In a potentiometer experiment, it is found that no current passes through the galvanometer when the

Q: Why does the balancing length decrease when the cell is shunted?

When the cell is shunted, it supplies current to the external resistor. This causes a potential drop across its internal resistance, making the terminal voltage less than the EMF. A smaller voltage requires a shorter balancing length on the potentiometer wire.

Q: What happens if the shunt resistance is removed completely?

Removing the shunt resistance is equivalent to an open circuit (infinite resistance). The potentiometer will again measure the true EMF of the cell, and the balancing length will return to its original, longer value.

Q: Does the potential gradient of the potentiometer wire change when we shunt the cell?

No, the potential gradient depends only on the primary circuit (the driver battery and the potentiometer wire itself). Shunting the cell in the secondary circuit does not affect the primary circuit at the null point.

Q: Why is no current flowing through the galvanometer at the null point?

At the null point, the potential difference across the balancing length of the wire is exactly equal and opposite to the potential difference provided by the cell circuit. Since the net potential difference is zero, no current flows through the galvanometer branch.

Q: Can we measure the internal resistance of a cell using a voltmeter instead of a potentiometer?

A voltmeter draws a small amount of current to operate, which means it measures the terminal voltage, not the true EMF. A potentiometer, at the null point, draws zero current from the cell, allowing it to measure the true EMF accurately, making it superior for finding internal resistance.

Question

In a potentiometer experiment, it is found that no current passes through the galvanometer when the terminals of the cell are connected across 52 cm of the potentiometer wire. If the cell is shunted by a resistance of $5 \Omega$, a balance is found when the cell is connected across 40 cm of the wire. Find the internal resistance of the cell.

Options:

(A) $1 \Omega$

(B) $1.5 \Omega$

(C) $2 \Omega$

(D) $2.5 \Omega$

Sigma Percentile · Accepted Answer

### The Magic of the Potentiometer

Imagine you have a battery, and you want to know its true strength—its Electromotive Force (EMF). If you connect a standard voltmeter, it draws a tiny bit of current, which means you're only measuring the terminal voltage, not the true EMF. Enter the **potentiometer**, an elegant device that measures voltage without drawing any current at all!

In this problem, we are using a potentiometer to find the internal resistance of a cell. The process involves finding a "null point" on a long wire where the galvanometer shows zero deflection. At this magical point, the potential difference across the wire exactly balances the potential difference of our cell circuit.

### Case 1: Measuring the True EMF

First, we connect the cell directly to the potentiometer without any extra resistance. The switch to our shunt resistor is open.

When we find the null point, the cell is in an open circuit. It's not supplying any current. Therefore, the potential difference across the balancing length $l_1$ is exactly equal to the true EMF of the cell, $E$.

Mathematically, we write this as:
$$E = k l_1$$

We are given that the balancing length $l_1$ is $52	ext{ cm}$. Plugging this in, we get our first master equation:
$$E = k 	imes 52$$

### Case 2: Measuring the Terminal Voltage

Now, we introduce a twist. We shunt the cell with a resistance $R = 5\,\Omega$. By closing the switch, we create a local closed circuit where current flows from the cell through the $5\,\Omega$ resistor.

Because the cell is now supplying current, it experiences a potential drop across its own internal resistance $r$. The potentiometer will now measure the **terminal voltage** $V$, which is strictly less than the EMF.

The new balancing length $l_2$ is given as $40	ext{ cm}$. This gives us our second equation:
$$V = k 	imes 40$$

Notice how the balancing length decreased from $52	ext{ cm}$ to $40	ext{ cm}$? This perfectly aligns with the fact that $V < E$.

### The Master Equation for Internal Resistance

How do we link EMF, terminal voltage, and internal resistance? From Ohm's law and circuit theory, we know that $V = E - ir$. With a bit of algebraic manipulation, we can derive a beautiful, direct formula for the internal resistance $r$:
$$r = R \left( \frac{E}{V} - 1 ight)$$

This formula is incredibly powerful because it relies only on the ratio of $E$ to $V$, meaning the potential gradient $k$ will completely cancel out!

### Final Calculation: Bringing It Home

Let's substitute our expressions for $E$ and $V$ into the master equation. We also know our shunt resistance $R$ is $5\,\Omega$.

$$r = 5 \left( \frac{k 	imes 52}{k 	imes 40} - 1 ight)$$

The $k$'s cancel out, leaving us with a simple fraction:
$$r = 5 \left( \frac{52}{40} - 1 ight)$$

Let's simplify the fraction $\frac{52}{40}$. Both numbers are divisible by $4$:
$$\frac{52}{40} = \frac{13}{10} = 1.3$$

Substituting this back into our equation:
$$r = 5 (1.3 - 1)$$
$$r = 5 	imes 0.3$$
$$r = 1.5\,\Omega$$

And there we have it! The internal resistance of the cell is exactly **$1.5\,\Omega$**. The potentiometer has once again proven its precision and elegance.

Select Answer:

Visualized Solution

Potentiometer Principle

Case 1: Open Circuit

Balancing Length for EMF

Case 2: Closed Circuit

Balancing Length for Terminal Voltage

Internal Resistance Formula

Substituting Values

Simplifying the Fraction

Final Calculation

Food for Thought

The Sigma Insight: Electrical Instruments

The Magic of the Potentiometer

Case 1

Case 2

The Master Equation for Internal Resistance

Final Calculation

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Select Answer:

Visualized Solution

Potentiometer Principle

Case 1: Open Circuit

Balancing Length for EMF

Case 2: Closed Circuit

Balancing Length for Terminal Voltage

Internal Resistance Formula

Substituting Values

Simplifying the Fraction

Final Calculation

Food for Thought

The Sigma Insight: Electrical Instruments

The Magic of the Potentiometer

Case 1

Case 2

The Master Equation for Internal Resistance

Final Calculation

Similar Questions

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