Two equal capacitors are first connected in series and then in parallel. The ratio of the equivalent

Q: Why does the capacitance decrease in a series combination?

When you connect capacitors in series, you are effectively increasing the total distance between the outermost plates. Since capacitance is inversely proportional to the distance between plates ($C = \frac{\epsilon_0 A}{d}$), an increased effective distance results in a lower overall capacitance.

Q: Why does the capacitance increase in a parallel combination?

Connecting capacitors in parallel is like connecting their plates together to form larger plates. Since capacitance is directly proportional to the plate area ($C = \frac{\epsilon_0 A}{d}$), an increased effective area results in a higher overall capacitance.

Q: Are the formulas for capacitors the same as for resistors?

No, they are exactly the opposite! For resistors in series, you add them directly ($R_s = R_1 + R_2$). For capacitors in series, you add their reciprocals ($\frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2}$). The reverse is true for parallel combinations.

Q: What if the two capacitors had different values, say $C_1$ and $C_2$?

You would use the same fundamental formulas. $C_s = \frac{C_1 C_2}{C_1 + C_2}$ and $C_p = C_1 + C_2$. The ratio would then be $\frac{C_s}{C_p} = \frac{C_1 C_2}{(C_1 + C_2)^2}$.

Q: Is there a quick trick if there are $n$ identical capacitors?

Yes! For $n$ identical capacitors of capacitance $C$, the series equivalent is $\frac{C}{n}$ and the parallel equivalent is $nC$. Their ratio $\frac{C_s}{C_p}$ is always $\frac{1}{n^2}$.

Question

Two equal capacitors are first connected in series and then in parallel. The ratio of the equivalent capacities in the two cases will be

Options:

(A) 1 : 2

(B) 2 : 1

(C) 4 : 1

(D) 1 : 4

Sigma Percentile · Accepted Answer

This is a classic and fundamental problem that tests your understanding of how capacitors behave when connected in different configurations. Let's break it down step-by-step and see the beautiful symmetry in the formulas.

### Analyzing the Setup

Imagine you have two identical capacitors. Let's assume the capacitance of each capacitor is $C$. We are going to connect them in two different ways: first in series, and then in parallel. Our goal is to find the ratio of their equivalent capacitances in these two scenarios.

### The Series Combination

First, let's connect them end-to-end, which is a series combination. Do you remember the formula for the equivalent capacitance in series? The reciprocal of the equivalent capacitance is the sum of the reciprocals of the individual capacitances.

Mathematically, this is written as:

$$\frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2}$$

Since both of our capacitors have the same capacitance $C$, we can substitute this into our formula:

$$\frac{1}{C_s} = \frac{1}{C} + \frac{1}{C} = \frac{2}{C}$$

To find $C_s$, we simply take the reciprocal of both sides:

$$C_s = \frac{C}{2}$$

Notice that connecting capacitors in series actually *decreases* the overall capacitance. It's as if you are increasing the distance between the plates of a single equivalent capacitor.

### The Parallel Combination

Now, let's look at the second case. We connect the same two capacitors in parallel, meaning their top plates are connected together, and their bottom plates are connected together. In a parallel configuration, the equivalent capacitance is simply the algebraic sum of the individual capacitances.

$$C_p = C_1 + C_2$$

Substituting our values, we get:

$$C_p = C + C = 2C$$

Connecting capacitors in parallel *increases* the overall capacitance. You can think of this as effectively increasing the total plate area.

### Final Calculation

Finally, we need to find the ratio of the equivalent capacities in the two cases, which is $C_s$ divided by $C_p$.

$$	ext{Ratio} = \frac{C_s}{C_p}$$

Substituting the expressions we found:

$$	ext{Ratio} = \frac{\frac{C}{2}}{2C}$$

When we simplify this fraction, the $C$ terms cancel out beautifully:

$$	ext{Ratio} = \frac{1}{4}$$

So, the ratio of the series equivalent capacitance to the parallel equivalent capacitance is **$1 : 4$**.

### The Way Forward

Here is a powerful shortcut for your competitive exams. What if you had $n$ identical capacitors instead of just two?

If you connect $n$ identical capacitors of capacitance $C$ in series, the equivalent capacitance is $C_s = \frac{C}{n}$. 
If you connect them in parallel, the equivalent capacitance is $C_p = nC$.

Therefore, the ratio $\frac{C_s}{C_p}$ will always be $\frac{1}{n^2}$. In our problem, $n=2$, so the ratio was $\frac{1}{2^2} = \frac{1}{4}$. Keep this generalization in mind to save precious seconds during the exam!

Animated Solution for Physics - Electrostatics: Two equal capacitors are first connected in series and then in parallel. The ratio of the equivalent capacities in the two cases will be

Select Answer:

Visualized Solution

Initial Setup

Series Combination Formula

Equivalent Series Capacitance

Parallel Combination Formula

Equivalent Parallel Capacitance

Ratio of Capacitances

Generalization

The Sigma Insight: Combination of Capacitors

Analyzing the Setup

The Series Combination

The Parallel Combination

Final Calculation

The Way Forward

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