Question Video: Combining Capacitors in Series | Nagwa Question Video: Combining Capacitors in Series | Nagwa

# Question Video: Combining Capacitors in Series Physics • Third Year of Secondary School

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Two capacitors, πΆβ and πΆβ, are connected in series, where πΆβ > πΆβ. Which of the following statements correctly relates the total capacitance, πΆ total, to πΆβ and πΆβ? [A] πΆ_(total) = πΆβ + πΆβ [B] πΆ_(total) = πΆβπΆβ [C] πΆ_(total) = (πΆβ + πΆβ)Β² [D] πΆ_(total) < πΆβ < πΆβ [E] πΆβ < πΆ_(total) < πΆβ

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### Video Transcript

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

In this question, we are told that there are two capacitors, πΆ one and πΆ 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 πΆ one plus one over πΆ two.

Now letβs 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 πΆ one by πΆ two over πΆ two and multiply one over πΆ two by πΆ one over πΆ one, we find that the right-hand side is equal to πΆ two over πΆ one πΆ two plus πΆ one over πΆ one πΆ two.

Adding the right-hand terms together, we find that one over πΆ total is equal to πΆ one plus πΆ two over πΆ one πΆ two. If we take the reciprocal of both sides, we find that πΆ total is equal to πΆ one πΆ two over πΆ one plus πΆ 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 πΆ total is less than πΆ two or greater than πΆ two.

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

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

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

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