A potentiometer wire AB having length L and resistance 12r is joined to a cell D of EMF epsilon and

Q: Why don't we consider the internal resistance $3r$ of cell C in our calculations?

At the null point, the galvanometer shows zero deflection, meaning absolutely no current flows through the secondary circuit. Since current is zero, the potential drop across the internal resistance $3r$ ($V = iR$) is also zero. Therefore, the full EMF of cell C is balanced by the wire.

Q: What is the role of the primary cell D?

Cell D acts as the driver cell. Its job is to provide a constant, steady current through the potentiometer wire $AB$. This steady current establishes a uniform potential gradient along the wire, which is essential for measuring unknown EMFs.

Q: Why does the potential gradient matter so much?

The potential gradient ($k$) is the heart of the potentiometer. It tells us exactly how much voltage drops for every unit length of the wire. Once we know $k$, we can map any balancing length directly to a voltage value using the simple relation $V = kx$.

Q: What if the EMF of cell C was greater than the EMF of cell D?

If the EMF of cell C was greater than the total potential drop across the entire wire $AB$ (which is slightly less than the EMF of cell D due to its internal resistance), a balance point would never be found on the wire. The galvanometer would show deflection in the same direction regardless of where you touch the jockey.

Q: Does the resistance of the galvanometer affect the position of the null point?

No, it does not. The null point is defined by the condition of zero current in the galvanometer branch. Since no current flows, the resistance of the galvanometer does not cause any voltage drop and thus has zero impact on the balancing length.

Question

A potentiometer wire $AB$ having length $L$ and resistance $12r$ is joined to a cell $D$ of EMF $\varepsilon$ and internal resistance $r$. A cell $C$ having emf $\frac{\varepsilon}{2}$ and internal resistance $3r$ is connected. The length $AJ$ at which the galvanometer as shown in figure shows no deflection is

Options:

(A) $\frac{5}{12}L$

(B) $\frac{11}{12}L$

(C) $\frac{13}{24}L$

(D) $\frac{11}{24}L$

Sigma Percentile · Accepted Answer

This is a brilliant problem based on the potentiometer, a classic and high-yield topic for JEE. Let's break down the setup and understand the physics behind it.

### Analyzing the Setup

We have two distinct circuits interacting with each other. The upper one is the **primary circuit**, containing a cell $D$ with EMF $\varepsilon$ and internal resistance $r$. It is connected across a uniform potentiometer wire $AB$ of length $L$ and resistance $12r$.

Below it is the **secondary circuit**, containing a cell $C$ with EMF $\frac{\varepsilon}{2}$ and internal resistance $3r$, connected to a sensitive galvanometer.

Now, recall the core principle of a potentiometer. At the null point, when the galvanometer shows zero deflection, the potential drop across the balancing length $AJ$ of the wire is exactly equal to the EMF of the secondary cell.

Look closely at the secondary circuit: the internal resistance $3r$ of cell $C$ plays absolutely no role here! Why? Because at the null point, no current flows through the secondary circuit, meaning there is zero potential drop across its internal resistance.

### The Master Equation

Let's move forward by calculating the current $i$ in the primary circuit. The total resistance of the primary loop is the cell's internal resistance $r$ plus the wire's resistance $12r$, which gives $13r$. By Ohm's law, the current $i$ is simply:

$i = \frac{\varepsilon}{13r}$

Next, we need to find the **potential gradient** ($k$), which is the potential drop per unit length of the wire. The total potential drop across wire $AB$ is the current $i$ times its resistance $12r$.

$V_{AB} = i 	imes 12r = \left(\frac{\varepsilon}{13r}ight) 	imes 12r = \frac{12\varepsilon}{13}$

Dividing this total voltage by the length $L$ gives us the potential gradient $k$:

$k = \frac{12\varepsilon}{13L}$

### Final Calculation

Now, let's set up our balancing equation. The potential drop across the unknown length $x$ is $k 	imes x$. This must be equal to the EMF of cell $C$, which is $\frac{\varepsilon}{2}$. Let's substitute the value of $k$ into this equation:

$\frac{12\varepsilon}{13L} 	imes x = \frac{\varepsilon}{2}$

Don't make a silly mistake here; carefully solve for $x$. The $\varepsilon$ terms cancel out beautifully on both sides. By cross-multiplying, we get:

$x = \frac{\varepsilon}{2} 	imes \frac{13L}{12\varepsilon}$

$x = \frac{13L}{24}$

We have our final answer! The balancing length $AJ$ is **$\frac{13}{24}L$**, which perfectly matches option (c).

As a thought experiment, visualize what would happen if we connected a shunt resistance in parallel with cell $C$. Current would then flow through cell $C$, and we would have to balance its *terminal voltage* instead of its EMF, causing the null point to shift to a smaller length. Make sure to revise that concept as well!

Select Answer:

Visualized Solution

\text{Circuit Analysis}

\text{Potentiometer Principle}

\text{Primary Circuit Current}

\text{Potential Gradient}

\text{Balancing Equation}

\text{Solving for } x

\text{Final Answer}

\text{The Way Forward}

The Sigma Insight: Electrical Instruments

Analyzing the Setup

The Master Equation

Final Calculation

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