### Video Transcript

Two capacitors, πΆ one and πΆ two,
are connected in series, where πΆ one is greater than πΆ two. Which of the following statements
correctly relates the total capacitance, πΆ total, to πΆ one and πΆ two? (A) πΆ total equals πΆ one plus πΆ
two. (B) πΆ total equals πΆ one πΆ
two. (C) πΆ total equals πΆ one plus πΆ
two squared. (D) πΆ total less than πΆ two less
than πΆ one. (E) πΆ two less than πΆ total less
than πΆ one.

In this question, we are told that
there are two capacitors, πΆ one and πΆ two, that are connected in series. We are also told that the
capacitance of capacitor one is larger than the capacitance of capacitor two. We are asked to choose the
statement that relates the total capacitance of the pair of them to their individual
capacitances.

We can recall that when capacitors
are connected in series, the total capacitance of that series of capacitors is given
by this equation. One over the total capacitance is
equal to one over the capacitance of the first capacitor plus one over the
capacitance of the second capacitor and so on, until we reach the last of the
capacitors in the series. Since we are dealing with two
capacitors connected in series, this equation simplifies to one over the total
capacitance is equal to one over πΆ one plus one over πΆ two.

Now letβs get an expression for the
total capacitance. We can simplify this equation by
considering the right-hand side first. We want to have both fractions with
the same denominator so that we can add them together. If we multiply one over πΆ one by
πΆ two over πΆ two and multiply one over πΆ two by πΆ one over πΆ one, we find that
the right-hand side is equal to πΆ two over πΆ one πΆ two plus πΆ one over πΆ one πΆ
two.

Adding the right-hand terms
together, we find that one over πΆ total is equal to πΆ one plus πΆ two over πΆ one
πΆ two. If we take the reciprocal of both
sides, we find that πΆ total is equal to πΆ one πΆ two over πΆ one plus πΆ two. We have now arrived at an equation
that links the capacitances of two capacitors in series to their total
capacitance.

We can see that options (A), (B),
and (C) do not match the equation for the total capacitance that we have just
calculated, so these three options are incorrect. So, this leaves us with options (D)
and (E), which compare the sizes of the capacitances. The difference between the two
options is working out whether πΆ total is less than πΆ two or greater than πΆ
two.

If we can get πΆ two to have the
same denominator as πΆ total, then we can compare the numerators for both to see
which one is larger. We can do this by multiplying πΆ
two by πΆ one plus πΆ two over πΆ one plus πΆ two to give us πΆ one πΆ two plus πΆ
two squared over πΆ one plus πΆ two.

We can now see that the numerator
for πΆ two is πΆ one πΆ two plus πΆ two squared, while the numerator for πΆ total is
just πΆ one πΆ two. This means that πΆ two is greater
than πΆ total, meaning that option (E) is incorrect. This leaves us with option (D). We have just found out that πΆ two
is greater than πΆ total, but we are also told in the question that πΆ one is larger
than πΆ two. So, πΆ one is also larger than πΆ
total.

Therefore, for two capacitors πΆ
one and πΆ two connected in series, πΆ one is the largest, πΆ two is smaller than πΆ
one, and πΆ total is the smallest. This corresponds with option (D),
πΆ total less than πΆ two less than πΆ one. So, this is the correct answer.