Easy V = IR problem, just help understand it.

[tnutty]

1. The problem statement, all variables and given/known data
When a 9.0V - battery is temporarily short-circuited, a 400mA- current flows.

What is the internal resistance of the battery?

Ans :R_int = 9/(400*10^-3)

Can someone explain why this is the answer, I understand its just V/I, but what exactly
does this mean? I am new to circuits so I am not sure what the term shot-circuit exactly
mean. I am also not sure what a internal resistance of the battery mean, just plain resistance?

[ehild]

To short-circuit a battery means that you connect the poles with a (short) piece of wire with negligible resistance. If you try it, you get even sparks, and your battery becomes hot, and flat in a very short time.
Now, internal resistance: When you connect something, for example a lamp to the battery, the current flows from one pole af the battery through the load (lamp) to the other pole and then through the battery to the first pole again, making a closed loop. But the battery itself has resistance. The electrodes themselves have, and there is some electrolyte material between them, and also some other material that absorbs gases, all of them have resistance. So we imagine a real battery as an ideal voltage source connected in series with a resistor Ri - the internal resistance of the battery. This internal resistance has to be taken into account if you calculate current. When you load the battery with a resistor R, R and Ri are connected in series, so I=E/(R+Ri). When the battery is short-circuited, R=0.

ehild