Physics

Chapter 9: Current and Resistance

Imron Rosyadi

Chapter Overview

This chapter covers the fundamental concepts of electrical current, resistance, and their applications.

We will explore:

• The definition and direction of electrical current.
• A model for conduction in metals, including drift velocity.
• The concepts of resistivity and resistance.
• Ohm’s Law and its applications.
• Electrical energy and power.
• The fascinating phenomenon of superconductors.

9.1 Electrical Current: Learning Objectives

By the end of this section, you will be able to:

• Describe an electrical current.
• Define the unit of electrical current.
• Explain the direction of current flow.

We’ll start by defining what electrical current is and how it’s measured. We’ll also cover the convention for its direction.

What is Electrical Current?

Electrical current is the rate at which charge flows.

• Charges move in response to an electric field created by a voltage difference.
• In conductors, charges experience a force from the electric field.
• They lose kinetic energy to the material, reaching a drift velocity.

Note

Think of it like an object falling through air: it speeds up, but then air resistance causes it to reach a constant terminal velocity. Similarly, electrons in a conductor are accelerated by the electric field but collide with atoms, maintaining a constant average drift velocity.

Historically, we first dealt with static charges. Now, we’re focusing on charges in motion, particularly through materials. The key here is the rate of flow. If many charges move quickly, you have a large current. If few charges move slowly, you have a small current. The concept of “drift velocity” is important because, while electric signals travel very fast, individual electrons move quite slowly due to collisions.

Defining Current and the Ampere

The average electrical current (I) is defined as:

\[I = \frac{\Delta Q}{\Delta t}\]

Where:

• \(\Delta Q\) is the net charge passing through a cross-sectional area.
• \(\Delta t\) is the time interval.

The SI unit for current is the ampere (A).

\[1 \, \text{A} = 1 \, \frac{\text{C}}{\text{s}}\]

Important

One ampere means one coulomb of charge passes a point per second.

This formula tells us that current is essentially how much charge goes by in a certain amount of time. It’s like measuring the flow rate of water in a pipe. André-Marie Ampère, a French physicist, is honored with this unit. It’s good to remember the definition: 1 Ampere = 1 Coulomb per second.

Instantaneous Current

For situations where the charge flow might not be constant, we define instantaneous electrical current as the time derivative of the charge:

\[I = \lim_{\Delta t \to 0} \frac{\Delta Q}{\Delta t} = \frac{dQ}{dt}\]

This allows us to analyze currents that change over time.

Figure 9.2: The rate of flow of charge is current. An ampere is the flow of one coulomb of charge through an area in one second. A current of one amp would result from \(6.25 \times 10^{18}\) electrons flowing through the area A each second.

Just like with velocity being the derivative of position, instantaneous current is the derivative of charge with respect to time. This is useful for circuits where current isn’t steady, such as when a capacitor is charging or discharging.

Example 9.1: Calculating Average Current

Problem: A truck battery moves 720 C of charge in 4.00 s to start an engine.

1. What is the average current?
2. How long does it take for 1.00 C of charge to flow?

Strategy:

1. Use \(I = \Delta Q / \Delta t\).
2. Rearrange \(I = \Delta Q / \Delta t\) to solve for \(\Delta t\).

This example applies the basic definition of average current to a practical scenario – starting a car engine, which requires a significant current.

Example 9.1: Solution

Part (a):

Given \(\Delta Q = 720 \, \text{C}\) and \(\Delta t = 4.00 \, \text{s}\).

\[I = \frac{\Delta Q}{\Delta t} = \frac{720 \, \text{C}}{4.00 \, \text{s}} = 180 \, \text{C/s} = 180 \, \text{A}\]

Part (b):

Given \(\Delta Q = 1.00 \, \text{C}\) and \(I = 180 \, \text{A}\).

\[\Delta t = \frac{\Delta Q}{I} = \frac{1.00 \, \text{C}}{180 \, \text{C/s}} = 5.56 \times 10^{-3} \, \text{s} = 5.56 \, \text{ms}\]

Tip

A large current means a large amount of charge moves in a short time.

The 180 A current highlights that starting an engine requires a very rapid flow of charge. The short time for 1 C to flow shows the speed at which charges are delivered when current is high.

Example 9.2: Calculating Instantaneous Current

Problem: A charge flows through a wire, modeled as \(Q(t) = Q_M(1 - e^{-t/\tau})\). Find the current through the wire.

Figure 9.3: A graph of the charge moving through a cross-section of a wire over time.

Strategy: Use \(I = dQ/dt\).

This example introduces a time-dependent charge function, leading to a time-dependent current. It’s a good way to see how calculus applies to electrical concepts.

Example 9.2: Solution

Given \(Q(t) = Q_M(1 - e^{-t/\tau})\).

\[I = \frac{dQ}{dt} = \frac{d}{dt} [Q_M(1 - e^{-t/\tau})]\] \[I = Q_M \frac{d}{dt} [1 - e^{-t/\tau}] = Q_M (0 - e^{-t/\tau} \cdot (-\frac{1}{\tau}))\] \[I = \frac{Q_M}{\tau} e^{-t/\tau}\]

Figure 9.4: A graph of the current flowing through the wire over time.

Note

The current decreases exponentially, typical for charging/discharging capacitors.

The result shows that the current is initially at its maximum and then decays exponentially. This kind of behavior is commonly observed in RC circuits, which we will study later.

Current in a Circuit

For current to flow, there must be a complete path (circuit).

Figure 9.5: (a) Simple circuit with headlight, battery, and switch. (b) Schematic with open switch. (c) Schematic with closed switch and current flow.

• Conventional current flow: From positive terminal to negative terminal.
  • This is the direction positive charge would flow.
• In metal wires, electrons (negative charges) actually move in the opposite direction.
• The electric field drives this motion, not increasing kinetic energy, but maintaining a drift velocity.

The concept of a complete circuit is fundamental. Without it, charges can’t continuously flow. We differentiate between conventional current (historical convention of positive charge flow) and actual electron flow (negative charges moving). While electrons move from negative to positive, the effect on the circuit is the same as positive charges moving from positive to negative, so the convention works. The battery provides the potential energy to keep charges moving against the “friction” of the material.

Direction of Current Flow

Figure 9.6: (a) Positive charges move in the direction of the electrical field (conventional current). (b) Negative charges move opposite to the electrical field. Conventional current is opposite to negative charge movement.

• Benjamin Franklin’s convention (1700s): Current flows from “positive” to “negative” electrical fluid.
• We now know current in metals is primarily carried by electrons (negative charges).
• Electrons move from the negative terminal to the positive terminal.
• However, the conventional current direction remains positive to negative.
• An electrical field is always present in conductors carrying current, maintaining charge motion.

This is a common point of confusion for students. It’s important to stress that “conventional current” is a historical convention that we stick to, even though the actual charge carriers in metals (electrons) move in the opposite direction. The physical effects of the current are the same regardless of which direction we imagine the positive charges are moving.

9.2 Model of Conduction in Metals: Learning Objectives

By the end of this section, you will be able to:

• Define the drift velocity of charges moving through a metal.
• Define the vector current density.
• Describe the operation of an incandescent lamp.

Here we’ll delve deeper into how charges actually move within a conductor, leading to the concepts of drift velocity and current density, and an application in light bulbs.

Drift Velocity

Electrical signals travel very rapidly (\(\sim 10^8 \, \text{m/s}\)), but individual electrons move very slowly (\(\sim 10^{-4} \, \text{m/s}\)).

• Signal speed: Due to rapid propagation of electric field changes.
• Electron speed: Electrons collide with atoms and other electrons, causing a “zig-zag” path.

Figure 9.8: A typical path of one electron. Collisions cause random motion, but an applied electric field gives a net drift velocity opposite to the field.

The drift velocity (\(\vec{v}_d\)) is the average velocity of the free charges.

Note

Collisions transfer energy from electrons to atoms, often increasing temperature. This requires continuous power input to maintain current (except in superconductors).

It’s crucial to distinguish between the speed of the electrical signal (which is very fast, nearly the speed of light) and the average speed of the actual charge carriers (electrons), which is quite slow. The analogy of an incompressible fluid is helpful here: when you push water into one end of a full pipe, water comes out the other end almost instantly, even though the individual water molecules don’t travel that fast through the pipe. The energy dissipated as heat is why wires get warm and light bulbs glow.

Current and Drift Velocity Relationship

Consider a segment of wire with cross-sectional area \(A\).

• Number density of free charges: \(n\) (number of charges/volume)
• Charge on each carrier: \(q\)

In time \(dt\), charges in a volume \(A v_d dt\) move out. The total charge \(dQ\) in this volume is \(q n A v_d dt\).

The current \(I\) is then:

\[I = \frac{dQ}{dt} = q n A v_d\]

Rearranging for drift velocity:

\[v_d = \frac{I}{n q A}\]

Tip

The drift velocity is inversely proportional to the number density of free charges and the cross-sectional area.

Figure 9.10: All the charges in the shaded volume of this wire move out in a time dt, having a drift velocity of magnitude vd.

This derivation is important as it connects macroscopic current to microscopic properties like charge density and drift velocity. It highlights that even with a significant current, if there are many charge carriers (high \(n\)) or a large area (\(A\)), the individual carriers don’t need to move very fast.

Example 9.3: Calculating Drift Velocity

Problem: Calculate the drift velocity of electrons in a 12-gauge copper wire (diameter = 2.053 mm) carrying a 20.0-A current. Copper has one free electron per atom, density of \(8.80 \times 10^3 \, \text{kg/m}^3\), and atomic mass of 63.54 g/mol.

Strategy:

1. Calculate \(n\) (number density of free electrons).
2. Calculate \(A\) (cross-sectional area).
3. Use \(v_d = I / (n q A)\).

This is a multi-step calculation that combines concepts from chemistry (molar mass, Avogadro’s number) with physics (current, charge). It gives a realistic value for drift velocity, often surprisingly small.

Example 9.3: Solution

1. Number density of free electrons (\(n\)):

   \(n = \frac{1 \, \text{e}^-}{\text{atom}} \times \frac{6.02 \times 10^{23} \, \text{atoms}}{\text{mol}} \times \frac{1 \, \text{mol}}{63.54 \, \text{g}} \times \frac{1000 \, \text{g}}{\text{kg}} \times \frac{8.80 \times 10^3 \, \text{kg}}{1 \, \text{m}^3}\)

   \(n = 8.34 \times 10^{28} \, \text{e}^-/\text{m}^3\)

2. Cross-sectional area (\(A\)):

   \(r = 2.053 \, \text{mm} / 2 = 1.0265 \times 10^{-3} \, \text{m}\)

   \(A = \pi r^2 = \pi (1.0265 \times 10^{-3} \, \text{m})^2 = 3.31 \times 10^{-6} \, \text{m}^2\)

3. Drift velocity (\(v_d\)):

   \(v_d = \frac{I}{n q A} = \frac{20.00 \, \text{A}}{(8.34 \times 10^{28} \, \text{/m}^3)(-1.60 \times 10^{-19} \, \text{C})(3.31 \times 10^{-6} \, \text{m}^2)}\)

   \(v_d = -4.54 \times 10^{-4} \, \text{m/s}\)

Warning

The minus sign indicates electrons move opposite to conventional current.

The small magnitude of the drift velocity is the key takeaway here. It reinforces that the electrical signal’s rapid travel is not due to individual electron speed but rather the domino effect of charge repulsion.

Current Density (\(\vec{J}\))

Current is a scalar, but current density is a vector quantity (\(\vec{J}\)).

• It describes the flow of charge through an infinitesimal area, including local magnitude and direction.
• Units: Amperes per meter squared (\(\text{A/m}^2\)).
• Direction: Direction of net flow of positive charges.

Relationship between current and current density:

\[dI = \vec{J} \cdot d\vec{A} = J dA \cos\theta\]

\[I = \int_A \vec{J} \cdot d\vec{A}\]

Magnitude of current density:

\[J = \frac{I}{A} = n |q| v_d\]

So, \(\vec{J} = n q \vec{v}_d\).

Figure 9.12: The current density J is a vector whose direction is the net flow of positive charges.

Current density is a more detailed way to describe current, especially when the current isn’t uniform across a cross-section or when we need to consider the direction more explicitly. It’s proportional to the current and inversely proportional to the area. If charge \(q\) is positive, \(\vec{v}_d\) is in the same direction as \(\vec{E}\). If \(q\) is negative (like electrons), \(\vec{v}_d\) is opposite to \(\vec{E}\), but \(\vec{J}\) is still in the direction of \(\vec{E}\).

Example 9.4: Calculating Current Density

Problem: A lamp with a 100-W bulb uses 0.87 A of current. It’s wired with 10-gauge copper wire (diameter = 2.588 mm). Find the magnitude of the current density.

Strategy:

1. Calculate the cross-sectional area \(A\) of the wire.
2. Use the formula \(J = I/A\).

This example directly applies the definition of current density. It helps to understand how the current is distributed over the wire’s cross-section.

Example 9.4: Solution

1. Cross-sectional area (\(A\)):

   \(r = 2.588 \, \text{mm} / 2 = 1.294 \times 10^{-3} \, \text{m}\)

   \(A = \pi r^2 = \pi (1.294 \times 10^{-3} \, \text{m})^2 = 5.26 \times 10^{-6} \, \text{m}^2\)

2. Current density (\(J\)):

   Given \(I = 0.87 \, \text{A}\).

   \[J = \frac{I}{A} = \frac{0.87 \, \text{A}}{5.26 \times 10^{-6} \, \text{m}^2} = 1.65 \times 10^5 \, \text{A/m}^2\]

Tip

For a given current, a larger wire diameter (larger area) results in a smaller current density.

The current density is quite large, but it makes sense given the relatively small cross-sectional area of a wire. This concept is fundamental for understanding how wires are rated and why different gauges are used for different applications.

9.3 Resistivity and Resistance: Learning Objectives

By the end of this section, you will be able to:

• Differentiate between resistance and resistivity.
• Define the term conductivity.
• Describe the electrical component known as a resistor.
• State the relationship between resistance of a resistor and its length, cross-sectional area, and resistivity.
• State the relationship between resistivity and temperature.

Here, we introduce the concepts that quantify how materials oppose current flow, and how this property is used in circuit components.

Resistivity (\(\rho\)) and Conductivity (\(\sigma\))

When a voltage is applied, an electric field (\(\vec{E}\)) causes a current density (\(\vec{J}\)).

For many materials (like metals at a given temperature), \(\vec{J}\) is proportional to \(\vec{E}\):

\[\vec{J} = \sigma \vec{E}\]

• Electrical conductivity (\(\sigma\)): A measure of a material’s ability to conduct electricity.
  • Units: \((\Omega \cdot \text{m})^{-1}\).
• Resistivity (\(\rho\)): A measure of how strongly a material opposes current flow.
  • It is the reciprocal of conductivity: \(\rho = 1/\sigma\).
  • Units: \(\Omega \cdot \text{m}\).

\[E = \rho J\]

Important

• Conductors: High \(\sigma\), low \(\rho\).
• Insulators: Low \(\sigma\), high \(\rho\).
• Semiconductors: Intermediate \(\rho\), strongly dependent on impurities and temperature.

Resistivity is an intrinsic property of the material itself (like density). Conductivity is just the inverse. Good conductors have low resistivity (e.g., copper, silver), while insulators have high resistivity (e.g., glass, rubber). This foundational relationship (\(J=\sigma E\) or \(E=\rho J\)) is the microscopic form of Ohm’s Law, which we’ll discuss soon.

Temperature Dependence of Resistivity

The resistivity of most conducting metals increases with increasing temperature.

• Increased atomic vibrations impede electron motion.

For many materials, this dependence is approximately linear:

\[\rho = \rho_0 (1 + \alpha \Delta T)\]

Where:

• \(\rho\): Resistivity at temperature \(T\).
• \(\rho_0\): Resistivity at a reference temperature \(T_0\) (usually \(20.00^\circ \text{C}\)).
• \(\alpha\): Temperature coefficient of the material.
• \(\Delta T = T - T_0\).

Note

For semiconductors, \(\alpha\) is often negative, meaning resistivity decreases with increasing temperature.

This temperature dependence is important in many applications. For example, incandescent light bulbs have higher resistance when hot. It’s also the basis for thermistors, which are temperature-dependent resistors used in thermometers.

Resistance (R)

Resistance is a measure of how difficult it is for current to pass through a specific wire or component.

• It depends on the material’s resistivity and the component’s geometry.

For a cylindrical conductor of length \(L\), cross-sectional area \(A\), and resistivity \(\rho\):

\[R = \rho \frac{L}{A}\]

• Units: Ohm (\(\Omega\)).
• Higher resistance means lower current for a given voltage.

Figure 9.13: A potential is applied to a segment of a conductor. Voltage is proportional to current.

Unlike resistivity (an intrinsic material property), resistance is a property of the object. A long, thin wire will have higher resistance than a short, thick wire of the same material. This formula is extremely practical for designing wires and resistors.

Resistors

Resistors are common electronic components used to:

• Reduce current flow.
• Provide a voltage drop.

Symbols:

Figure 9.14: (a) ANSI symbol; (b) IEC symbol.

Figure 9.16: Resistor color bands indicate resistance and tolerance.

Note

Resistors are often made of carbon or metal film. Their value is indicated by color codes.

Resistors are one of the most basic and ubiquitous components in electrical circuits. They allow us to control current and voltage in specific parts of a circuit. The color code is a practical way to identify their resistance value without needing a meter.

Example 9.5: Current Density, Resistance, and Electric Field

Problem: Calculate the current density, resistance, and electrical field for a 5-m length of 12-gauge copper wire (diameter = 2.053 mm) carrying \(I = 10 \, \text{mA}\).

Strategy:

1. Calculate cross-sectional area \(A\).
2. Calculate current density \(J = I/A\).
3. Calculate resistance \(R = \rho L / A\).
4. Calculate electrical field \(E = \rho J\). (Resistivity of copper \(\rho = 1.68 \times 10^{-8} \, \Omega \cdot \text{m}\))

This example ties together current density, resistance, and the electric field within a conductor, using the material’s resistivity.

Example 9.5: Solution

1. Cross-sectional area (\(A\)):

   \(r = 2.053 \, \text{mm} / 2 = 1.0265 \times 10^{-3} \, \text{m}\)

   \(A = \pi r^2 = \pi (1.0265 \times 10^{-3} \, \text{m})^2 = 3.31 \times 10^{-6} \, \text{m}^2\)

2. Current density (\(J\)):

   \(J = \frac{I}{A} = \frac{10 \times 10^{-3} \, \text{A}}{3.31 \times 10^{-6} \, \text{m}^2} = 3.02 \times 10^3 \, \text{A/m}^2\)

3. Resistance (\(R\)):

   \(R = \rho \frac{L}{A} = (1.68 \times 10^{-8} \, \Omega \cdot \text{m}) \frac{5.00 \, \text{m}}{3.31 \times 10^{-6} \, \text{m}^2} = 0.025 \, \Omega\)

4. Electrical field (\(E\)):

   \(E = \rho J = (1.68 \times 10^{-8} \, \Omega \cdot \text{m}) (3.02 \times 10^3 \, \text{A/m}^2) = 5.07 \times 10^{-5} \, \text{V/m}\)

Tip

Copper has very low resistance, making it an excellent conductor.

The resistance of this 5-meter segment of copper wire is very small, which explains why copper is so effective for wiring. The electric field is also quite small, which is typical for good conductors.

Temperature Dependence of Resistance

Since \(R = \rho L / A\), and resistivity \(\rho\) depends on temperature, resistance \(R\) also depends on temperature:

\[R = R_0 (1 + \alpha \Delta T)\]

Where:

• \(R\): Resistance at temperature \(T\).
• \(R_0\): Original resistance (usually at \(20.00^\circ \text{C}\)).
• \(\alpha\): Temperature coefficient of the material.
• \(\Delta T = T - T_0\).

Figure 9.17: Digital thermometers often use thermistors, whose resistance changes with temperature.

This temperature dependence is crucial. For instance, in an incandescent light bulb, the filament’s resistance increases dramatically as it heats up, affecting the current. This principle is also used in resistance thermometers and thermistors.

Example 9.6: Calculating Resistance Change

Problem: A tungsten filament at \(20^\circ \text{C}\) has a resistance of \(0.350 \, \Omega\). What is its resistance if the temperature increases to \(2850^\circ \text{C}\)? (\(\alpha\) for tungsten = \(4.5 \times 10^{-3} \, (^\circ \text{C})^{-1}\))

Strategy: Use the formula \(R = R_0 (1 + \alpha \Delta T)\).

This example shows a very large temperature change, characteristic of incandescent filaments, and the corresponding significant change in resistance.

Example 9.6: Solution

Given \(R_0 = 0.350 \, \Omega\), \(\alpha = 4.5 \times 10^{-3} \, (^\circ \text{C})^{-1}\).

\(\Delta T = T - T_0 = 2850^\circ \text{C} - 20^\circ \text{C} = 2830^\circ \text{C}\).

\[R = R_0 (1 + \alpha \Delta T)\]

\[R = (0.350 \, \Omega) [1 + (4.5 \times 10^{-3} \, (^\circ \text{C})^{-1})(2830 \, ^\circ \text{C})]\]

\[R = (0.350 \, \Omega) [1 + 12.735]\]

\[R = (0.350 \, \Omega) (13.735) = 4.80 \, \Omega\]

Important

The resistance increases by more than a factor of 10! This means initial current will be much higher than operating current.

This result explains why incandescent light bulbs often burn out at the moment they are turned on – the initial rush of current through the cold (low resistance) filament is much higher.

Resistance of Coaxial Cable

Problem: Determine the resistance of a coaxial cable of length \(L\), with inner conductor radius \(r_i\) and outer conductor radius \(r_o\), filled with an insulator of resistivity \(\rho\).

Figure 9.18: Coaxial cable structure.

Strategy: The resistance formula \(R = \rho L/A\) cannot be used directly. Instead, consider infinitesimal cylindrical shells and integrate.

This problem introduces an important application of resistance calculation that requires integration, as the “area” for leakage current changes with radius. This is a common technique when dealing with non-uniform geometries.

Example 9.7: Solution (Coaxial Cable)

Consider a cylindrical shell of thickness \(dr\) at radius \(r\). The area for radial current flow is \(A = 2\pi r L\).

The differential resistance \(dR\) is:

\[dR = \rho \frac{dr}{A} = \rho \frac{dr}{2\pi r L}\]

Integrate from the inner radius \(r_i\) to the outer radius \(r_o\):

\[R = \int_{r_i}^{r_o} \frac{\rho}{2\pi L} \frac{1}{r} dr = \frac{\rho}{2\pi L} \int_{r_i}^{r_o} \frac{1}{r} dr\]

\[R = \frac{\rho}{2\pi L} [\ln(r)]_{r_i}^{r_o} = \frac{\rho}{2\pi L} (\ln(r_o) - \ln(r_i))\]

\[R = \frac{\rho}{2\pi L} \ln \left(\frac{r_o}{r_i}\right)\]

Note

This resistance accounts for leakage current between conductors, causing signal attenuation.

This result shows that the resistance between the two conductors of a coaxial cable depends on its length, the radii of the conductors, and the resistivity of the insulating material. This “leakage” current is typically undesirable as it degrades the signal.

9.4 Ohm’s Law: Learning Objectives

By the end of this section, you will be able to:

• Describe Ohm’s law.
• Recognize when Ohm’s law applies and when it does not.

Now we bring together voltage, current, and resistance into a single, very important relationship: Ohm’s Law. We’ll also discuss its limitations.

Description of Ohm’s Law

Ohm’s Law describes a relationship between voltage, current, and resistance for many materials.

• Experimentally observed by Georg Simon Ohm.
• States that current (\(I\)) in a metal wire is directly proportional to the voltage (\(V\)) applied.

\[V = IR\]

Where:

• \(V\): Voltage (Volts)
• \(I\): Current (Amperes)
• \(R\): Resistance (Ohms)

Important

• Ohmic materials/components: Obey Ohm’s law (linear V-I relationship).
• Nonohmic materials/components: Do not obey Ohm’s law (non-linear V-I relationship).

Ohm’s Law is often considered the most fundamental relationship in circuit analysis. It’s an empirical law, meaning it’s based on observation, not derived from first principles. It’s a simplification, and not all materials or devices follow it perfectly, but for many common materials like metals, it’s an excellent approximation. It’s a macroscopic description, flowing from the microscopic relationship \(E=\rho J\).

Ohm’s Experiment

Georg Ohm’s experiment: Measured current and voltage across various circuits.

Figure 9.19: Experimental setup to determine if a resistor is ohmic. Voltmeter in parallel, ammeter in series.

Figure 9.20: Plot of voltage versus current for an ohmic resistor. The linear relationship confirms Ohm’s Law.

Note

The slope of the V-I graph is the resistance \(R\).

Ohm’s original experiments showed this linear relationship, which forms the basis of the law. The experimental setup shown is standard for verifying Ohm’s law or characterizing a resistor. The direction of current doesn’t change the linear relationship, only the sign of V and I.

Example 9.8: Measuring Resistance

Problem: A carbon resistor at \(20^\circ \text{C}\) is connected to a 9.00-V battery. Current is 3.00 mA.

1. What is the resistance at room temperature?
2. If temperature increases to \(60^\circ \text{C}\), what is the new current? (\(\alpha\) for carbon = \(-0.0005 \, (^\circ \text{C})^{-1}\))

Strategy:

1. Use Ohm’s law: \(R = V/I\).
2. Use temperature dependence of resistance: \(R = R_0 (1 + \alpha \Delta T)\), then Ohm’s law again for new current.

This example combines Ohm’s law with the temperature dependence of resistance, showing how temperature can affect circuit behavior. Notice that carbon has a negative temperature coefficient.

Example 9.8: Solution

Part (a): Resistance at room temperature (\(R_0\))

\(R_0 = \frac{V}{I} = \frac{9.00 \, \text{V}}{3.00 \times 10^{-3} \, \text{A}} = 3.00 \times 10^3 \, \Omega = 3.00 \, \text{k}\Omega\)

Part (b): Current at \(60^\circ \text{C}\)

\(\Delta T = 60^\circ \text{C} - 20^\circ \text{C} = 40^\circ \text{C}\).

New resistance \(R\):

\(R = R_0 (1 + \alpha \Delta T) = 3.00 \times 10^3 \, \Omega (1 + (-0.0005 \, (^\circ \text{C})^{-1})(40^\circ \text{C}))\)

\(R = 3.00 \times 10^3 \, \Omega (1 - 0.02) = 3.00 \times 10^3 \, \Omega (0.98) = 2.94 \, \text{k}\Omega\)

New current \(I\):

\(I = \frac{V}{R} = \frac{9.00 \, \text{V}}{2.94 \times 10^3 \, \Omega} = 3.06 \times 10^{-3} \, \text{A} = 3.06 \, \text{mA}\)

Note

For carbon, resistance decreases with increasing temperature, leading to a slightly higher current.

The 2% change in current might seem small, but in sensitive electronic circuits, such changes can be significant. This is why temperature control (e.g., fans in computers) is important for electronics.

Nonohmic Devices

Many devices do not obey Ohm’s law. Their current-voltage relationship is non-linear.

Example: Diode

• A semiconductor device that allows current flow in only one direction.
• Anode positive, cathode negative = forward bias (conducts).
• Anode negative, cathode positive = reverse bias (very little current).

Figure 9.22: I-V curve for a diode. Clearly non-linear, especially in reverse bias and near forward voltage threshold.

The diode is a classic example of a nonohmic device. Its behavior is highly non-linear, with distinct forward and reverse bias characteristics. This non-linearity is fundamental to how semiconductor devices work and is why they are essential for modern electronics.

9.5 Electrical Energy and Power: Learning Objectives

By the end of this section, you will be able to:

• Express electrical power in terms of the voltage and the current.
• Describe the power dissipated by a resistor in an electric circuit.
• Calculate the energy efficiency and cost effectiveness of appliances and equipment.

This section connects current, voltage, and resistance to the concepts of energy and power, which are critical for understanding how much work electrical systems do and how much they cost to operate.

Power in Electric Circuits

Electrical energy is converted into other forms (e.g., thermal, light).

• Kinetic energy from accelerating electrons is converted to thermal energy via collisions.

Electric Power (\(P\)): Rate at which energy is transferred.

\[P = IV\]

Where:

• \(P\): Power (Watts, W)
• \(I\): Current (Amperes, A)
• \(V\): Voltage (Volts, V)

Using Ohm’s Law (\(V=IR\)), we can derive other forms for power dissipated by a resistor:

\[P = I^2 R\]

\[P = \frac{V^2}{R}\]

Important

Power dissipated by a resistor is converted into heat and/or light.

Figure 9.23: Higher power bulbs glow brighter (a) but CFLs (b) achieve similar brightness with less power.

These are fundamental power equations. They are used constantly in electrical engineering. Emphasize that power is the rate of energy transfer. The three forms of the power equation (\(IV\), \(I^2R\), \(V^2/R\)) are all equivalent for ohmic devices and are chosen based on what variables are known. The \(I^2R\) term represents Joule heating, which is why wires get hot.

Example 9.9: Calculating Power in Electric Devices

Problem: A DC winch motor is rated at 20.00 A and 115 V. It lifts a 4900.00 N object 10.00 m in 30.00 s at constant speed.

1. What is the power consumed by the motor?
2. What is the power used in lifting the object?
3. Estimate the resistance of the motor (assuming power difference is dissipated as heat).

Strategy:

1. \(P = IV\).
2. \(P = Fv\), where \(v = d/t\).
3. \(P_{dissipated} = I^2 R\).

This example breaks down the power consumed by a motor into useful work and dissipated heat, illustrating the concept of efficiency.

Example 9.9: Solution

Part (a): Power consumed by the motor (\(P_{in}\))

\(P_{in} = IV = (20.00 \, \text{A})(115 \, \text{V}) = 2300.00 \, \text{W}\)

Part (b): Power used lifting the object (\(P_{out}\))

Speed \(v = \frac{10.00 \, \text{m}}{30.00 \, \text{s}} = 0.333 \, \text{m/s}\)

Force \(F = 4900.00 \, \text{N}\) (equal to weight for constant speed)

\(P_{out} = Fv = (4900.00 \, \text{N})(0.333 \, \text{m/s}) = 1617.00 \, \text{W}\) (approx \(1633.33 \text{W}\) if using more decimal places for \(v\))

Part (c): Resistance of the motor

Power dissipated as heat \(P_{heat} = P_{in} - P_{out} = 2300.00 \, \text{W} - 1617.00 \, \text{W} = 683.00 \, \text{W}\) Using \(P_{heat} = I^2 R\):

\(R = \frac{P_{heat}}{I^2} = \frac{683.00 \, \text{W}}{(20.00 \, \text{A})^2} = \frac{683.00 \, \text{W}}{400.00 \, \text{A}^2} = 1.71 \, \Omega\)

The difference between input and output power represents the inefficiency, often lost as heat due to the motor’s internal resistance. This is why motors get warm during operation.

The Cost of Electricity

You pay for energy used, not power. Energy \((E)\) is power \((P)\) multiplied by time \((t)\):

\[E = Pt\]

• Common unit for electric bills: kilowatt-hour (kW·h).
  • \(1 \, \text{kW·h} = 3.6 \times 10^6 \, \text{J}\).

Energy efficiency is key to saving money and reducing environmental impact.

• LEDs and CFLs are much more efficient than incandescent bulbs.

Figure 9.26: Summary of relationships between P, I, V, R for ohmic devices.

It’s important for students to understand the difference between power and energy for practical reasons, especially regarding utility bills. Emphasize the shift towards more energy-efficient lighting (CFLs, LEDs) due to their lower power consumption for the same light output and longer lifespan.

9.6 Superconductors: Learning Objectives

By the end of this section, you will be able to:

• Describe the phenomenon of superconductivity.
• List applications of superconductivity.

We conclude the chapter with a look at a truly remarkable phenomenon: superconductivity, where materials lose all electrical resistance at very low temperatures.

The Resistance of Mercury

In 1911, Heike Kamerlingh Onnes discovered that the resistance of mercury abruptly dropped to zero at a critical temperature (\(T_c\)) of \(4.2 \, \text{K}\) (\(-269.2^\circ \text{C}\)).

Figure 9.27: Resistance of mercury drops to zero at \(T_c = 4.2 \, \text{K}\) (superconducting phase).

• This phenomenon is called superconductivity.
• Zero resistance means no \(I^2R\) power losses.
• Currents, once established, can persist indefinitely without a voltage source.

This discovery was groundbreaking because it meant that at sufficiently low temperatures, some materials could conduct electricity with absolutely no energy loss. This has huge implications for energy transmission and storage. Note that many good normal conductors (like copper, silver) are not superconductors.

Other Superconducting Materials

• Research has found other materials with higher critical temperatures (\(T_c\)).
  • Niobium-nitride: \(T_c = 16 \, \text{K}\)
  • Vanadium-silicon: \(T_c = 17.5 \, \text{K}\)
• High-temperature superconductors (HTS):
  • Discovered in 1986, ceramic compounds like YBCO (yttrium barium copper oxide).
  • \(T_c = 92 \, \text{K}\) (\(-181^\circ \text{C}\)), which is above the boiling point of liquid nitrogen (\(77 \, \text{K}\)).
  • Liquid nitrogen is much cheaper for cooling than liquid helium.

Important

Superconductors also exhibit the Meissner effect: they exclude magnetic fields from their interior. This leads to magnetic levitation!

Figure 9.28: A magnet levitating over a superconductor due to the Meissner effect.

The quest for higher \(T_c\) superconductors is a major area of research. Reaching temperatures accessible with liquid nitrogen (\(77 K\)) was a huge leap, making some applications more feasible. The Meissner effect is a dramatic and visually striking property, illustrating the unique interaction of superconductors with magnetic fields.

Types of Superconductors

Type I Superconductors:

• About 30 pure metals.
• Exhibit zero resistivity and Meissner effect below \(T_c\).
• Superconductivity destroyed by relatively weak magnetic fields (\(H_c\)).
• Limited practical applications due to low \(H_c\).
• Described by BCS theory.

Type II Superconductors:

• Ceramics (like YBCO) and alloys.
• Much higher critical magnetic fields (\(H_c\)).
• Can carry much higher current densities.
• Include “high-temperature” superconductors.

Tip

BCS Theory (Bardeen, Cooper, Schrieffer): Explains Type I superconductivity. Electrons form “Cooper pairs” through lattice vibrations (phonons), which behave as bosons and can condense into a common quantum state, allowing current without resistance.

Understanding the two types helps explain why some materials are more practical for applications. The BCS theory, while complex, provides the quantum mechanical basis for zero resistance. The idea of electrons pairing up and behaving like bosons is a fascinating quantum effect.

Applications of Superconductors

Superconductors have numerous cutting-edge applications:

1. Superconducting Magnets:
   • 10 times stronger than electromagnets.
   • Used in Magnetic Resonance Imaging (MRI) for high-quality medical images.
   • Used in particle accelerators (e.g., CERN’s LHC) to guide particle beams.
2. Maglev Trains:
   • Utilize superconducting magnets for levitation, eliminating friction.
   • Achieve extremely high speeds (e.g., Japanese Maglev train, 603 km/h).
3. SQUIDs (Superconducting QUantum Interference Devices):
   • Extremely sensitive magnetometers.
   • Measure subtle magnetic fields (e.g., brain activity, geological surveys).
   • Based on Josephson junctions.

Figure 9.29: A SQUID device, utilizing Josephson junctions.

These applications demonstrate the transformative potential of superconductivity. From medical diagnostics to high-speed transport and fundamental physics research, the properties of zero resistance and the Meissner effect are exploited for significant technological advancements. The ongoing challenge is to find materials that superconduct at higher, more practical temperatures.

Key Takeaways

1. Electrical Current (\(I\)) is the rate of charge flow (\(\Delta Q / \Delta t\)), measured in Amperes (\(\text{C/s}\)). Conventional current flows from positive to negative.
2. Drift Velocity (\(v_d\)) is the slow average velocity of charge carriers, while electrical signals travel much faster due to electric field propagation. \(I = nqAv_d\).
3. Current Density (\(\vec{J}\)) is a vector quantity representing current per unit area, \(\vec{J} = nq\vec{v}_d\).
4. Resistivity (\(\rho\)) is an intrinsic material property measuring opposition to current flow (\(E = \rho J\)), while Resistance (\(R\)) is a property of a component (\(R = \rho L/A\)). Both are temperature-dependent.
5. Ohm’s Law (\(V = IR\)) describes a linear relationship between voltage and current for ohmic materials. Not all devices (e.g., diodes) are ohmic.
6. Electrical Power (\(P\)) is the rate of energy transfer. It can be expressed as \(P = IV = I^2R = V^2/R\). Energy consumed is \(E = Pt\), measured in kilowatt-hours (kW·h).
7. Superconductivity is the phenomenon of zero electrical resistance below a critical temperature (\(T_c\)), along with the Meissner effect (magnetic field exclusion). Superconductors have applications in strong magnets (MRI), Maglev trains, and SQUIDs.

Key Equations

Equation	Description
\(I = \Delta Q / \Delta t\)	Average electrical current
\(I = dQ / dt\)	Instantaneous electrical current
\(I = nqAv_d\)	Current in terms of drift velocity
\(\vec{J} = nq\vec{v}_d\)	Current density vector
\(E = \rho J\)	Microscopic Ohm’s Law (resistivity and electric field)
\(\rho = \rho_0 (1 + \alpha \Delta T)\)	Temperature dependence of resistivity
\(R = \rho L/A\)	Resistance of a cylindrical conductor
\(R = R_0 (1 + \alpha \Delta T)\)	Temperature dependence of resistance
\(V = IR\)	Ohm’s Law (macroscopic)
\(P = IV\)	Electrical power
\(P = I^2R\)	Power dissipated by a resistor
\(P = V^2/R\)	Power dissipated by a resistor
\(E = Pt\)	Electrical energy consumed

Key Terms

Term	Definition
Electrical Current (I)	The rate at which electric charge flows, measured in amperes (A).
Ampere (A)	The SI unit of electrical current, equal to one coulomb per second (C/s).
Conventional Current	The direction in which positive charge would flow, from positive to negative terminals.
Drift Velocity (\(\vec{v}_d\))	The average velocity of charge carriers in a material due to an electric field.
Current Density (\(\vec{J}\))	A vector quantity defined as the current per unit cross-sectional area, indicating the local magnitude and direction of charge flow.
Resistivity (\(\rho\))	An intrinsic property of a material quantifying its opposition to the flow of electric current, measured in ohm-meters (\(\Omega \cdot \text{m}\)).
Conductivity (\(\sigma\))	The reciprocal of resistivity, measuring a material’s ability to conduct electricity, measured in \((\Omega \cdot \text{m})^{-1}\).
Resistance (R)	A measure of how difficult it is to pass current through a specific component or wire, measured in ohms (\(\Omega\)).
Ohm (\(\Omega\))	The SI unit of resistance, equal to one volt per ampere (V/A).
Resistor	An electrical component designed to introduce resistance into a circuit.
Ohm’s Law	The empirical relationship stating that current is directly proportional to voltage (\(V=IR\)) for ohmic materials.
Ohmic Material	A material or component that obeys Ohm’s law, exhibiting a linear voltage-current relationship.
Nonohmic Material	A material or component that does not obey Ohm’s law, exhibiting a non-linear voltage-current relationship (e.g., diodes).
Electrical Power (P)	The rate at which electrical energy is transferred or dissipated, measured in watts (W).
Kilowatt-hour (kW·h)	A unit of electrical energy commonly used for billing, equal to 3.6 million joules.
Superconductivity	A phenomenon where certain materials exhibit zero electrical resistance and expel magnetic fields below a critical temperature.
Critical Temperature (\(T_c\))	The temperature below which a material becomes a superconductor.
Meissner Effect	The expulsion of a magnetic field from the interior of a superconductor as it transitions into the superconducting state.
Cooper Pairs	Pairs of electrons that interact through lattice vibrations in superconductors, enabling current flow without resistance (BCS theory).
SQUID	Superconducting QUantum Interference Device, a highly sensitive magnetometer based on superconductivity.
Josephson Junction	A weak link between two superconductors through which a supercurrent can flow.