Animated Solution for Physics - Electrostatics: Seven capacitors each of capacitance $2\mu \text{F}$ are connected in a configuration to obtain an effective capacitance $\frac{10}{11}\mu \text{F}$. Which of the following combination will achieve the desired result ?

Imagine you are given a handful of identical electronic components and asked to build a specific, seemingly random value out of them. This is exactly what this problem asks us to do. We have $7$ identical capacitors, each with a capacitance of $C=2\mu \text{F}$. Our goal is to arrange them in a specific configuration to achieve an equivalent capacitance of exactly $C_{\text{eq}}=\frac{10}{11}\mu \text{F}$.

At first glance, trying out every possible combination of $7$ capacitors in series and parallel would be a nightmare. There are simply too many permutations! Instead of guessing blindly, we need to reverse-engineer the target value.

Let's take a very close look at the target fraction: $\frac{10}{11}$.

Does this structure remind you of any standard physics formula? Think about the formula for two capacitors connected in series. If we have two capacitors $C_1$ and $C_2$ in series, their equivalent capacitance is given by:

Now, let's rewrite our target fraction to match this structure. Notice that the denominator $11$ can be written as $10+1$. And the numerator $10$ can be written as $10\times 1$. Let's substitute this back:

This is a perfect match! It tells us a beautiful secret about the circuit's architecture: the final circuit is simply a series combination of two distinct groups of capacitors. One group must have an equivalent capacitance of $C_1=10\mu \text{F}$, and the other group must have an equivalent capacitance of $C_2=1\mu \text{F}$.

Now the problem is broken down into two much simpler sub-problems. We need to build a $10\mu \text{F}$ capacitor and a $1\mu \text{F}$ capacitor using only our basic $2\mu \text{F}$ building blocks.

Block 1: The $10\mu \text{F}$ Group How do we get a larger capacitance from smaller ones? We connect them in parallel! In a parallel combination, capacitances simply add up. Since each capacitor is $2\mu \text{F}$, we can achieve $10\mu \text{F}$ by connecting five of them in parallel:

This uses up $5$ of our $7$ available capacitors.

Block 2: The $1\mu \text{F}$ Group How do we get a smaller capacitance from larger ones? We connect them in series! When $n$ identical capacitors of capacitance $C$ are connected in series, their equivalent capacitance becomes $C/n$. To get $1\mu \text{F}$ from $2\mu \text{F}$ capacitors, we just need to connect two of them in series:

This uses exactly $2$ capacitors.

Let's tally up our components. We used $5$ capacitors for the first block and $2$ capacitors for the second block. That's exactly $5+2=7$ capacitors, which is exactly the number we were given!

The final arrangement is a group of $5$ capacitors in parallel, connected in series with a group of $2$ capacitors in series. This elegant reverse-engineering approach leads us directly to the correct configuration without any blind guessing.