/*! 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 44 A battery has emf \(\mathcal{E}\... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A battery has emf \(\mathcal{E}\) and internal resistance \(r=2.00 \Omega\). A \(12.0 \Omega\) resistor is connected to the battery, and the resistor consumes electrical power at a rate of \(96.0 \mathrm{~J} / \mathrm{s}\). What is the emf of the battery?

Short Answer

Expert verified
The emf of the battery is calculated through the formulas for Electric Power and Ohm's law. After finding the value of the current \(I\) with the Power formula, it is substituted in Ohm's law along with the resistance values to find the emf of the battery.

Step by step solution

01

Calculate the Current

Use the power dissipated in the external resistor to calculate the current, using the formula \(P = I^2 R\), where \(P\) is the power, \(I\) is the current and \(R\) is the resistance. In the given problem, \(P = 96.0 W\) and \(R = 12.0 \Omega\). Arrange the equation to find \(I\), giving \[I = \sqrt{P/R}\].
02

Substitute values into the Current formula

Substitute \(P = 96.0 W\) and \(R = 12.0 \Omega\) into the formula \[I = \sqrt{P/R}\] to find the current.
03

Calculate the emf

Use Ohm's law, which states that the emf equals the current times the total resistance, to calculate the emf. The formula is \(\mathcal{E} = I(R+r)\). Substitute the calculated current \(I\) and given resistance values \(R = 12.0 \Omega\) and \(r = 2.0 \Omega\) into the formula to find \(\mathcal{E}\)
04

Substitute values into the emf formula

Substitute the found current and the resistance values into the formula \(\mathcal{E} = I(R+r)\) to find the emf.

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.

Internal Resistance
In the world of electronics, when we discuss a battery or any other electrical source, we often refer to an ideal version with no flaws. However, in reality, every battery comes with some level of internal resistance. This resistance is a hindrance within the battery itself that opposes the flow of electric charge.

Imagine it as a form of friction that electrons encounter within the battery's materials. This internal resistance, often denoted as r, impacts how the battery can deliver electrical energy to an external circuit. The higher the internal resistance, the larger the energy that's converted to heat inside the battery, rather than being used to do useful work in the external circuit.

Understanding internal resistance helps us in calculating a battery's electromotive force (emf) by taking into account the voltage drop due to internal resistance – key to discovering the true output of the battery under load.
Ohm's Law
Ohm's law is a fundamental principle in the field of electrical circuits, expressing the relationship between voltage, current, and resistance. It states that the current (I) flowing through a conductor between two points is directly proportional to the voltage (V) across the two points, and inversely proportional to the resistance (R) of the conductor. The formula is succinctly expressed as V = IR.

This law is crucial when working with electrical circuits as it allows us to calculate any one of the three variables if we know the other two. In the context of our battery exercise, Ohm's law helps in determining the electromotive force (emf) of the battery by accounting for the current flowing through and the total resistance encountered (which includes both external and internal resistance).
Electrical Power Dissipation
When electrons flow through a resistor, they collide with the particles that make up the resistor, and energy is lost in the form of heat. This loss of energy is known as electrical power dissipation, and it's why devices get warm when they're used. In an electrical circuit, power dissipation can be calculated using the formula P = IV, where P is the power in watts, I is the current in amperes, and V is the voltage in volts.

There's also another useful variation of this formula: P = I^2R, which was used in the given exercise to calculate the current, based on the power dissipated in the resistor and the resistance. Power dissipation is an essential concept for understanding energy efficiency and thermal management in electronic systems.
Direct Current Circuits
When we talk about direct current (DC) circuits, we're referring to a system where electric charge flows in one consistent direction. It's the type of electrical power supplied by batteries, DC power supplies, and similar sources. Unlike alternating current (AC) circuits, where the flow of charge periodically changes direction, DC provides a constant voltage or current.

Key components in DC circuits include sources like batteries, resistors, capacitors, and sometimes, in more complex circuits, inductors and diodes. Ohm's law plays a pivotal role in understanding and designing these circuits, as it allows for precise calculations of current, voltage, and resistance throughout the system. The performance of a DC circuit is also affected by factors like internal resistance, which, as we've learned, impacts the actual voltage available to external 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

A battery has emf \(30.0 \mathrm{~V}\) and internal resistance \(r .\) A \(9.00 \Omega\) resistor is connected to the terminals of the battery, and the voltage drop across the resistor is \(27.0 \mathrm{~V}\). What is the internal resistance of the battery?

Displacement Current in a Dielectric. Suppose that the parallel plates in Fig. 29.23 have an area of \(3.00 \mathrm{~cm}^{2}\) and are separated by a 2.50 -mm- thick sheet of dielectric that completely fills the volume between the plates. The dielectric has dielectric constant \(4.70 .\) (You can ignore fringing effects.) At a certain instant, the potential difference between the plates is \(120 \mathrm{~V}\) and the conduction current \(i_{\mathrm{C}}\) equals \(6.00 \mathrm{~mA} .\) At this instant, what are (a) the charge \(q\) on each plate; (b) the rate of change of charge on the plates; (c) the displacement current in the dielectric?

Two cylindrical cans with insulating sides and conducting end caps are filled with water, attached to the circuitry shown in Fig. \(\mathbf{P} 25.69,\) and used to determine salinity levels. The cans are identical, with radius \(r=5.00 \mathrm{~cm}\) and length \(L=3.00 \mathrm{~cm} .\) The battery supplies a potential of \(10.0 \mathrm{~V},\) has a negligible internal resistance, and is connected in series with a resistor \(R=15.0 \Omega .\) The left cylinder is filled with pure distilled water, which has infinite resistivity. The right cylinder is filled with a saltwater solution. It is known that the resistivity of the saltwater solution is determined by the relationship \(\rho=\left(s_{0} / s\right) \Omega \cdot \mathrm{m},\) where \(s\) is the salinity in parts per thousand \((\mathrm{ppt})\) and \(s_{0}=6.30\) ppt. (a) The ammeter registers a current of \(484 \mathrm{~mA}\). What is the salinity of the saltwater solution? (b) The left cylinder acts as a capacitor. Use Eq. (24.19) for its capacitance. How much charge is present on its upper plate? Note that pure water has a dielectric constant of \(80.4 .\) (c) At what rate is energy dissipated by the saltwater? (d) For what salinity level would the \(15.0 \Omega\) resistor dissipate half the power supplied by the battery?

A copper wire has a square cross \(2.3 \mathrm{~mm}\) on a side. The wire is \(4.0 \mathrm{~m}\) long and carries a current of \(3.6 \mathrm{~A}\). The density of free electrons is \(8.5 \times 10^{28} / \mathrm{m}^{3} .\) Find the magnitudes of (a) the current density in the wire and (b) the electric field in the wire. (c) How much time is required for an electron to travel the length of the wire?

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?

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.