Animated Solution for Physics - Electrostatics Potential and Capacitance: Consider the combination of two capacitors $C_1$ and $C_2$, with $C_2>C_1$, when connected in parallel, the equivalent capacitance is $15/4$ time the equivalent capacitance of the same connected in series. Calculate the ratio of capacitors $\frac{C_2}{C_1}$.

Imagine you are given two capacitors, $C_1$ and $C_2$. When you connect them in parallel, their equivalent capacitance is simply their sum:

When you connect them in series, the equivalent capacitance is the product over the sum:

The problem states a very specific condition: the parallel capacitance is $\frac{15}{4}$ times the series capacitance.

Let's translate this condition into a mathematical equation:

To solve this, we need to clear the fractions. By cross-multiplying, we bring the $(C_1+C_2)$ term to the left side:

Now, let's expand the perfect square on the left side:

Distributing the $4$ gives us:

Bringing all terms to one side to form a homogeneous quadratic equation:

We are asked to find the ratio $\frac{C_2}{C_1}$. Let's define a new variable $x=\frac{C_2}{C_1}$. To introduce $x$ into our equation, we divide the entire equation by $C_1^2$:

This simplifies to a standard quadratic equation in terms of $x$:

Checking the Discriminant

Let's apply the quadratic formula to solve for $x$:

Wait a minute! The term inside the square root (the discriminant) is negative. This means the roots of this equation are imaginary numbers.

Since capacitance is a physical property that must be a real, positive number, it is physically impossible for two real capacitors to satisfy the condition given in the problem. The question is mathematically flawed! In competitive exams like JEE, such questions are usually dropped, and bonus marks are awarded to all students.

The AM-GM Inequality

Could we have spotted this error without doing all the algebra? Yes!

Let's look at the ratio of parallel to series capacitance for any two positive capacitors:

By the AM-GM (Arithmetic Mean - Geometric Mean) inequality, the sum of a positive number and its reciprocal is always greater than or equal to $2$. Therefore, $\frac{C_1}{C_2}+\frac{C_2}{C_1}\ge 2$.

Adding $2$ to both sides, we get:

This is a universal truth for any two real, positive capacitors: their parallel capacitance is always at least 4 times their series capacitance.

The problem claimed that $C_p=\frac{15}{4}{C}_s$, which means $\frac{C_p}{C_s}=3.75$. Since $3.75$ is strictly less than $4$, the condition is fundamentally impossible. Knowing this trick saves you precious time during the exam!