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A cell when balanced with potentiometer gave a balance length of \(50 \mathrm{~cm}\). Then, a \(4.5 \Omega\) external resistance is introduced in the circuit. Now it is balanced on \(45 \mathrm{~cm}\). the internal resistance of the cell is (a) \(0.25 \Omega\) (b) \(0.5 \Omega\) (c) \(1.0 \Omega\) (d) \(1.5 \Omega\)

Short Answer

Expert verified
The internal resistance of the cell is \(0.5\, \Omega\).

Step by step solution

01

Understand the Problem

We are given a problem involving a potentiometer, a cell, and the changes in balance length when an external resistance is added. We need to find the internal resistance of the cell.
02

Apply the Formula for Internal Resistance

The relationship between internal resistance \( r \), external resistance \( R \), and the balance lengths can be derived from the condition of balance in a potentiometer experiment: \( \frac{E}{l_1} = \frac{E}{(l_2 + r)} \). Here, \( l_1 \) and \( l_2 \) are the lengths associated with balance conditions. Simplifying gives us \( r = \left( \frac{l_1 - l_2}{l_2} \right) \times R \).
03

Plug in the Values

Using the given values: \( l_1 = 50 \text{ cm} \), \( l_2 = 45 \text{ cm} \), and \( R = 4.5 \Omega \), substitute into the formula: \[ r = \left( \frac{50 - 45}{45} \right) \times 4.5 \Omega \].
04

Calculate the Internal Resistance

Calculate the expression: \( \frac{50 - 45}{45} = \frac{5}{45} = \frac{1}{9} \). Multiply by the external resistance: \( r = \frac{1}{9} \times 4.5 = 0.5 \Omega \).
05

Confirm the Calculation

We verify that the steps and calculations have been followed correctly, confirming that the internal resistance of the cell is indeed \( 0.5 \Omega \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Internal Resistance
In a potentiometer experiment, internal resistance refers to the inherent resistance to the flow of electric charge inside a cell. Every cell inherently has some resistance within it, not readily visible, but vital for understanding the cell's performance.
When a cell is in operation, this internal resistance plays a crucial role. It impacts how the voltage behaves as the current flows through the cell. Essentially, this resistance can decrease the cell's efficiency, reducing the voltage output that reaches an external circuit.
The formula for determining internal resistance in the context of a potentiometer experiment is: \ \( r = \left( \frac{l_1 - l_2}{l_2} \right) \times R \). Here, \( l_1 \) and \( l_2 \) represent the balance lengths, and \( R \) is the external resistance added to the circuit. Knowing this helps us understand how internal resistance can affect the potential difference and the performance of electrical circuits.
Role of External Resistance
External resistance is any resistance added intentionally to a circuit, outside of the cell itself. This resistance is critical in experiments with a potentiometer, as it helps us measure other aspects like the internal resistance effectively.
One can think of external resistance as a controllable factor in the setup. By varying the resistance, you can observe different behaviors in the circuit, thus gaining insights into the cell’s characteristics.
  • It helps in determining the cell's efficiency by affecting the balance length when introduced.
  • Allows for modifications in the circuit’s total resistance and thus the flow of current.
In the given problem, the external resistance of \(4.5 \, \Omega\) was used to determine the internal resistance by observing changes in balance lengths at specific resistance levels.
Understanding Balance Length
Balance length is the length on a potentiometer wire at which the potential difference across it exactly matches the electromotive force of the cell being measured.
In the given exercise, two sets of balance lengths are compared to determine how the voltage across the cell is affected by additional resistance. This forms the core principle behind the use of a potentiometer: obtaining a null deflection by matching potentials.
When an external resistance is added, you might observe a shift in this balance length, as seen from the initial \(50 \, \text{cm}\) to \(45 \, \text{cm}\) with added resistance. Changes in balance length give us vital clues to the internal properties of the cell. This null point method, where the current is in theory zero, ensures the most accurate reading of a cell's EMF.
Nature of a Cell in Circuit Experiments
In the world of circuits, a cell is the basic energy source, capable of converting stored chemical energy into electrical energy. Each cell is characterized by its EMF (electromotive force), internal resistance, and its ability to drive current through a circuit.
The energy output and efficiency of this cell can be tested using experiments like the potentiometer experiment. In such setups, the cell's EMF acts as a reference to measure potential differences across various sections of a circuit.
Utilizing the potentiometer setup, where you don’t draw current from the cell, allows for a clear measurement of EMF. It's insightful for understanding how the cell’s internal characteristics, such as internal resistance, influence overall performance.

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Most popular questions from this chapter

Two wires of resistance \(R_{1}\) and \(R_{2}\) have temperature coefficient of resistance \(\alpha_{1}\) and \(\alpha_{2}\), respectively. These are joined in series. The effective temperature coefficient of resistance is (a) \(\frac{\left(\alpha_{1}+\alpha_{2}\right)}{2}\) (b) \(\sqrt{\alpha_{1} \alpha_{2}}\) (c) \(\frac{\left(\alpha_{1} R_{1}+\alpha_{2} R_{2}\right)}{R_{1}+R_{2}}\) (d) \(\frac{\sqrt{R_{1} R_{2} \alpha_{1} \alpha_{2}}}{\sqrt{R_{1}^{2}+R_{2}^{2}}}\)

The length of a wire of a potentiometer is \(100 \mathrm{~cm}\), 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 \(l=30 \mathrm{~cm}\) from the positive end, the emf of the battery is (a) \(0.2 E\) (b) \(0.3 E\) (c) \(0.4 E\) (d) \(0.5 E\)

A current \(I\) flows through a uniform wire of diameter \(d\), when the mean drift velocity is \(v_{d}\) The same current will flow through a wire of diameter \(d / 2\) made of the same material if the mean drift velocity of the electrons is (a) \(v / 4\) (b) \(v / 2\) (c) \(4 v\) (d) \(2 v\)

When there is an electric current through a conducting wire along its length, then an electric field must exist (a) outside the wire but normal to it (b) outside the wire but around it (c) inside the wire but parallel to it (d) inside the wire but normal to it

\(n\) cells of emfs \(E_{1}, E_{2}, E_{3}, \ldots, E_{n}\) and internal resistance \(r_{1}, r_{2}, r_{3}, \ldots, r_{n}\) are connected in series to form a closed circuit with zero external resistance. For each cell the ratio of emf to internal resistance is \(K\), where \(K\) is a constant; then current in the circuit is (a) \((1 / K)\) (b) \(K\) (c) \(K^{2}\) (d) \(\left(1 / K^{2}\right)\)

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