/*! 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 61 The potential difference across ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 25: Problem 61

The potential difference across the terminals of a battery is \(8.40 \mathrm{~V}\) when there is a current of \(1.50 \mathrm{~A}\) in the battery from the negative to the positive terminal. When the current is \(3.50 \mathrm{~A}\) in the reverse direction, the potential difference becomes \(10.20 \mathrm{~V}\). (a) What is the internal resistance of the battery? (b) What is the emf of the battery?

Short Answer

Expert verified
The internal resistance of the battery is approximately \(0.4 \, \Omega\) and the emf of the battery is approximately \(9.0 \, V\).

Step by step solution

01

Setup the expressions for voltage

Use Ohm's law (\(V = IR\)) to set up an expression for the voltage across the battery in each case. When the current flows from negative to positive terminal (let's call this case 1), the voltage \(V1\) is equal to the emf (\(E\)) minus the product of the current (\(I1 = 1.50 A\)) and the internal resistance (\(r\)): \(V1 = E - I1 * r\). When the current flows in the reverse direction (case 2), the voltage \(V2\) is equal to the emf plus the product of the current (\(I2 = 3.50 A\)) and the internal resistance: \(V2 = E + I2 * r\).
02

Solve for internal resistance

You can solve for the internal resistance (\(r\)) by subtracting equation from Step 1, case 1 from equation from Step 1, case 2: \( V2 - V1= (E + I2 * r) - (E - I1 * r) = I2 * r + I1 * r = (I1 + I2) * r\). You can now solve for \(r\): \( r = (V2 - V1) / (I1 + I2)\). Plug into this formula the given values: \( r = (10.20 V - 8.40 V) / (1.50 A + 3.50 A) \).
03

Find the emf of the battery

Once you have found \(r\), you can substitute it back into either of the original equations from Step 1 to find the emf \(E\). Using the equation from case 1, \(E = V1 + I1 * r\), plug in the given and found values: \(E = 8.40 V + 1.50 A * r\).

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.

Ohm's Law
Ohm's Law is one of the fundamental principles in electricity. It states that the current flowing through a conductor between two points is directly proportional to the voltage across the two points. The formula given by Ohm's Law is:
  • \( V = IR \)
Where:
  • \( V \) is the voltage (in volts).
  • \( I \) is the current (in amperes).
  • \( R \) is the resistance (in ohms).
Ohm's Law helps us understand how voltage, current, and resistance interact in an electrical circuit. For example, if you increase the resistance while keeping the voltage constant, the current will decrease. This simple relationship is incredibly useful for solving problems in circuits and is often used to calculate unknown variables when other quantities are known. In our exercise, it's instrumental in setting up equations to solve for internal resistance and electromotive force.
Internal Resistance
Internal resistance is a property of all batteries and power supplies. It represents the resistance to the flow of electric current within the battery itself. When current flows through the battery, the internal resistance causes some of the energy to be lost as heat. This can affect the terminal voltage of the battery.

In formula terms, when you measure a lower terminal voltage than expected, it's likely due to the internal resistance. The modified version of Ohm's Law for a battery becomes:
  • \( V = E - Ir \)
Where:
  • \( E \) is the electromotive force (emf).
  • \( r \) is the internal resistance.
  • \( I \) is the current.
Our exercise used this concept to differentiate between the emf and actual terminal voltages for different current directions, enabling the computation of the internal resistance.
Electromotive Force (emf)
The electromotive force (emf) of a battery is the maximum potential difference the battery can provide. Think of it as the energy supplied by the battery to each coulomb of charge.

Emf is defined as the voltage across the terminals of the battery when no current is flowing. In real-life circumstances, when a current flows, the presence of internal resistance means that the terminal voltage is usually less than the emf.
  • \( E = V + Ir \)
Where:
  • \( E \) is the emf.
  • \( V \) is the terminal voltage.
  • \( I \) is the current flowing through the battery.
  • \( r \) is the internal resistance.
In the exercise scenario, understanding emf allowed us to determine the net energy delivered to the electric circuit when currents varied in magnitude and direction, affirming the relationship between the internal components of the battery.

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 silver wire \(2.6 \mathrm{~mm}\) in diameter transfers a charge of \(420 \mathrm{C}\) in 80 min. Silver contains \(5.8 \times 10^{28}\) free electrons per cubic meter. (a) What is the current in the wire? (b) What is the magnitude of the drift velocity of the electrons in the wire?

In a physics laboratory experiment, a coil with 200 turns enclosing an area of \(12 \mathrm{~cm}^{2}\) is rotated in \(0.040 \mathrm{~s}\) from a position where its plane is perpendicular to the earth's magnetic field to a position where its plane is parallel to the field. The earth's magnetic field at the lab location is \(6.0 \times 10^{-5} \mathrm{~T}\). (a) What is the magnetic flux through each turn of the coil before it is rotated? After it is rotated? (b) What is the average emf induced in the coil?

On your first day at work as an electrical technician, you are asked to determine the resistance per meter of a long piece of wire. The company you work for is poorly equipped. You find a battery, a voltmeter, and an ammeter, but no meter for directly measuring resistance (an ohmmeter). You put the leads from the voltmeter across the terminals of the battery, and the meter reads \(12.6 \mathrm{~V}\). You cut off a \(20.0 \mathrm{~m}\) length of wire and connect it to the battery, with an ammeter in series with it to measure the current in the wire. The ammeter reads 7.00 A. You then cut off a \(40.0 \mathrm{~m}\) length of wire and connect it to the battery, again with the ammeter in series to measure the current. The ammeter reads 4.20 A. Even though the equipment you have available to you is limited, your boss assure you of its high quality: The ammeter has very small resistance, and the voltmeter has very large resistance. What is the resistance of 1 meter of wire?

Compact Fluorescent Bulbs. Compact fluorescent bulbs are much more efficient at producing light than are ordinary incandescent bulbs. They initially cost much more, but they last far longer and use much less electricity. According to one study of these bulbs, a compact bulb that produces as much light as a \(100 \mathrm{~W}\) incandescent bulb uses only \(23 \mathrm{~W}\) of power. The compact bulb lasts 10,000 hours, on the average, and costs \(\$ 11.00,\) whereas the incandescent bulb costs only \(\$ 0.75\), but lasts just 750 hours. The study assumed that electricity costs \(\$ 0.080\) per kilowatt-hour and that the bulbs are on for \(4.0 \mathrm{~h}\) per day. (a) What is the total cost (including the price of the bulbs) to run each bulb for 3.0 years? (b) How much do you save over 3.0 years if you use a compact fluorescent bulb instead of an incandescent bulb? (c) What is the resistance of a "100 W" fluorescent bulb? (Remember, it actually uses only \(23 \mathrm{~W}\) of power and operates across \(120 \mathrm{~V} .\) )

A slender rod, \(0.240 \mathrm{~m}\) long, rotates with an angular speed of \(8.80 \mathrm{rad} / \mathrm{s}\) about an axis through one end and perpendicular to the rod. The plane of rotation of the rod is perpendicular to a uniform magnetic field with a magnitude of \(0.650 \mathrm{~T}\). (a) What is the induced emf in the rod? (b) What is the potential difference between its ends? (c) Suppose instead the rod rotates at \(8.80 \mathrm{rad} / \mathrm{s}\) about an axis through its center and perpendicular to the rod. In this case, what is the potential difference between the ends of the rod? Between the center of the rod and one end?
See all solutions

Recommended explanations on Physics Textbooks

Famous Physicists

Read Explanation

Torque and Rotational Motion

Read Explanation

Kinematics Physics

Read Explanation

Classical Mechanics

Read Explanation

Fields in Physics

Read Explanation

Engineering 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.