/*! 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 7 A wire of resistance \(5.0 \Omeg... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A wire of resistance \(5.0 \Omega\) is connected to a battery whose emf 8 is \(2.0 \mathrm{~V}\) and whose internal resistance is \(1.0 \Omega .\) In \(2.0 \mathrm{~min},\) how much energy is (a) transferred from chemical form in the battery, (b) dissipated as thermal energy in the wire, and (c) dissipated as thermal energy in the battery?

Short Answer

Expert verified
Energy transferred: 79.2 J, dissipated in wire: 65.34 J, in battery: 13.07 J.

Step by step solution

01

Finding the Total Resistance

The total resistance in the circuit consists of the resistance of the wire and the internal resistance of the battery. Therefore, the total resistance \( R_{\text{total}} \) is the sum of these two resistances: \[ R_{\text{total}} = R_{\text{wire}} + R_{\text{internal}} = 5.0 \Omega + 1.0 \Omega = 6.0 \Omega. \]
02

Calculating the Current

Using Ohm's law \( V = IR \), where \( V \) is the emf of the battery, \( I \) is the current, and \( R \) is the total resistance, we can solve for the current \( I \): \[ I = \frac{V}{R_{\text{total}}} = \frac{2.0 \mathrm{~V}}{6.0 \Omega} = 0.33 \mathrm{~A}. \]
03

Calculating Energy Transferred from Battery

The energy transferred from chemical form in the battery, also known as the energy supplied by the battery, is calculated using \( E = IVt \), where \( t \) is the time in seconds. First convert minutes to seconds: \( 2.0 \text{ minutes} = 120 \text{ seconds} \). Now, \[ E_{\text{total}} = (2.0 \mathrm{~V})(0.33 \mathrm{~A})(120 \mathrm{~s}) = 79.2 \mathrm{~J}. \]
04

Energy Dissipated in the Wire

The energy dissipated as thermal energy in the wire can be found using \( E = I^2Rt \), where \( R \) is the resistance of the wire alone. Therefore, \[ E_{\text{wire}} = (0.33 \mathrm{~A})^2(5.0 \Omega)(120 \mathrm{~s}) = 65.34 \mathrm{~J}. \]
05

Energy Dissipated in the Battery

Similarly, the energy dissipated as thermal energy in the internal resistance of the battery is \( E = I^2Rt \), where \( R \) is now the internal resistance of the battery. So, \[ E_{\text{battery}} = (0.33 \mathrm{~A})^2(1.0 \Omega)(120 \mathrm{~s}) = 13.07 \mathrm{~J}. \]

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 91Ó°ÊÓ!

Key Concepts

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

Ohm's Law
Ohm's Law is a fundamental principle in the study of electric circuits. It describes the relationship between voltage (\( V \)), current (\( I \)), and resistance (\( R \)). The formula is simple: \( V = IR \). This means:
  • Voltage is the potential difference provided by a power source like a battery.
  • Current is the flow of electric charge through the circuit.
  • Resistance is how much the circuit resists the flow of current.
Given these relationships, you can easily determine any one of these elements if the other two are known. For instance, if you know the voltage across a circuit and the resistance, Ohm's Law lets you find the current: \( I = \frac{V}{R} \). This relationship is fundamental in determining how circuits operate and how energy is distributed across them.
In the given problem, using Ohm's Law, the current in the circuit was calculated based on the total resistance and the emf of the battery. This illustrates how the law helps us understand the flow of electricity in a practical setup.
Resistance
Resistance is the property of a material that resists the flow of electric current, turning some of that electrical energy into heat. It is measured in ohms (\( \Omega \)). The resistance of a wire or a component in a circuit determines how much current will flow through it for a given voltage. Higher resistance leads to lower current, assuming the voltage is constant.
  • Factors influencing resistance include the material's nature, length, and cross-sectional area. For example, longer wires or those with smaller cross-sections tend to have higher resistance.
  • Resistors are specific components designed to bring a particular resistance into a circuit.
In the exercise, the total resistance was calculated by adding the resistance of the wire to the internal resistance of the battery, which gave a total of \( 6.0 \Omega \). By understanding resistance, you gain insight into how different materials and configurations impact current flow in electric circuits.
Energy Dissipation
When current flows through a resistance, electrical energy is transformed into heat. This process is called energy dissipation. This is why electrical appliances often heat up when in use. The energy dissipated in a resistor can be calculated using the formula: \( E = I^2Rt \), where \( E \) is the energy in joules, \( I \) is the current in amperes, \( R \) is the resistance in ohms, and \( t \) is the time in seconds.
  • Understanding dissipation helps in designing circuits to ensure they handle generated heat efficiently and safely.
  • Real-world scenario: This dissipation explains why wires can get hot if too much current flows through them.
In the exercise, energy dissipation was calculated separately for the wire and the battery. Each had different resistances, hence different energy dissipations: \( 65.34 \text{ J} \) in the wire and \( 13.07 \text{ J} \) in the battery. These calculations help in understanding how energy supplied by the battery translates into heat, leading to thermal effects in the components.

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

When resistors 1 and 2 are connected in series, the equivalent resistance is \(16.0 \Omega .\) When they are connected in parallel, the equivalent resistance is \(3.0 \Omega .\) What are (a) the smaller resistance and (b) the larger resistance of these two resistors?

A certain car battery with a \(12.0 \mathrm{~V} \mathrm{emf}\) has an initial charge of \(120 \mathrm{~A} \cdot \mathrm{h}\). Assuming that the potential across the terminals stays constant until the battery is completely discharged, for how many hours can it deliver energy at the rate of \(100 \mathrm{~W} ?\)

Nine copper wires of length \(l\) and diameter \(d\) are connected in parallel to form a single composite conductor of resistance \(R\). What must be the diameter \(D\) of a single copper wire of length \(l\) if it is to have the same resistance?

A total resistance of \(3.00 \Omega\) is to be produced by connecting an unknown resistance to a \(12.0 \Omega\) resistance. (a) What must be the value of the unknown resistance, and (b) should it be connected in series or in parallel?

The following table gives the electric potential difference \(V_{T}\) across the terminals of a battery as a function of current \(i\) being drawn from the battery. (a) Write an equation that represents the relationship between \(V_{T}\) and \(i .\) Enter the data into your graphing calculator and perform a linear regression fit of \(V_{T}\) versus \(i\). From the parameters of the fit, find (b) the battery's emf and (c) its internal resistance. $$ \begin{array}{llllllll} \hline i(\mathrm{~A}): & 50.0 & 75.0 & 100 & 125 & 150 & 175 & 200 \\ V_{T}(\mathrm{~V}): & 10.7 & 9.00 & 7.70 & 6.00 & 4.80 & 3.00 & 1.70 \\ \hline \end{array} $$

See all solutions

Recommended explanations on Physics Textbooks

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.