Animated Solution for Physics - Current Electricity: The value of current in the $6\Omega$ resistance is

Have you ever looked at a complex circuit with multiple batteries and resistors and felt a wave of intimidation? I know that feeling. It looks like a tangled web of wires, and finding the current in just one specific branch seems like an impossible puzzle.

But what if I told you there is a master key? A single, elegant technique that can unlock almost any circuit problem you encounter in JEE Physics?

That master key is Nodal Analysis.

Today, we are going to use this incredible tool to find the current flowing through the $6\Omega$ resistor in our given circuit. Let's dive in and see the magic unfold!

The Reference Node

Imagine you are an explorer mapping out a new territory. The first thing you need is a base camp—a reference point from which all heights are measured.

In circuit analysis, this base camp is our reference node or ground. We typically choose the bottom-most wire connecting all the branches because it simplifies our math immensely.

Let's assign a potential of $0\text{ V}$ to this bottom wire.

Now, look at the top junction where all three branches meet. This is the busy intersection of our circuit. We don't know the voltage here yet, so let's call it $V$.

Our entire mission now boils down to finding the value of this single variable, $V$. Once we have $V$, the rest of the circuit will reveal its secrets to us.

Kirchhoff's Current Law

To find $V$, we need a rule that governs this junction. Enter Kirchhoff's Current Law (KCL).

KCL is beautifully simple: it states that the total current entering a junction must equal the total current leaving it. Or, even simpler, the algebraic sum of all currents leaving a junction is exactly zero.

Think of it like water flowing through pipes. Water can't magically appear or disappear at a joint; whatever flows in must flow out.

Let's assume that currents $I_1$, $I_2$, and $I_3$ are all leaving our top node $V$ and traveling down the three branches.

According to KCL, we can write:

Now, we need to express these currents in terms of our unknown voltage $V$. We do this using good old Ohm's Law, which tells us that current is the potential difference divided by resistance ($I=\frac{\Delta V}{R}$).

Let's look at the left branch. The current $I_1$ flows from node $V$, through the $20\Omega$ resistor, and into the positive terminal of the $140\text{ V}$ battery. The potential difference is $(V-140)$. So, the current is:

Next, let's examine the right branch. The current $I_2$ flows from node $V$, through the $5\Omega$ resistor, and into the positive terminal of the $90\text{ V}$ battery. The potential difference is $(V-90)$. So, the current is:

Finally, the middle branch. The current $I_3$ flows from node $V$, straight through the $6\Omega$ resistor, down to our $0\text{ V}$ ground. The potential difference is simply $(V-0)$. So, the current is:

Now, let's plug these expressions back into our KCL equation. We get:

Take a moment to appreciate this equation. We have transformed a complex physical circuit into a single, solvable algebraic equation. This is the true power of physics!

I know fractions can be annoying to deal with. So, let's get rid of them!

We need to find the Least Common Multiple (LCM) of our denominators: $20$, $5$, and $6$. A quick mental check tells us that the LCM is $60$.

Let's multiply the entire equation by $60$. This is a classic mathematical move to clear the board and make our calculation smooth and error-free.

Simplifying the terms, we get:

Now, it's just basic algebra. Let's expand the brackets carefully. Watch out for those minus signs!

Let's group the $V$ terms together and the constant numbers together:

Moving the constant to the other side:

Dividing both sides by $25$:

Boom! We have found the potential at our top junction. It is exactly $60\text{ V}$.

Now that we have our master key, $V=60\text{ V}$, we can find anything we want in this circuit.

The question asks for the current in the $6\Omega$ resistance, which is our middle branch. We already defined this current as $I_3$.

Substitute our hard-earned value of $V$:

And there is our final answer! The current flowing through the $6\Omega$ resistor is $10\text{ A}$.

Notice how a seemingly complex problem melted away into simple algebra just by applying Nodal Analysis systematically. This is why mastering this technique is an absolute game-changer for your JEE preparation. Keep practicing, stay curious, and never let a complex circuit intimidate you again!