Question

The length of a potentiometer wire is \(600 \mathrm{~cm}\) and it carries a current of \(40 \mathrm{~m} \mathrm{~A}\) for cell of emf \(2 \mathrm{~V}\) and internal resistance \(10 \Omega\), the null point is found to be at \(500 \mathrm{~cm}\) on connecting a voltmeter across the cell, the balancing length is decreased by \(10 \mathrm{~cm}\) The voltmeter reading will be. (A) \(1.96 \mathrm{~V}\) (B) \(1.8 \mathrm{~V}\) (C) \(1.64 \mathrm{~V}\) (D) \(0.96 \mathrm{~V}\)

Short answer

The voltmeter reading is approximately 1.61 V, which is closest to option (C) 1.64 V.

Step-by-step solution

Calculate voltage across the entire potentiometer wire

First, we need to calculate the voltage across the entire 600 cm potentiometer wire. To do this, we can use Ohm's Law: \(V = IR\) Here, V is voltage, I is current, and R is resistance. We are given I = 40 mA (or 0.04 A) and the total resistance of the potentiometer wire can be found by: \(R_{total} = \frac{rL}{l}\) Where r is the internal resistance of the cell (10 Ω), L is the total length of the potentiometer wire (600 cm), and l is the null point (500 cm).

Calculate the total resistance of the potentiometer wire

Now we will calculate the total resistance of the potentiometer wire: \(R_{total} = \frac{(10 Ω)(600 cm)}{500 cm}\) \(R_{total} = 12 Ω\)

Calculate the potential difference across the entire wire

Now that we have the total resistance of the potentiometer wire, we can calculate the potential difference across the entire wire: \(V_{total} = IR_{total}\) \(V_{total} = (0.04 A)(12 Ω)\) \(V_{total} = 0.48 V\)

Calculate the potential difference across the new balancing length

Since the balancing length decreased by 10 cm, the new balancing length is 490 cm. Now we can calculate the potential difference across the new balancing length using the proportion of the original null point to the new balancing length, and multiplying it by the potential difference across the entire wire: \(V_{new} = \frac{490 cm}{600 cm} \times V_{total}\) \(V_{new} = \frac{490 cm}{600 cm} \times 0.48 V\) \(V_{new} = 0.392 V\)

Calculate the voltmeter reading

Finally, we can calculate the voltmeter reading. The voltmeter reading will be equal to the emf of the cell minus the potential difference across the new balancing length: \(V_{voltmeter} = E - V_{new}\) \(V_{voltmeter} = 2 V - 0.392 V\) \(V_{voltmeter} ≈ 1.61 V\) The voltmeter reading is approximately 1.61 V, which is closest to option (C) 1.64 V.

Key concepts

EMF

Electromotive Force or EMF is a fundamental concept in understanding how circuits work. It represents the maximum potential difference that a battery or cell can provide when it is not connected to any external circuit. Essentially, it's the measure of energy available to push electrons through a conductor. Unlike regular voltage, EMF is considered an open-circuit measurement because it assumes no current flows, which means no energy is lost across internal resistance.
This is crucial for understanding any circuit equation, as it sets the total energy scale. For example, in the given exercise, the cell's EMF is noted to be 2 V, indicating the total energy potential the cell has before encountering any components like resistors or the potentiometer.

Null Point

The null point in a potentiometer circuit is a position along the wire where the electrical voltage (potential difference) is zero when balanced against the voltage of another circuit part.
This means no current flows through the galvanometer, ensuring accuracy in measurements. In this scenario, when we have found a point on the wire where the galvanometer shows zero current, it signifies that the voltage drop across the wire length up to this null point is precisely equal to the EMF of the test cell in use.
This becomes a critical step to simplifying calculations because it lets us directly relate wire length proportions to voltage without relying on complex variables.

Balancing Length

Balancing length is another key feature in potentiometer setups, referring to the segment of the wire over which the voltage measurement is taken. When current travels through the wire, each centimeter can be considered to produce a small, linear drop in voltage.
The balance is achieved when this drop perfectly equals the EMF or another known voltage being tested. Initially, a length of 500 cm was sufficient to achieve this balance given the setup parameters. However, when external conditions shift, like the introduction of a voltmeter, balancing requires adjustments. In the exercise, this led to a change in balancing length, requiring recalculations to maintain voltage matching at 490 cm.

Voltmeter

A voltmeter is an indispensable instrument used to measure electrical potential difference between two points. Unlike the straightforward method of using a potentiometer alone, a voltmeter provides real-time readings with minimal adjustment needs. However, inserting a voltmeter can subtly alter a circuit's behavior due to its own internal resistance, albeit designed to be quite high to minimize this effect.
In this exercise, after the introduction of a voltmeter, the balancing changed and influenced the observed voltage, interpreting also a drop from the ideal EMF (2 V) to the recorded EMF, which was slightly lower. The final reading of 1.64 V indicates real-world factors like minor resistive losses, evident when complex circuits interplay.