Animated Solution for Physics - Current Electricity: The voltage across the $10\Omega$ resistor in the given circuit is $x$ volt. The value of $x$ to the nearest integer is .......

Welcome, future engineers and physicists, to another thrilling exploration of electrical circuits! Today, we are tackling a classic problem from the JEE Main 2021 exam. At first glance, circuit diagrams can look like a confusing maze of lines and squiggles. But fear not! By applying a few fundamental principles, we can untangle this maze and reveal the elegant logic hidden within.

Imagine you are a tiny electron leaving the positive terminal of our $170\text{ V}$ battery. Your journey begins as you travel along the wire. The first obstacle you encounter is the $10\Omega$ resistor. Every single electron must pass through this resistor; there is no alternate route. This tells us that the $10\Omega$ resistor is on the main line of our circuit.

However, right after passing through this $10\Omega$ resistor, you arrive at a crossroads—a junction. Here, the path splits into two distinct branches. One branch contains a $50\Omega$ resistor, and the other contains a $20\Omega$ resistor. As an electron, you must choose one of these paths. After navigating through either the $50\Omega$ or the $20\Omega$ resistor, both paths merge back together before returning to the negative terminal of the battery.

This splitting and recombining of the current path is the hallmark of a parallel connection. Because both the $50\Omega$ and $20\Omega$ resistors are connected across the exact same two nodes, they share the exact same potential difference. Identifying this parallel structure is our first major victory in solving this problem!

Now that we've identified the parallel block, our strategy is to simplify the circuit. We want to replace this complex parallel combination with a single, equivalent resistor. Think of it like replacing two narrow toll booths with one wider toll booth that allows the same flow of traffic.

The formula for the equivalent resistance $R_p$ of two resistors in parallel is a beautiful piece of algebraic symmetry:

This formula is incredibly handy because it allows us to calculate the equivalent resistance directly without having to deal with the reciprocals until the very end. Let's substitute our specific values into this master equation. We have $R_1=50\Omega$ and $R_2=20\Omega$.

Now, let's perform the arithmetic. In the numerator, $50$ multiplied by $20$ gives us $1000$. In the denominator, $50$ plus $20$ is $70$.

By canceling out a factor of $10$ from both the numerator and the denominator, we simplify this to:

A crucial tip for competitive exams: Resist the urge to convert this fraction into a decimal! A decimal like $14.28\Omega$ is messy, prone to rounding errors, and difficult to manipulate in further calculations. Keeping it as the pristine fraction $\frac{100}{7}\Omega$ will make our lives much easier in the next steps.

With our parallel block simplified, let's visualize the new, equivalent circuit. The complex web has vanished. In its place, we have a simple series circuit consisting of the original $10\Omega$ resistor and our new $\frac{100}{7}\Omega$ equivalent resistor. This entire series combination is powered by the $170\text{ V}$ battery.

Our ultimate goal is to find the voltage drop across the $10\Omega$ resistor. In a series circuit, the total voltage from the battery is divided among the resistors in proportion to their resistance. The larger the resistance, the larger the share of the total voltage it claims.

To find this specific voltage drop, we can deploy one of the most powerful tools in circuit analysis: the Voltage Division Rule. This rule states that the voltage $V_x$ across a specific resistor $R_x$ in a series circuit is equal to the total voltage $V_{total}$ multiplied by the ratio of that resistance to the total equivalent resistance of the series circuit.

Let's adapt this formula for our specific situation. We want the voltage across the $10\Omega$ resistor, which we'll call $V_{10}$. The total voltage is $170\text{ V}$. The resistance we are interested in is $10\Omega$. The total resistance of our series circuit is the sum of the $10\Omega$ resistor and our equivalent parallel resistance, $\frac{100}{7}\Omega$.

We have successfully translated the physical circuit into a pure mathematical expression. Now, it's time for the final execution. Let's focus on simplifying the denominator first. We need to add $10$ and $\frac{100}{7}$.

To add a whole number and a fraction, we need a common denominator. We can rewrite $10$ as $\frac{70}{7}$.

Now, let's substitute this simplified denominator back into our main voltage division equation:

Take a moment to appreciate the beauty of this expression. Do you see what's about to happen? The numbers were not chosen at random; they were meticulously crafted by the exam setters to reward students who keep their values as fractions!

The $170$ in the main numerator and the $170$ in the denominator will perfectly cancel each other out. The $7$ in the denominator of the denominator will flip up to the numerator.

After the satisfying cancellation of the $170$s, we are left with a beautifully simple multiplication:

And there we have it! The voltage across the $10\Omega$ resistor is exactly $70\text{ V}$. The problem asks for the value of $x$, so our final answer is simply $70$.

By systematically breaking down the circuit—first identifying the parallel components, simplifying them, and then applying the voltage division rule—we transformed a potentially intimidating problem into a smooth, logical sequence of steps. Keep practicing these fundamental techniques, and you'll be mastering complex circuits in no time!