Answer:
<em>Hooking up the bulb to the battery in a series arrangement will draw the least amount of current.</em>
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.
<em>The least amount of energy will be drawn when the arrangement provides the maximum resistance possible.</em>
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
<em>I = V/2R</em>
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
<em>I = 2V/R</em>
<em></em>
<em>we see that arranging the bulbs in parallel draws 4 times the current compared to arranging the bulb in series</em>
From the above, <em>we see that arranging the bulbs in series provides the maximum resistance, which means a lesser amount of current is drawn from the battery</em>
