Question 20.1: (a) What is the current involved when a truck battery sets i...

Solution for (a)
Entering the given values for charge and time into the definition of current gives

I=\frac{\Delta Q}{\Delta t}=\frac{720 C }{4.00 s }=180 C / s                (20.3)

= 180 A.

Discussion for (a)
This large value for current illustrates the fact that a large charge is moved in a small amount of time. The currents in these “starter motors” are fairly large because large frictional forces need to be overcome when setting something in motion.
Solution for (b)
Solving the relationship I = ΔQ / Δt for time Δt , and entering the known values for charge and current gives

\Delta t=\frac{\Delta Q}{I}=\frac{1.00 C }{0.300 \times 10^{-3} C / s }                 (20.4)

=3.33 \times 10^{3} s.

Discussion for (b)
This time is slightly less than an hour. The small current used by the hand-held calculator takes a much longer time to move a smaller charge than the large current of the truck starter. So why can we operate our calculators only seconds after turning them on? It’s because calculators require very little energy. Such small current and energy demands allow handheld calculators to operate from solar cells or to get many hours of use out of small batteries. Remember, calculators do not have moving parts in the same way that a truck engine has with cylinders and pistons, so the technology requires smaller currents.