terminals and if that does not solve the problem charge the battery with a battery charger. If it will not take a charge then replace the battery.
A current needs a charge carrier. In biological tissue, such charge carriers would usually be ions, from dissolved salts.
When you charge more than your current credit limit
It depends how they are connected. If the terminals of two 12 volt batteries batteries are connected to the same circuit (in parallel) then there is no change in the flow of current- it will just last for twice as long a time. But if two opposing terminals of the batteries are connected to each other first and the free terminals then connected to the circuit (in series) then there would be a total voltage of 24 volts and twice as much current would flow. The two batteries would last for the same length of time as one battery on its own.
Oh, dude, let's break it down. So, if you have a current of 10mA, which is 0.01A, and you want to deliver 30c of charge, you just divide the charge by the current. So, 30c divided by 0.01A equals 3000 seconds. Easy peasy, right?
Yes, your can charge the battery with the radio on. But why would you want to do this? It will just take longer to charge with something drawing current from the battery.
The amount of charge accumulated in a dielectric charged by a current can be calculated using Q = I * t, where Q is the charge, I is the current, and t is the time. In this case, Q = 4 A * 5 s = 20 C. Therefore, 20 Coulombs of charge would be accumulated in the dielectric.
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 in the boot, where the spare tire would naturally live, so I assume the terminals would be there!