The length of a wire of a potentiometer is 100 cm and the emf of its standard cell is E volt. It is

Q: Why doesn't the $0.5\ \Omega$ internal resistance affect the final answer?

At the balance point (null point), the galvanometer shows zero deflection, meaning no current flows through the secondary circuit. Since voltage drop across a resistor is $V = IR$, and $I = 0$, the voltage drop across the internal resistance is zero. The potentiometer measures the true emf.

Q: What exactly is a potential gradient?

The potential gradient is the potential drop per unit length of the potentiometer wire. It tells you how many volts are dropped for every centimeter you move along the wire. It is calculated as the total voltage across the wire divided by its total length ($E/L$).

Q: Why must the positive terminals of both batteries be connected to the same point A?

If they were connected with opposite polarities, their potentials would add up rather than oppose each other. To find a balance point where the potential difference is zero, the potentials must oppose each other, which requires connecting the same polarities (usually positive) to the starting point.

Q: What if the primary battery had some internal resistance?

If the primary battery had internal resistance, the total voltage drop across the potentiometer wire would be less than $E$. You would first need to calculate the actual voltage across the wire using Ohm's law before finding the potential gradient.

Q: Can a potentiometer measure an emf greater than the primary battery's emf ($E$)?

No. The maximum potential difference that can be balanced is the total potential drop across the entire length of the potentiometer wire, which is at most $E$. If the unknown emf is greater than $E$, a balance point will never be found on the wire.

Question

The length of a wire of a potentiometer is $100	ext{ cm}$ and the emf of its standard cell is $E$ volt. It is employed to measure the emf of a battery whose internal resistance is $0.5\ \Omega$. If the balance point is obtained at $l = 30	ext{ cm}$ from the positive end, the emf of the battery is

Options:

(A) $\frac{30E}{100.5}$

(B) $\frac{30E}{100 - 0.5}$

(C) $\frac{30(E - 0.5I)}{100}$, where $I$ is the current in the potentiometer wire

(D) $\frac{30E}{100}$

Sigma Percentile · Accepted Answer

### The Magic of the Potentiometer

Imagine you have a magical measuring tape that doesn't measure length, but measures electrical potential. That's exactly what a potentiometer is! In this problem, we are tasked with finding the electromotive force (emf) of a battery using a standard potentiometer setup.

Let's break down the physical reality of the circuit. We have a primary circuit consisting of a standard cell with an emf $E$ connected across a long, uniform wire of length $L = 100	ext{ cm}$. This primary circuit establishes a steady, uniform voltage drop across the entire length of the wire.

### The Principle of Potential Gradient

The core principle of a potentiometer is beautifully simple: if a wire has a uniform cross-section and a constant current flows through it, the potential drop across any segment of the wire is directly proportional to the length of that segment.

Mathematically, we define a **potential gradient** ($k$), which is the potential drop per unit length of the wire. 
$$k = \frac{E}{L}$$

### The Null Point and Internal Resistance

Now, we connect our unknown battery (let's call its emf $V$) in the secondary circuit, along with a galvanometer. We slide the jockey along the wire until the galvanometer shows **zero deflection**. This magical spot is called the **null point** or balance point.

Here is the crucial catch that often trips students up: the problem states that the unknown battery has an internal resistance of $0.5\ \Omega$. Does this affect our reading?

**Absolutely not!**

Why? Because at the null point, the galvanometer shows zero deflection, meaning **no current** is flowing through the secondary circuit. According to Ohm's law, the voltage drop across a resistor is $V = IR$. If the current $I$ is zero, the voltage drop across the internal resistance is also zero. The potentiometer measures the true, open-circuit emf of the battery, completely unaffected by its internal resistance.

### The Final Calculation

Since no current flows, the emf of the unknown battery $V$ is exactly balanced by the potential drop across the balancing length $l$ of the potentiometer wire.

$$V = k 	imes l$$

Substituting our potential gradient $k = \frac{E}{L}$:

$$V = \left(\frac{E}{L}ight) 	imes l$$

We are given the total length $L = 100	ext{ cm}$ and the balancing length $l = 30	ext{ cm}$. Plugging these values into our master equation:

$$V = \frac{E}{100} 	imes 30$$

$$V = \frac{30E}{100}$$

And there we have it! The elegance of the potentiometer lies in its ability to measure emf without drawing any current, rendering internal resistance completely irrelevant at the balance point.

Select Answer:

Visualized Solution

Visualizing the Potentiometer Setup

Principle of Potentiometer

The Null Point Condition

Formulating the Equation

Substituting the Values

Final Answer

The Sigma Insight: Measuring Instruments

The Magic of the Potentiometer

The Principle of Potential Gradient

The Null Point and Internal Resistance

The Final Calculation

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