Imagine that you want to make sure the battery for your string of lights will last as long as possible. A battery will last longer if it powers a circuit with low current. How could you hook up a battery and 2 light bulbs so the least amount of current flows through the battery

Hooking up the bulb to the battery in a series arrangement will draw the least amount of current.

Explanation:

In this case now, the bulb will serve as the load on the battery (resistance).

For the current to last longer, the least amount of energy must be drawn.

The least amount of energy will be drawn when the arrangement provides the maximum resistance possible.

Let us take the resistance of each bulb as 'R'

If we arrange the bulbs in series, then, the total resistance will be

Rt = R + R = 2R

at a EMF of V from the battery, current I through the battery will be

I = V/2R

If we arrange the bulbs in parallel, then , the total resistance will be

1/Rt = 1/R + 1/R

1/Rt = 2/R

therefore

Rt = R/2

at an EMF of V from the battery, the current I that will be drawn through the battery will be

I = 2V/R

we see that arranging the bulbs in parallel draws 4 times the current compared to arranging the bulb in series

From the above, we see that arranging the bulbs in series provides the maximum resistance, which means a lesser amount of current is drawn from the battery

0.11 m/s and 0.28 m/s should be correct