Explanation:
It is given that two loads have 0.75 Ampere current each. And, they contain 2500 milli ampere per hour Ni-Cd battery.
As both the loads are connected in parallel. Hence, total current will be calculated as follows.
I = [tex]I_{1} + I_{2}[/tex]
= 0.75 A + 0.75 A
= 1.5 A
= [tex]1.5 A \times \frac{1000 mA}{1 A}[/tex]
= 1500 mA
Relation between time and capacity of battery is as follows.
Capacity = Current × time (in hour)
therefore, time = [tex]\frac{Capacity}{Current}[/tex]
= [tex]\frac{2500 mA. h}{1500 A}[/tex]
= 1.667 hr
Thus, we can conclude that the battery provide power to the load up to 1.667 hours.
