Current is very confusing

I presume you ask why the voltage across the terminals of a battery (the terminal voltage) drops when the battery is delivering a large current.

The answer is because there is internal resistance present inside the battery.

The equation describing the situation where the battery is delivering a current is:

emf = I.R + I.r
where emf is the electromotive force of the battery
I is the current the battery is delivering
r is the internal resistance of the battery
R is the total resistance of the external circuit.

The terminal voltage of the battery (i.e. the voltage that you can measure by using a voltmeter) is given by the term (I.R), which is equal to [emf-I.r]
i.e. IR = emf-Ir
The term I.r is sometimes called the [lost voltage] because it indicates the voltage loss bue to the internal resistance of the battery.

Since the production of electrical energy in a battery is due to chemical reaction that takes place inside the battery. Any chemical reaction will generate heat. The larger the current produced, the more severe the reaction is and more heat will be produced. Therefore, when the battery is delivering a large current, the battery becomes hot and this results in the internal resistance of the battery to increase.

Thus the term (I.r) will increase when a large current is delivered by the battery because of the increase of both I and r.

Hence, the terminal voltage (= emf-I.r) will decrease. A decrease of terminal voltage leads to the decrease of power delivered by the battery.
]Note: power produced by a battery = (terminal voltage)^2/external resistance]