Animated Solution for Physics - Electrostatics: Seven capacitors, each of capacitance $2\mu \text{F}$ are to be connected in a configuration to obtain an effective capacitance of $\left(\frac{6}{13}\right)\mu \text{F}$. Which of the combinations shown in figures below will achieve the desired value?

The problem asks us to find the correct configuration of 7 identical capacitors, each with a capacitance of $2\mu \text{F}$, to achieve an equivalent capacitance of $\frac{6}{13}\mu \text{F}$.

The first thing to notice is the target value: $\frac{6}{13}\mu \text{F}$. Since $\frac{6}{13}$ is less than 1, and our individual capacitors are $2\mu \text{F}$, the equivalent capacitance is significantly smaller than a single unit.

In circuit theory, connecting capacitors in series reduces the overall capacitance, while connecting them in parallel increases it. The small target value strongly suggests that our circuit must have a dominant series component. If we look at the options provided, they all follow a specific topology: a single parallel bank of capacitors connected in series with a string of individual capacitors.

Instead of calculating the equivalent capacitance for each option one by one, we can use a powerful algebraic shortcut. Let's assume a general configuration based on the options: - Let $m$ be the number of capacitors connected in parallel. - Let $n$ be the number of individual capacitors connected in series with this parallel group.

We know the total number of capacitors is 7, so:

Now, let's write the expression for the equivalent capacitance ($C_{eq}$) of this general setup. The $m$ capacitors in parallel will have an equivalent capacitance of:

This parallel group ($2m$) is in series with $n$ individual capacitors (each $2\mu \text{F}$). Using the series combination formula, the reciprocal of the total equivalent capacitance is the sum of the reciprocals:

We are given that the final equivalent capacitance $C_{eq}$ must be $\frac{6}{13}\mu \text{F}$. Let's substitute this into our equation:

To make it easier to compare, let's take a common denominator on the right side:

This is where the magic happens. By simply comparing the denominators on both sides, we can find $m$:

If $m=3$, and we know the total number of capacitors is 7, then $n$ must be:

We have our values, but we must verify them by checking the numerators. Does $1+mn$ equal 13? Let's plug in our values:

The math aligns perfectly! Our winning configuration requires 3 capacitors in parallel, connected in series with 4 individual capacitors.

Looking at the given options, Option (c) perfectly matches this exact arrangement. We have successfully cracked the circuit without having to test every single option!