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Chapter 18: Problem 119

A cell can be balanced against \(110 \mathrm{~cm}\) and \(100 \mathrm{~cm}\) of potentiometer wire, respectively with and without being short circuited through a resistance of \(10 \Omega\). Its internal resistance is (a) \(2.0 \mathrm{ohm}\) (b) zero (c) \(1.0 \mathrm{ohm}\) (d) \(0.5 \mathrm{ohm}\)

Short Answer

Expert verified
The internal resistance of the cell is \(1.0\, \Omega\).

Step by step solution

01

Understanding the Problem

The problem involves a cell that can balance against different lengths of potentiometer wire. When short circuited through a resistance of 10 惟 and when not short circuited, it balances at lengths of 110 cm and 100 cm, respectively. We need to determine the internal resistance of the cell based on these observations.
02

Setting Up the Equations

Let the emf of the cell be \( E \) and the internal resistance be \( r \). The length of the potentiometer wire is proportional to the potential difference, so the potential difference when not short circuited (balancing length 110 cm) and when short circuited through 10 惟 resistance (balancing length 100 cm) can be expressed as:\[ \frac{E}{110} = kE \quad \text{and} \quad \frac{E}{100} = k \frac{E}{r + 10} \]Where \( k \) is a constant of proportionality.
03

Analyzing the Equations

Since \( k \) is the same in both cases, divide the two equations:\[ \frac{E}{110} = \frac{E}{100} \frac{r + 10}{r} \]Simplifying:\[ \frac{1}{110} = \frac{1}{100} \frac{r + 10}{r} \]This simplifies further to:\[ \frac{100}{110} = \frac{r + 10}{r} \]
04

Solving for Internal Resistance

Cross-multiply and simplify the equation:\[ 100r = 110(r + 10) \]\[ 100r = 110r + 1100 \]Move terms involving \( r \) to one side:\[ 1100 = 10r \]Divide by 10 to isolate \( r \):\[ r = 1.0 \Omega \]
05

Verify the Solution

Check the calculated internal resistance by substituting back into the understanding of the problem. If \( r = 1.0 \Omega \), the ratios of lengths are consistent with the potential drop across the known resistor and internal resistance. The calculation holds true, confirming our answer.

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Key Concepts

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

Potentiometer
A potentiometer is an intriguing electrical device that helps measure the electromotive force (emf) of a cell without drawing current. It comprises a long, uniform wire where a known voltage is applied, usually over a precise length. By comparing the potential drop with an unknown potential from a cell, we can determine the cell's emf.
  • The wire length is proportional to the potential difference.
  • Balances when the potential difference equals the emf of the cell.
  • Operates on the principle without affecting the circuit's current flow.
This ensures a practical method for measuring voltage without loading the cell. In the given exercise, different balancing lengths help determine the internal resistance of a cell, using the properties of a potentiometer.
Ohm's Law
Ohm's Law is a fundamental principle in electronics and physics, defining the relationship between voltage, current, and resistance in an electrical circuit. The law is expressed as \[ V = IR \], where:
  • \( V \) is the voltage across the circuit.
  • \( I \) is the current flowing through it.
  • \( R \) is the resistance of the circuit.
Understanding Ohm's Law is key to solving many practical circuit problems, including those involving internal resistance and potential differences. It predicts how changes in the circuit affect voltage drops and assists in determining unknown quantities such as internal resistance in the exercise's cell problem.
Emf of Cell
The electromotive force (emf) of a cell is crucial to understanding how cells work. It represents the energy supplied per charge and effectively differs from terminal voltage when current flows.
  • Emf is measured when no current flows through the circuit.
  • Typically higher than the actual voltage across terminals while delivering current.
In the scenario given, as the cell remains non-short circuited, the emf dictates the initial state measure, highlighting its ultimate potential. Additionally, it serves as a constant factor across both balancing length states in the context of the potentiometer.
Resistive Circuit
Resistive circuits, like the one inferred from the exercise, include components that resist current flow, converting electrical energy into heat. They follow Ohm's Law and embody varying resistance depending on composition.
  • Composed typically of resistors, wiring, and power sources.
  • In the exercise, resistance appears via the given resistor and cell's internal resistance.
Understanding these resistive components is crucial when calculating potential differences especially while the circuit is short circuited. The short circuit affects the balancing length since the experience of resistance modifies circuit potential.

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Most popular questions from this chapter

The temperature of a metal wire rises when an electric current passes through it because (a) collision of metal atoms with each other releases heat energy (b) collision of conduction electrons with each other releases heat energy (c) When the conduction electrons fall from higher energy level to lower energy level heat energy is released (d) Collision of conduction electrons with the atoms of metal give them energy which appears as heat

Assume that each atom of copper contributes one electron. If the current flowing through a copper wire of \(1 \mathrm{~mm}\) diameter is \(1.1 \mathrm{~A}\), the drift velocity of electrons will be (Density of \(\mathrm{Cu}=9 \mathrm{~g} \mathrm{~cm}^{-3}\), at. wt. of \(\mathrm{Cu}=63\) ) (a) \(0.3 \mathrm{~mm} / \mathrm{s}\) (b) \(0.5 \mathrm{~mm} / \mathrm{s}\) (c) \(0.1 \mathrm{~mm} / \mathrm{s}\) (d) \(0.2 \mathrm{~mm} / \mathrm{s}\)

The electron drift speed is small and the charge of the electron is also small but still, we obtain a large current in a conductor. This is due to (a) the conducting property of the conductor (b) the resistance of the conductor is small (c) the electron number density of the conductor is small (d) the electron number density of the conductor is enormous.

In a Wheatstone's network, \(P=2 \Omega ; Q=2 \Omega ; R=2 \Omega\) and \(S=3 \Omega\). the resistance with which \(S\) is to be shunted in order that the bridge may be balanced is (a) \(1 \Omega\) (b) \(2 \Omega\) (c) \(4 \Omega\) (d) \(6 \Omega\)

In a meter bridge experiment, null point is obtained at \(20 \mathrm{~cm}\) from one end of the wire when resistance \(X\) is balanced against another resistance \(Y\). If \(X
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