/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 110 A \(2 \Omega\) resistor is conne... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 18: Problem 110

A \(2 \Omega\) resistor is connected in series with an \(R \Omega\) resistor. This combination is connected across a cell. When the potential difference across \(2 \Omega\) resistor is balanced on a potentiometer wire, null point is obtained at a length of \(300 \mathrm{~cm}\). When the same procedure is repeated for \(R \Omega\) resistor, null point is obtained at a length of \(350 \mathrm{~cm}\). Value of \(R\) is (a) \(2.33 \Omega\) (b) \(4.6 \Omega\) (c) \(3.33 \Omega\) (d) \(5 \Omega\)

Short Answer

Expert verified
The value of \( R \) is \( 2.33 \Omega \).

Step by step solution

01

Understand the Setup

The problem involves a series circuit with two resistors, a 2惟 resistor, and an R惟 resistor. A potentiometer wire is used to measure the potential difference across each resistor separately by obtaining null points. The potentiometer wire length is proportional to the potential difference across the resistor when the null point is achieved.
02

Recall Potentiometer Principle

In a potentiometer, the potential drop across the known resistor for a null point gives the balance equation for potential difference, which is directly proportional to the length of the wire (l) at null point i.e., V 鈭� l.
03

Write Equations for Null Points

For the 2惟 resistor, let the potential difference across it be V鈧�, then the equation for the null point is: \[ V_1 imes L_1 = k \times 2 \Omega \times I \] For the R惟 resistor, let the potential difference across it be V鈧�, then: \[ V_2 imes L_2 = k \times R \times I \]where L鈧� = 300 cm and L鈧� = 350 cm.
04

Set up Proportional Equations from Potentiometer Readings

The potential difference across each resistor can be expressed as:For 2惟 resistor: \[ V_1 = k \times L_1 = k \times 300 \]For R惟 resistor:\[ V_2 = k \times L_2 = k \times 350 \]
05

Use the Voltage Proportionality

Since the total voltage supplied to the resistors is the same, apply the proportionality of voltages across the two resistors:\[ \frac{V_1}{V_2} = \frac{2}{R} \] Substitute expressions for V鈧� and V鈧� using link lengths:\[ \frac{300}{350} = \frac{2}{R} \]
06

Solve for R

Solve the equation from above:\[ \frac{300}{350} = \frac{2}{R} \]Cross-multiply to get:\[ 300 R = 700 \]\[ R = \frac{700}{300} = \frac{7}{3} = 2.33 \Omega \]
07

Verify the Answer

The calculated value of \( R \) is \( 2.33 \Omega \), which matches option (a).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Resistor in Series
In a series circuit, resistors are connected end-to-end. This ensures that the current flows through each resistor one after another. The unique feature of resistors in series is that they share the same current. However, the total or equivalent resistance of the circuit is the sum of all individual resistances.

So, if you have a 2惟 resistor in series with an unknown R惟 resistor in our setup, the overall resistance is their sum:
  • Equivalent resistance = 2惟 + R惟
When resistors are arranged in series, they provide a great setup to study potential differences through components like a potentiometer, given their straightforward nature of direct current sharing.
Null Point
The null point in a potentiometer setup refers to the specific point on the wire where the potential difference across a section of the wire and a connected part of the circuit are equal. At this point, no current flows through the galvanometer connected to the potentiometer, indicating a perfect balance.

Finding the null point in a potentiometer experiment is crucial, as it serves as an indication that the potential difference across a component (like a resistor) matches the potential difference at that specific spot on the wire:
  • It confirms balance between two potential differences.
  • It allows the potential difference to be measured accurately.
The null point becomes more relevant when measuring potential differences across different resistors in series, as seen in the exercise where the lengths of 300 cm and 350 cm correspond to specific potential differences.
Proportionality of Voltage
One of the fundamental principles in working with potentiometers is that the potential difference is directly proportional to the length of the wire at the null point. This relationship is expressed as:
  • Voltage ( ext{V} ) 鈭� Length ( ext{L} )
In our exercise, this means the larger the length at the null point, the larger the potential difference across the resistor.
  • A length of 300 cm indicates a specific voltage across the 2惟 resistor.
  • A length of 350 cm indicates a different voltage across the R惟 resistor.
This proportionality is a key concept when calibrating and measuring precise voltages without losing out on accuracy due to calibration errors or voltage drops that may affect readings otherwise.
Current Electricity
Current electricity refers to the flow of electric charge through conductors. In our setup with the series resistors, the same current flows through both resistors. Current, represented by 'I', is crucial for determining potential differences using Ohm's Law, which states:
  • PV ( I 脳 R) = Current ( I ) 脳 Resistance ( R )
This means:
  • The same current flows through the 2惟 and R惟 resistors.
  • The potential difference across each is influenced by this shared current.
Understanding current electricity in conjunction with resistors in series helps explain how potential differences are distributed and how they can be accurately measured using potential proportionality principles.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The temperature coefficient of resistance of a wire is \(0.00125^{\circ} \mathrm{C}^{-1}\). At \(300 \mathrm{~K}\) its resistance is one ohm. The resistance of the wire will be \(2 \mathrm{ohm}\) at (a) \(1154 \mathrm{~K}\) (b) \(1100 \mathrm{~K}\) (c) \(1400 \mathrm{~K}\) (d) \(1127 \mathrm{~K}\)

An energy source will supply a constant current into the load if its internal resistance is (a) zero (b) non zero but less than the resistance of the load (c) equal to resistance of the load (d) very large as compared to the load resistance

The resistance of a bulb filaments is \(100 \Omega\) at a temperature of \(100{ }^{\circ} \mathrm{C}\). If its temperature coefficient of resistance be \(0.005 l{ }^{\circ} \mathrm{C}^{-1}\), its resistance will become \(200 \Omega\) at a temperature of (a) \(200^{\circ} \mathrm{C}\) (b) \(300^{\circ} \mathrm{C}\) (c) \(400^{\circ} \mathrm{C}\) (d) \(500^{\circ} \mathrm{C}\)

A teacher asked a student to connect \(N\) cells each of \(\operatorname{emf} e\) in series to get a total emf of \(\mathrm{Ne}\). While connecting, the student, by mistake, reversed the polarity of \(n\) cells. The total emf of the resulting series combination is (a) \(e\left(N-\frac{n}{2}\right)\) (b) \(e(N-n)\) (c) \(e(N-2 n)\) (d) \(e N\)

Two wires of same metal have same length but their cross-sections are in the ratio \(3: 1\). They are joined in series. The resistance of the thicker wire is \(10 \Omega\). The total resistance of the combination will be (a) \((5 / 2) \Omega\) (b) \((40 / 3) \Omega\) (c) \(40 \Omega\) (d) \(100 \Omega\)
See all solutions

Recommended explanations on Physics Textbooks

Physics of Motion

Read Explanation

Fundamentals of Physics

Read Explanation

Measurements

Read Explanation

Electrostatics

Read Explanation

Work Energy and Power

Read Explanation

Kinematics Physics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.