Animated Solution for Physics - Current Electricity: In order to measure the internal resistance $r_1$ of a cell of emf $E$, a meter bridge of wire resistance $R_0=50\Omega$, a resistance $R_0/2$, another cell of emf $E/2$ (internal resistance $r$) and a galvanometer $G$ are used in a circuit, as shown in the figure. If the null point is found at $l=72\text{cm}$, then the value of $r_1=\_\_\_\_\_\Omega$.

Imagine you are an electrical detective, and this circuit is your crime scene. At first glance, it looks like a tangled web of resistors and batteries. But the moment you spot the word "null point," everything changes. The null point is the ultimate clue—it tells us that a specific part of the circuit is perfectly balanced, like a seesaw resting in perfect equilibrium. Let's embark on a thrilling journey to decode this meter bridge and uncover the hidden internal resistance, $r_1$.

Before we dive into the math, we must understand the landscape. Look closely at the diagram. We have a main outer loop powered by a primary cell of EMF $E$. This loop contains the unknown internal resistance $r_1$, a known resistor $R_0/2$, and the long meter bridge wire of resistance $R_0$.

Then, we have a middle branch connected between a specific node (let's call it Node C, located between $r_1$ and $R_0/2$) and a movable jockey on the wire. This branch contains a smaller cell of EMF $E/2$, its internal resistance $r$, and a sensitive galvanometer $G$.

The problem states that at a length $l=72\text{ cm}$, the galvanometer shows zero deflection. This is our golden ticket. When the galvanometer reads zero, it means absolutely no current is flowing through that middle branch.

Why does the current stop? It stops because the electrical pressure (potential) pushing from one side is exactly matched by the pressure pushing from the other. Specifically, the potential difference between Node C and the jockey point J on the wire must perfectly balance the EMF of the cell in that branch.

Mathematically, this means:

Because no current flows into the middle branch, the main current $i$ generated by the primary cell $E$ flows entirely and uninterrupted through the outer loop.

Let's find this main current. The total resistance of the outer loop is the sum of all its components in series:

Using Ohm's Law, the main current is simply the total EMF divided by this total resistance:

Now comes the most crucial step: tracing the potential drop from Node C to the jockey point J along the path of the main current.

Imagine you are an electron traveling from Node C. First, you must pass through the resistor $R_0/2$. Then, you reach the right end of the meter bridge wire (let's call it point B). From point B, you travel along the wire to reach the jockey point J.

We know the jockey is at $72\text{ cm}$ from the left end. Since a standard meter bridge wire is $100\text{ cm}$ long, the distance from the right end B to the jockey J is $100-72=28\text{ cm}$.

Because the wire has uniform resistivity, its resistance is directly proportional to its length. Therefore, the resistance of this $28\text{ cm}$ segment is:

The total resistance you encounter on your journey from C to J is the sum of these two resistances:

The potential drop across this path is simply the main current $i$ multiplied by this total resistance:

We now have two expressions for the potential difference $V_C-V_J$. Let's equate them to form our master equation:

Now, substitute the expression for the main current $i$ that we derived earlier:

Look at how beautifully the physics aligns! The unknown EMF $E$ appears on both sides of the equation. It doesn't matter if the primary battery is 10 Volts or 100 Volts; the balance point depends purely on the ratio of the resistances. We can confidently cancel $E$ from both sides.

With $E$ gone, we are left with a clean algebraic equation:

Let's cross-multiply to solve for $r_1$:

Subtracting $1.5R_0$ from both sides, we isolate our target:

We have successfully expressed the unknown internal resistance in terms of the known wire resistance. The problem states that $R_0=50\Omega$. Let's plug it in for the final strike:

And there we have it! By carefully tracing the path of the current and understanding the profound implications of a null point, we have deduced that the internal resistance of the primary cell is exactly $3\Omega$. This problem is a masterpiece of circuit analysis, reminding us that complex networks can always be unraveled by patiently applying fundamental laws.