whats the maximum power thats delivered by a 12V battery of internal resistance equal to 2ohm?

Answer 1

Trick question

You want the power delivered to a load connected to the battery.

Max Power that can be delivered is when the load = the internal resitance of the battery. Rinternal = 2 ohms, then Rload = 2 ohms

P=V*I (Voltage x Current)

I =V/R (R=Rinternal + Rload = 4 ohms)

I = 12/4 = 3 Amps

Voltage to load is 6V
Vload = V x Rload/(Rinternal + Rload)
Vload=12V x 2ohm/(2ohm+2ohm)=6V

Pload = Vload x I load = 6V x 3A = 18Watts


So your answer is 18 Watts is the maximum power that can be delivered to the load on the battery

To prove Rload=Rinternal for max power to load, try graphing a few values of Pload with Rload greater and less then Rinternal = 2 ohms to confirm

Answer 2

Maximum power into a load occurrs when the load equals the source resistance.

So Pmax occurs when a two ohm load R is placed across the battery, and that power is:

Pmax = R V^2 / (R + 2)^2 = 144/8 = 18, or 18 watts.

Also noteworthy, an equal amount of power is dissipated within the battery when matched.

You can derrive the max power by plotting the delivered Load Power as you change the Load Resistor. Just calculate the load voltage and current. The Load power will peak at R=2, any other load resistor will receive less power than Pmax.

Note also, that if you place a short across the battery you will cause the battery to dissipate the most power it can, but it will be useless power because it will not be delivered to a load (R=0).

So to summarize: In the case of the matched load, 36 W is supplied by the battery, 18W is delivered to the load and 18W is "wasted" in the battery.

In the case of the short, 72 W is dissapated (and wasted) in the battery, with zero Watts in the load.

Resistive power dissipation results in a measurable heating, but unless you are just trying to heat up a room, power dissapated in a battery is usually considered wasted power. Its usually the power delivered to the load that is important.
But the power dissipated in the battery should also be acknowledged. This self-heating is what got Sony into trouble recently that resulted in a major recall of millions of batteries.
HTH.

Answer 3

Do you mean when the only resistance in the circuit is the internal resistance of the battery? If so:-
I= 12/2 A = 6A
P= I² r = 6² x 2 W = 72 W

Answer 4

Assuming short-circuit load power can be found simply by

V^2 / R = 12^2 / 2 = 144 / 2 = 72W

However this power is within the battery and wasted mainly as heat.

Answer 5

6 amps is the answer using Ohms law (amps = Voltage/Resistence)
As in Voltage ,12 volts , divided by Resistance ,2 ohms ,equals 6 amps
this does not include the resistance of cable used ,it is the max

Answer 6

I think the question is askin for the power, so 1 extra step:
I = V/R
I = 12/2
I = 6 A

P= IV
P= 6 x 12
P = 72 W

Hope it helps!