/*! 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 30 A Nonideal Ammeter. Unlike the i... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 30

A Nonideal Ammeter. Unlike the idealized ammeter any real ammeter has a nonzero resistance. (a) An ammeter with resistance \(R_{\mathrm{A}}\) is connected in series with a resistor \(R\) and a battery of emf \(E\) and internal resistance \(r .\) The current measured by the ammeter is \(I_{A}\). Find the current through the circuit if the ammeter is removed so that the battery and the resistor form a complete circuit. Express your answer in terms of \(I_{\mathrm{A}}, r, R_{\mathrm{A}}\), and \(R\). The more "ideal" the ammeter, the smaller the difference between this current and the current \(I_{A}\). (b) If \(R=3.80 \Omega\), \(\mathrm{E}=7.50 \mathrm{~V}\), and \(r=0.45 \Omega\), find the maximum value of the ammeter resistance \(R_{\mathrm{A}}\) so that \(I_{\mathrm{A}}\) is within \(1.0 \%\) of the current in the circuit when the ammeter is absent. (c) Explain why your answer in part (b) represents a maximum value.

Short Answer

Expert verified
(b) Maximum \( R_A \approx 0.083 \Omega \); (c) Beyond this, ammeter affects current beyond 1%.

Step by step solution

01

Determine Current with Ammeter

When the ammeter is connected in series with the resistor and battery, the total resistance in the circuit is given by the sum of the three resistances: \( R + R_A + r \). The EMF of the battery is \(E\) and the current \(I_A\) measured by the ammeter is described by Ohm's law, \( I_A = \frac{E}{R + R_A + r} \).
02

Determine Current without Ammeter

When the ammeter is removed, the circuit only includes the battery and the resistor. The total resistance now is \( R + r \). The current in this simpler circuit is given by \( I = \frac{E}{R + r} \).
03

Express Current without Ammeter in terms of Known Variables

Express \( I \) in terms of \( I_A \) and other given quantities as follows. From Step 1, substitute \( I_A \) to get \( E = I_A (R + R_A + r) \). Substitute this into the expression from Step 2, yielding: \( I = \frac{I_A (R + R_A + r)}{R + r} \).
04

Solve for Maximum Ammeter Resistance in Part (b)

Rearrange the relation from Step 3 to obtain the ammeter resistance that ensures \( I_A \) is within 1% of \( I \). Set \( I - I_A \leq 0.01 I \). Substitute the previous equation to get the functional form \( \frac{I_A (R + R_A + r)}{R + r} - I_A \leq 0.01 \frac{E}{R+r} \). Further simplify and solve for \( R_A \) to derive the criteria for maximum resistance.
05

Input Numerical Values to Calculate Maximum Resistance

Plug the known values into the inequality from Step 4. \( R = 3.80 \text{ }\Omega\), \( E = 7.50 \text{ }V\), \( r = 0.45 \text{ }\Omega\). Solve for \( R_A \) and find \( R_A \leq 0.08314 \text{ }\Omega \).
06

Interpretation and Justification

The solution in Part (b) defines the upper bound for \( R_A \) so that the ammeter's resistance does not significantly alter the circuit. If \( R_A \) exceeds this value, the ammeter's influence becomes non-trivial, resulting in a current measure deviation larger than 1%.

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.

Ammeter Resistance
An ammeter is a device used to measure the current flowing through an electric circuit. Unlike an ideal ammeter, which would have zero resistance, real-world ammeters include a small resistance, denoted as \( R_A \). Adding an ammeter to a circuit introduces an additional resistance that can affect the total circuit resistance. In a typical setup where an ammeter is in series with other resistive components, the total resistance is the sum of all resistances, which can slightly alter the current measured by \( I_A \). The smaller the ammeter's resistance, \( R_A \), the closer the measured current \( I_A \) will be to the original current without the ammeter. Key points to keep in mind:
  • A lower ammeter resistance allows for more accurate current measurements.
  • An increase in \( R_A \) causes the measured current to differ more from the actual current.
  • The resistance tolerance depends on the desired measurement accuracy, often specified in percentage terms.
Ohm's Law
Ohm's Law is a fundamental principle in circuit analysis, describing the relationship between voltage, current, and resistance. It is articulated as \( V = IR \), where \( V \) is the voltage across the circuit, \( I \) is the current flowing through the circuit, and \( R \) is the resistance. When applying Ohm's Law in the context of an ammeter connected in series, the total resistance impacts the current measurement:
  • With the ammeter: \( I_A = \frac{E}{R + R_A + r} \), where \( E \) is the electromotive force, and \( r \) is the internal resistance of the battery.
  • Without the ammeter: \( I = \frac{E}{R + r} \), simplifying the resistance equation.
Understanding how an ammeter alters the circuit's resistance and using Ohm's Law can help demonstrate these differences in current measurements.
Circuit Analysis
Conducting a circuit analysis involves assessing how different components, like resistors and ammeters, contribute to the overall behavior of an electric circuit. When performing circuit analysis, consider:- **Component Connections**: How components, such as the battery, resistor \( R \), and ammeter \( R_A \) are wired in the circuit, typically in series or parallel.- **Total Resistance**: Calculate the total resistance based on the arrangement of components: - In a series circuit, resistances add up (\( R + R_A + r \)). - In a parallel circuit, the reciprocal of the total resistance is the sum of the reciprocals of each component's resistance.- **Current Direction**: Understanding current flow can assist in predicting how changes in the setup, such as removing or adding an ammeter, affect the overall circuit current.By carefully analyzing a circuit and considering measurement tools like ammeters, you can predict how those tools will impact results, allowing for more accurate interpretations and outcomes in practical applications.

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

A 14-gauge copper wire of diameter \(1.6 \mathrm{~mm}\) carries a current of \(12 \mathrm{~mA}\). (a) What is the potential difference across a \(2.00-\mathrm{m}\) length of the wire? (b) What would the potential difference in part (a) be if the wire were silver instead of copper, but all else were the same? Take \(\rho_{c \|}=1.76 \times 10^{-8} \Omega-m, \rho_{\text {siluer }}=1.43 \times 10^{-8} \Omega-m\)

A 1.50-m cylindrical rod of diameter \(0.500 \mathrm{~cm}\) is connected to a power supply that maintains a constant potential difference of \(15.0 \mathrm{~V}\) across its ends, while an ammeter measures the current through it. You observe that at room temperature \(\left(20.0^{\circ} \mathrm{C}\right.\) ) the ammeter reads \(18 \mathrm{~A}\), while at \(90.0^{\circ} \mathrm{C}\) it reads \(16.5 \mathrm{~A}\). You can ignore any thermal expansion of the rod. Find (a) the resistivity at \(20.0^{\circ} \mathrm{C}\) and (b) the temperature coefficient of resistivity at \(20^{\circ} \mathrm{C}\) for the material of the rod.

A ductile metal wire has resistance \(R\). What will be the resistance of this wire in terms of \(R\) if it is stretched to three times its original length, assuming that the density and resistivity of the material do not change when the wire is stretched? (Hint: The amount of metal does not change, so stretching out the wire will affect its cross-sectional area.)

Lightning Strikes. During lightning strikes from a cloud to the ground, currents as high as \(25,000 \mathrm{~A}\) can occur and last for about \(40 \mu \mathrm{s}\). How much charge is transferred from the cloud to the earth during such a strike?

The region between two concentric conducting spheres with radii \(a\) and \(b\) is filled with a conducting material with resistivity \(\rho .\) (a) Show that the resistance between the spheres is given by \(R=\frac{\rho}{4 \pi}\left(\frac{1}{a}-\frac{1}{b}\right)\) (b) Derive an expression for the current density as a function of radius, in terms of the potential difference \(V_{a b}\) between the spheres.
See all solutions

Recommended explanations on Physics Textbooks

Astrophysics

Read Explanation

Famous Physicists

Read Explanation

Dynamics

Read Explanation

Modern Physics

Read Explanation

Electromagnetism

Read Explanation

Circular Motion and Gravitation

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.