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Chapter 25: Problem 44

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.

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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 鈥� 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.

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Most popular questions from this chapter

A magnetic field of \(0.080 \mathrm{~T}\) is in the \(y\) -direction. The velocity of wire segment \(S\) has a magnitude of \(78 \mathrm{~m} / \mathrm{s}\) and components of \(18 \mathrm{~m} / \mathrm{s}\) in the \(x\) -direction, \(24 \mathrm{~m} / \mathrm{s}\) in the \(y\) -direction, and \(72 \mathrm{~m} / \mathrm{s}\) in the \(z\) -direction. The segment has length \(0.50 \mathrm{~m}\) and is parallel to the \(z\) -axis as it moves. (a) Find the motional emf induced between the ends of the segment. (b) What would the motional emf be if the wire segment was parallel to the \(y\) -axis?

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 typical small flashlight contains two batteries, each having an emf of \(1.5 \mathrm{~V},\) connected in series with a bulb having resistance \(17 \Omega .\) (a) If the internal resistance of the batteries is negligible, what power is delivered to the bulb? (b) If the batteries last for \(5.0 \mathrm{~h}\), what is the total energy delivered to the bulb? (c) The resistance of real batteries increases as they run down. If the initial internal resistance is negligible, what is the combined internal resistance of both batteries when the power to the bulb has decreased to half its initial value? (Assume that the resistance of the bulb is constant. Actually, it will change somewhat when the current through the filament changes, because this changes the temperature of the filament and hence the resistivity of the filament wire.

The resistivity of a semiconductor can be modified by adding different amounts of impurities. A rod of semiconducting material of length \(L\) and cross- sectional area \(A\) lies along the \(x\) -axis between \(x=0\) and \(x=L\). The material obeys Ohm's law, and its resistivity varies along the rod according to \(\rho(x)=\rho_{0} \exp (-x / L) .\) The end of the rod at \(x=0\) is at a potential \(V_{0}\) greater than the end at \(x=L\). (a) Find the total resistance of the rod and the current in the rod. (b) Find the electric-field magnitude \(E(x)\) in the rod as a function of \(x\). (c) Find the electric potential \(V(x)\) in the rod as a function of \(x\). (d) Graph the functions \(\rho(x), E(x),\) and \(V(x)\) for values of \(x\) between \(x=0\) and \(x=L\)

\(\mathrm{A}\) cell phone or computer battery has three ratings marked on it: a charge capacity listed in mAh (milliamp-hours), an energy capacity in Wh (watt-hours), and a potential rating in volts. (a) What are these three values for your cell phone? (b) Convert the charge capacity \(Q\) into coulombs. (c) Convert the energy capacity \(U\) into joules. (d) Multiply the charge rating \(Q\) by the potential rating \(V,\) and verify that this is equivalent to the energy capacity \(U\). (e) If the charge \(Q\) were stored on a parallel-plate capacitor with air as the dielectric, at the potential \(V,\) what would be the corresponding capacitance? (f) If the energy in the battery were used to heat \(1 \mathrm{~L}\) of water, estimate the corresponding change in the water temperature? (The heat capacity of water is \(4190 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K} .)\)
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