Consider the following Nobreak circuit, each battery has 13.8V (four batteries in the circuit), each battery has a internal resistance of 0.01 Ohms, each battery has the capacity of 60 Ah (Ampere hour). The machine consumes 280W, has a efficiency of 80% and Power Factor of 0.98.
Find the current supplied by the batteries and how much time the machine can work using the Nobreak

Question

Consider the following Nobreak circuit, each battery has 13.8V (four batteries in the circuit), each battery has a internal resistance of 0.01 Ohms, each battery has the capacity of 60 Ah (Ampere hour). The machine consumes 280W, has a efficiency of 80% and Power Factor of 0.98.
Find the current supplied by the batteries and how much time the machine can work using the Nobreak

anonymous · Answer

attachment

radar · Answer

With a given efficiency of 80% how much wattage will need to be supplied to deliver the 280 Watts?

.8x=280       Calculate x.

radar · Answer

That is a first step.

anonymous · Answer

i was thinking to find the current supplied by the batteries, i = 13.8/0.01 = 1380A, then the power supplied is P = 13.8*1380 = 19.044kW
Is that right?

radar · Answer

Next calculate the "Apparent Power" required sometimes this is referred to "VoltsAmperes".  The power factor is given as .98.  
.98VA=350     (350 is the amount of power needed to provide the 280 watts considering the efficiency given of 80%  That would be the 2nd step.
You work from back from the load (the machine).  It is the load that is the ultimate factor that determines the current.

anonymous · Answer

Ok, the apparent power (S) is:
S*0.98 = 350 => S = 357.1428VA

radar · Answer

Now look at the DC part.|dw:1404953749478:dw|