In this problem, we solve the problem that is a follow up on Example 21.2 from the book, where CD player consumption was analyzed in terms of the current. We calculate the period of time if the flow of $7.5 times 10^{18}$ passes trough the disk drive. We first remember the definition of the current:

$$
I=frac{Delta Q}{Delta t}
$$

expressing time interval from the previous relationship we have:

$$
Delta t=frac{Delta Q}{I}
$$

From the $textbf{Example 21.2.}$ we can take and assume that current passing is 0.22 A, but we need to calculate the total charge assumed for this amount of electrons and this magnitude of the current. We know that the ratio of total charge per one electron charge will give the number of electrons:

$$
N=frac{Delta Q}{e}
$$

So the total charge will be:

$$
Delta Q = N e
$$

Putting in the numbers we have:

$$
Delta Q = 7.5 times 10^{18} cdot 1.6times 10^{-19}
$$

We have a result of:

$$
Delta Q = 1.2 mathrm{~C}
$$

Now we have all the information’s to calculate the time interval, so getting back to the relationship we have written at the start we have:

$$
Delta t=frac{Delta Q}{I} = frac{1.2 mathrm{~C}}{0.22 mathrm{~A}}
$$

$$
Delta t=frac{Delta Q}{I} = boxed{color{#c34632}5.45 mathrm{~s}}
$$

$$
Delta t=5.45 mathrm{~s}
$$

In this problem, we calculate the current for the given charge and the time interval. We calculate the current with the known relation:

$$
I=frac{Delta Q}{Delta t}
$$

We have 3600 C and 17 min. Since we need to work with seconds, we convert minutes to the second. So 17 minutes will be 1020 seconds ( 1 min = 60 s). So putting in the numbers for our current relation we have:

$$
I=frac{3600 mathrm{~C}}{1020 mathrm{~s}}
$$

$$
boxed{color{#c34632}I=3.53 mathrm{~A}}
$$

$$
I=3.53 mathrm{~A}
$$

Here we calculate how much time is required for 14 C of charge to pass through the lightbulb. From our current expression:

$$
I=frac{Delta Q}{Delta t}
$$

expressing the time we have:

$$
Delta t=frac{Delta Q}{I}
$$

We put in the numbers:

$$
Delta t=frac{14 mathrm{~C}}{0.18 mathrm{~A}}
$$

We have the result of:

$$
boxed{color{#c34632}Delta t=77.78 mathrm{~s}}
$$

$$
Delta t=77.78 mathrm{~s}
$$