Chapter 02 · Current Electricity · Full book

Current Electricity

Core intuition: Chapter Roadmap

The previous two chapters held charges still. This chapter sets them moving, and in doing so trades the clean geometry of electrostatics for the messier, more practical world of circuits — batteries, wires, resistors, meters, and the steady flow that lights a room. The surprise is how little new physics is required: conservation of charge and conservation of energy, applied to a network, are very nearly the whole story. The chapter develops in seven movements:

Why charge moves, and what current is. Breaking electrostatic equilibrium; the field inside a live conductor; current as a flux of charge, and the drift of electrons that carries it.
Resistance, from the outside and the inside. Ohm's law as a ratio, not a law of nature; the Drude relaxation-time model that explains where resistance comes from and why it grows with temperature.
Resistivity and geometry. Separating the material (\(\rho\)) from the shape (\(L/A\)); how resistivity varies across metals, semiconductors, and insulators.
Building and solving networks. Series and parallel reduction; real cells with internal resistance and terminal voltage; Kirchhoff's two rules as bookkeeping for charge and energy.
Measuring without disturbing. Turning a galvanometer into an ammeter and a voltmeter — and why every real meter loads the circuit; then the null methods (Wheatstone bridge, metre bridge, potentiometer) that sidestep the problem entirely.
Power and heat. Where the energy goes: Joule heating, the three power formulas, and the maximum-power-transfer condition.
The first time-dependent circuit. Charging and discharging a capacitor through a resistor; the time constant \(\tau = RC\) as the circuit's own clock.

Perplexing Questions

Perplexing questions

Before we begin, let these questions sit with you. Do not reach for formulas. You will return to each one—with full answers—before this chapter ends.

The Crawling Electron: The free electrons in a copper wire drift at roughly \(0.1\,\mathrm{mm}\) per second under a household current. At that speed, an electron would take over two hours to travel one metre. Yet the light at the end of a one-metre wire turns on within a nanosecond of the switch being thrown. How can something travel at two hours per metre and arrive in a nanosecond?
The Vanishing Field: In the previous chapter, we proved that the electric field inside a conductor is exactly zero. The proof was rigorous. Yet every copper wire in a working circuit carries a non-zero electric field inside it—that field is, in fact, precisely what drives the current. Is the earlier proof wrong? Or are these two facts compatible?
The Battery's Secret: A battery is rated at \(9\,\mathrm{V}\). But when you connect a large load to it, the voltage you measure across its terminals is only \(7\,\mathrm{V}\) or \(6\,\mathrm{V}\). Where did the rest of the voltage go? The battery has not been damaged—it is doing exactly what it is supposed to do. What is happening?
The Series Bulb Paradox: A \(60\,\mathrm{W}\) bulb and a \(100\,\mathrm{W}\) bulb are connected in series across the mains supply. Most students predict the \(100\,\mathrm{W}\) bulb glows more brightly—after all, it is the more powerful bulb. In fact, the \(60\,\mathrm{W}\) bulb is brighter. How?
The Heating Wire: Two resistors are connected to the same battery. The first has twice the resistance of the second. Which one dissipates more power? The answer depends on how they are connected—and reverses completely depending on whether the connection is series or parallel. Why should the same two resistors rank differently under the same battery?
The Backward Battery: In a circuit with two batteries of different voltages, one of them may be receiving charge rather than supplying it. A battery that is being “charged” has current flowing into its positive terminal. How does the circuit “know” which battery is charging and which is discharging? And how can you tell from the equations alone?

Each question targets a trap that catches students every year.
By the end of this chapter, every one of them will be transparent.

Why Charges Move

In the previous chapter, every question we asked about charges assumed they were stationary. We computed forces, fields, potentials, and energies for charges that had been placed in position and left alone. The word electrostatics advertises this limitation: the physics of charges at rest.

But charges need not remain at rest. A conductor is filled with free electrons—roughly one per atom in a metal—each of which is free to wander through the lattice. Why, then, did we spend an entire chapter treating conductors as though their charges were fixed?

Because in electrostatic equilibrium, they are.

Electrostatic Equilibrium Revisited

Recall the central result from Section : inside a conductor in electrostatic equilibrium, the electric field is exactly zero. Any non-zero field would exert a force on the free electrons, accelerating them until they redistribute themselves to cancel the field. The charges rearrange until the field inside vanishes and every point in the conductor sits at the same potential.

This is a self-correcting process. It is also the end of the story—as long as nothing from outside maintains a potential difference across the conductor.

Breaking Equilibrium

Now imagine connecting the two ends of a copper wire to two bodies at different electrostatic potentials—say, the two plates of a charged capacitor. At the instant of connection, the potential at one end of the wire is higher than at the other. The wire is a conductor, so free electrons inside it experience an electric field directed from high potential to low potential.

These electrons begin to drift. They drift toward the higher-potential end, because the force on a negative charge is opposite to the electric field. As they accumulate there, they begin to neutralise the charge on the high-potential plate, and the potential difference between the plates falls.

Given enough time, the charges redistribute until both plates, the wire, and everything connected to them reach a common potential. The field inside the wire drops to zero. Equilibrium is restored, and charge flow ceases.

This is exactly what happens when you discharge a capacitor through a wire: a brief, transient current that dies away as equilibrium re-establishes itself.

The key observation is this: current flowed only as long as a potential difference existed. Once the system reached a uniform potential, the driving force for charge motion vanished.

Sustaining the Flow

If you want charges to flow continuously, you need something that maintains a potential difference despite the flow of charge that naturally acts to equalise it. This is the role of a source of electromotive force (EMF): a battery, a generator, a solar cell—any device that does work on charges to maintain a potential difference across its terminals.

Inside a battery, chemical reactions do work on positive charges, pushing them from the low-potential terminal (negative) to the high-potential terminal (positive) against the electric field. The energy comes from chemical potential energy stored in the reactants. This internal, non-electrostatic force is what prevents the system from reaching equilibrium.

Outside the battery, charges flow through the external circuit from high potential to low potential, driven by the electric field established by the battery. Inside the battery, charges are pushed from low to high potential by the non-electrostatic force. The result is a closed loop of steady charge flow—a steady current.

Core intuition: The Condition for Steady Current

A steady electric current requires:

A closed conducting path through which charge can circulate.
A source of EMF that maintains a potential difference, continuously doing work on the charge carriers to drive them around the loop.

Without a source of EMF, any initial current is transient: charges redistribute, the conductor reaches equilibrium, and current ceases. The transition from electrostatics to current electricity is precisely the transition from systems that reach equilibrium to systems where an external energy source prevents equilibrium from ever being achieved.

The Electric Field Inside a Current-Carrying Conductor

Here is where students must make a sharp conceptual break from the previous chapter.

In electrostatics, the electric field inside a conductor is zero. We proved this; we used it repeatedly; it became reflexive. Now, in a conductor carrying a steady current, the electric field inside is not zero.

There is no contradiction. The electrostatic result rested on the assumption that charges are in equilibrium—not accelerating, not flowing. When a battery maintains a potential difference across a wire, that assumption no longer holds. A non-zero field exists inside the wire, directed along it, and this field is precisely what drives the steady drift of charge carriers.

Misconception alert: “The Electric Field Inside a Conductor Is Always Zero”

The misconception: Students apply the electrostatic result \(\vec{E} = 0\) inside a conductor to all situations, including wires carrying current. Why it is wrong: The result \(\vec{E} = 0\) was derived under the condition of electrostatic equilibrium—no net motion of charge. In a current-carrying conductor, charges are drifting under the influence of an applied field. The field is small but non-zero, typically on the order of \(10^{-2}\,\mathrm{V/m}\) in household wiring. The correct statement: The electric field inside a conductor is zero in electrostatic equilibrium. When current flows, a non-zero \(\vec{E}\) exists inside the conductor, directed along the wire, and is the direct cause of the current. JEE/NEET trap: Problems that ask for the field “inside a conductor” are testing whether you check for equilibrium. A conductor in a circuit is not in electrostatic equilibrium.

From Field to Transport

In electrostatics, we studied what fields look like: their geometry, their flux, their relationship to potential. Now we study what fields do to matter.

The electric field inside a current-carrying conductor exerts a force on each free electron: \(\vec{F} = -e\vec{E}\). But the electron does not accelerate indefinitely. Within the lattice of a metal, an electron travels only a short distance before colliding with an ion or a lattice vibration (phonon), losing its acquired drift momentum. It is then re-accelerated by the field, drifts briefly, and collides again.

The result is not smooth, uniform acceleration but a slow, erratic shuffle superimposed on the vigorous thermal motion of the electron. The net effect, averaged over many collisions and many electrons, is a small average velocity in the direction of the electric force. This average velocity is called the drift velocity, and it is the microscopic origin of macroscopic electric current.

We will quantify this in Section , but the qualitative picture is essential: current is not a stream of electrons racing through a wire; it is an enormous number of electrons shuffling slowly and collectively, each contributing a tiny fraction of the total charge transport.

✓ Checkpoint · Why Charges Move

A copper wire connects the two plates of an isolated charged capacitor. Describe qualitatively what happens to the current in the wire as a function of time, and explain why the current eventually stops.
Current is maximum at the instant of connection (maximum potential difference) and decays exponentially to zero as the capacitor discharges. It stops because the charge redistribution eliminates the potential difference across the wire; once both plates are at the same potential, the electric field inside the wire is zero and there is no force to drive further charge flow.
A student claims: “The electric field inside the copper wire of a working flashlight circuit is zero, because copper is a conductor.” Identify the error and state the correct physics.
The student is applying the electrostatic equilibrium result (\(\vec{E} = 0\) inside a conductor) to a non-equilibrium situation. A flashlight circuit has a battery maintaining a potential difference, so a non-zero electric field exists inside the wire, directed along it. This field drives the steady current through the filament and wire.
Why is a source of EMF necessary for a steady current but not for a transient current?
A transient current is driven by an existing potential difference (e.g., a charged capacitor) that is consumed by the current itself. A steady current requires continuous replenishment of the potential difference, which demands an external energy source (EMF) that does work on charges to push them against the electric field, maintaining the potential difference indefinitely.

Electric Current

We have established that charges move through conductors when a potential difference is maintained. The next question is natural: how much charge flows, and how fast?

Not “how fast do the electrons move”—that question is subtler than it appears and belongs to Section . The question here is simpler and more fundamental: at any cross-section of a wire, how much charge passes through per unit time?

Defining Current

Consider a cross-section of a conductor. Charge carriers—electrons in a metal, ions in an electrolyte—pass through this cross-section as they drift under the influence of the applied electric field.

In a time interval \(\Delta t\), suppose a net charge \(\Delta q\) crosses the surface. The average current through that cross-section is: \[ I_{\text{avg}} = \frac{\Delta q}{\Delta t}. \] If the rate of charge flow varies with time, we define the instantaneous current as: \[ I = \frac{dq}{dt}. \]

Core intuition: Electric Current

Electric current \(I\) is the rate of flow of electric charge through a cross-section: \[ I = \frac{dq}{dt}. \] SI unit: ampere (A), where \(1\,\mathrm{A} = 1\,\mathrm{C/s}\). The ampere is one of the seven SI base units. In fact, the coulomb is defined in terms of the ampere: \(1\,\mathrm{C}\) is the charge transported by a current of \(1\,\mathrm{A}\) flowing for \(1\,\mathrm{s}\). Current is a scalar quantity in the context of circuit analysis. It does not obey the rules of vector addition. Although current has an associated direction along the wire, it does not transform as a vector under rotation of coordinates.

Sign Convention and Conventional Current

In a metallic conductor, the charge carriers are electrons, which carry negative charge. When an electric field \(\vec{E}\) points from left to right along a wire, the force on electrons is \(\vec{F} = -e\vec{E}\), directed from right to left. Electrons drift to the left.

But by convention, the direction of current is defined as the direction in which positive charge would move under the same field. Positive charge would drift from left to right—in the direction of the field.

So the conventional current direction is opposite to the actual electron drift direction.

This is not an error. It is a historical convention established by Benjamin Franklin, who guessed (incorrectly) that positive charge was the mobile carrier. The convention survived because the mathematics of circuit analysis is identical regardless of the sign of the carrier: a flow of negative charge to the left transports the same current as an equal flow of positive charge to the right.

When does the sign of the carrier matter? In situations where the physical identity of the moving particle matters—the Hall effect, electrolysis, semiconductor physics—the sign convention must be handled with care. In basic circuit analysis, conventional current is sufficient and universal.

Steady Current and Charge Conservation

A current is called steady (or direct) if \(I\) does not change with time at any cross-section. For a steady current: \[ q = It, \] where \(q\) is the total charge transported through the cross-section in time \(t\).

An important consequence of charge conservation: in a steady state, the current entering any junction must equal the current leaving it. If this were not the case, charge would accumulate (or deplete) at the junction, creating an ever-growing electric field that would redirect current until balance was restored. This is the physical basis of Kirchhoff's current law, which we will formalise later in this chapter.

Physical Scale

To build physical intuition for the magnitude of current, consider these reference points:

Situation	Typical current
Nerve impulse in a single neuron	\(\sim 10^{-9}\,\mathrm{A}\) (nA)
Wristwatch circuit	\(\sim 10^{-6}\,\mathrm{A}\) (\(\mu\)A)
LED indicator	\(\sim 20\,\mathrm{mA}\)
Household light bulb	\(\sim 0.4\,\mathrm{A}\)
Electric kettle	\(\sim 8\)–\(10\,\mathrm{A}\)
Car starter motor	\(\sim 200\,\mathrm{A}\)
Lightning bolt (peak)	\(\sim 10^4\)–\(10^5\,\mathrm{A}\)

A current of \(1\,\mathrm{A}\) transports \(1\,\mathrm{C}\) of charge per second. Since each electron carries \(1.6 \times 10^{-19}\,\mathrm{C}\), this corresponds to approximately \(6.2 \times 10^{18}\) electrons crossing a cross-section every second. The individual electron contributes almost nothing; current is an overwhelmingly collective phenomenon.

Misconception alert: “Current Is Used Up Around the Circuit”

The misconception: Students often believe that current “gets used up” as it passes through a resistor or a bulb, so the current “after” the component is less than the current “before” it. Why it is wrong: Current is a measure of how much charge flows past a point per second. Charge is conserved. In a series circuit with steady current, the same number of electrons that enter one end of a resistor must exit the other end in the same time interval—otherwise charge would pile up inside the resistor, which does not happen in steady state. What is consumed is not current but electrical energy. The potential drops across the resistor; the charge carriers emerge with less energy per unit charge than they entered with. That energy is converted to heat. JEE/NEET trap: A problem states that a bulb in series with a resistor “uses up current.” The correct response: current is the same everywhere in a series loop; only the potential (voltage) drops across each element.

✓ Checkpoint · Electric Current

The current through a wire is given by \(I(t) = (6t^2 + 2)\,\mathrm{A}\), where \(t\) is in seconds. Find the total charge that flows through a cross-section between \(t = 1\,\mathrm{s}\) and \(t = 3\,\mathrm{s}\).
\(q = \int_1^3 (6t^2 + 2)\,dt = \bigl[2t^3 + 2t\bigr]_1^3 = (54+6)-(2+2) = 56\,\mathrm{C}\).
Electrons in a copper wire drift to the left. State the direction of conventional current, and explain why.
Conventional current is to the right. By convention, current direction is defined as the direction a positive charge would drift. Since electrons (negative) drift left, a positive carrier under the same field would drift right. Equivalently, the net transport of negative charge to the left is the same as net transport of positive charge to the right.
A lightning bolt transfers \(3\,\mathrm{C}\) of charge to the ground in \(2\,\mathrm{ms}\). Estimate the average current.
\(I_{\text{avg}} = q/\Delta t = 3/(2\times10^{-3}) = 1500\,\mathrm{A} = 1.5\,\mathrm{kA}\).
In a simple series circuit, a student measures \(I = 2.0\,\mathrm{A}\) before a light bulb and \(I = 2.0\,\mathrm{A}\) after it. The student is surprised: “if the bulb uses energy, where does the energy come from if the current doesn't decrease?” Answer the student's question.
The current (charge flow per second) is the same before and after the bulb because charge is conserved. The energy comes from the potential drop across the bulb: charge enters at a higher potential and exits at a lower potential, losing electrical potential energy which is converted to heat and light. It is voltage, not current, that drops across the bulb.

Current Density and Drift Velocity

◈ Drift velocity & Ohm’s law

live bench

Set the current, length and area — watch the electrons crawl and read v_d, R, V and J update live, with the charge-traverse time as the payoff.

◈ loading the interactive bench…

↗ open in the Simulations page

Current tells us how much charge flows through a wire per second. But it says nothing about how that flow is distributed across the wire's cross-section, or about the microscopic behaviour of the charge carriers responsible for it.

Two wires may carry the same current of \(1\,\mathrm{A}\), but if one wire has ten times the cross-sectional area of the other, the charge carriers in the thinner wire must be moving faster—or be more densely packed—or both. To capture this distinction, we need a quantity that describes the intensity of current flow at a point: the current density.

Current Density

Define the current density \(\vec{J}\) at a point inside a conductor as a vector whose magnitude equals the current per unit cross-sectional area perpendicular to the flow, and whose direction is the direction of conventional current at that point.

For a wire of uniform cross-section \(A\) carrying a steady current \(I\) distributed uniformly across the cross-section: \[ J = \frac{I}{A}. \] SI unit: \(\mathrm{A/m^2}\).

More generally, for a non-uniform distribution, the current through an arbitrary surface \(S\) is: \[ I = \int_S \vec{J} \cdot d\vec{A}. \] This is the precise relationship: current is the flux of current density through a surface.

Core intuition: Current Density

The current density \(\vec{J}\) at a point inside a conductor has:

Magnitude: \(|\vec{J}|\) = current per unit area perpendicular to the flow.
Direction: the direction of conventional current (i.e., the direction positive carriers would drift, or opposite to the electron drift in metals).

Unlike current \(I\), which is a scalar property of a cross-section, \(\vec{J}\) is a vector field defined at every point within the conducting material. It is the local, microscopic counterpart of the macroscopic, global quantity \(I\).

The Microscopic Picture: Thermal Motion Versus Drift

To connect current density to the behaviour of individual electrons, we must confront a physical fact that surprises nearly every student encountering it for the first time.

Free electrons in a metal at room temperature are not sitting still, waiting for an electric field to push them. They are in constant, violent thermal motion, bouncing off lattice ions at speeds of order \(10^5\) to \(10^6\,\mathrm{m/s}\). Their thermal kinetic energy is roughly \(\tfrac{3}{2}k_BT\), and at room temperature this translates to speeds far exceeding anything associated with the electric field in a circuit.

But this thermal motion is random. At any instant, as many electrons move to the left as to the right, as many move up as down. Averaged over any macroscopic time interval, the net displacement due to thermal motion is zero. No net charge transport results from thermal motion alone.

Now apply an electric field \(\vec{E}\) along the conductor. Between collisions with the lattice, each electron is accelerated by the force \(\vec{F} = -e\vec{E}\). It picks up a small additional velocity in the direction of the force, then collides with a lattice ion and its velocity is randomised again. The electron is re-accelerated, drifts briefly, collides, and so on.

The net effect, averaged over many collision cycles, is a tiny shift in the average velocity of the electron population. This shift is the drift velocity \(\vec{v}_d\), and it is superimposed on the much larger random thermal velocity.

Typical magnitudes:

Quantity	Typical value in copper
Thermal speed of electrons	\(\sim 10^5\,\mathrm{m/s}\)
Drift velocity at \(1\,\mathrm{A}\) in household wire	\(\sim 10^{-4}\,\mathrm{m/s}\)

The drift velocity is roughly a billion times smaller than the thermal speed. If you could watch a single electron, you would see it careening wildly in random directions, and you would need extraordinary patience to notice the imperceptible bias in one direction that constitutes the drift.

Yet this tiny drift, multiplied by the staggering number of free electrons per unit volume (\(\sim 10^{28}\,\mathrm{m^{-3}}\) in copper), produces a perfectly measurable macroscopic current.

Deriving the Relation: \(I = nqAv_d\)

◇ Figure 1

Fig 1. Under an applied field \(\vec{E}\), electrons drift slowly opposite to \(\vec{E}\) while the conventional current points along \(\vec{E}\). The drift speed is tiny compared with the random thermal speed.

Birth of the formula: Current and Drift Velocity

Consider a cylindrical segment of a conductor with uniform cross-sectional area \(A\). Let the charge carriers have:

number density \(n\) (carriers per unit volume),
charge \(q\) per carrier (positive for conventional treatment; for electrons, \(q = e\) and the sign is handled separately),
drift velocity of magnitude \(v_d\) directed along the conductor.

In a time interval \(dt\), each carrier advances a distance \(v_d\,dt\). All carriers within a cylindrical volume of length \(v_d\,dt\) and cross-section \(A\) will cross the chosen cross-section during \(dt\). Volume of this cylinder: \[ dV = A\,v_d\,dt. \] Number of carriers in this volume: \[ dN = n\,dV = nA\,v_d\,dt. \] Total charge crossing the cross-section: \[ dq = q\,dN = nqA\,v_d\,dt. \] Therefore the current is: \[ \boxed{I = \frac{dq}{dt} = nqAv_d.} \] In vector form, the current density is: \[ \boxed{\vec{J} = nq\vec{v}_d,} \] where \(\vec{v}_d\) is the drift velocity vector. For electrons in a metal, \(q = -e\), so \(\vec{J}\) points opposite to \(\vec{v}_d\)—consistent with the convention that current flows opposite to electron drift.

Reading the equation. The result \(I = nqAv_d\) makes transparent what controls the magnitude of current. Current is large when:

the carrier density \(n\) is large (metals have \(n \sim 10^{28}\, \mathrm{m^{-3}}\); semiconductors have much smaller \(n\)),
the cross-sectional area \(A\) is large (thicker wire, more carriers per cross-section),
the drift velocity \(v_d\) is large (stronger applied field).

A wire can carry a large current with a very small drift velocity, provided \(n\) and \(A\) are large enough. This is exactly the situation in metallic conductors.

Misconception alert: “Drift Velocity Is the Speed of Electricity”

The misconception: Students calculate drift velocity (\(\sim 10^{-4}\,\mathrm{m/s}\)) and conclude that it takes hours for the “electricity” to travel from a switch to a light bulb across a room. But the bulb turns on almost instantly. This seems contradictory. Why it is wrong: Drift velocity is the average speed of an individual electron's net displacement through the conductor. It is not the speed at which the electric field, the current, or the energy propagates through the circuit. When you close a switch, the electric field is established throughout the wire at nearly the speed of light (\(\sim 2 \times 10^8\,\mathrm{m/s}\) in copper, slightly less than \(c\) due to the dielectric properties of the insulation). Every electron in the wire begins to drift almost simultaneously. It is the field that propagates fast, not the electrons. Analogy: Push one end of a long queue of people. The push propagates down the line almost instantly (each person nudges the next), but no individual person moves very far. The signal speed (push propagation) is vastly greater than the transport speed (how far each person moves). JEE/NEET trap: A problem asks “how long does it take for an electron to travel from the battery to the bulb?” and gives wire length and drift velocity. The answer is hours. This does not mean the bulb takes hours to turn on—that would confuse drift velocity with signal velocity.

✓ Checkpoint · Current Density and Drift Velocity

A silver wire (one free electron per atom, \(\rho = 10{,}500\,\mathrm{kg/m^3}\), \(M = 108\,\mathrm{g/mol}\)) of cross-sectional area \(2.0 \times 10^{-6}\,\mathrm{m^2}\) carries a current of \(5.0\,\mathrm{A}\). Calculate the drift velocity.
\(n = \rho N_A / M = 10500 \times 6.022\times10^{23} / (0.108) \approx 5.85 \times 10^{28}\,\mathrm{m^{-3}}\). \(v_d = I/(neA) = 5.0 / (5.85\times10^{28} \times 1.6\times10^{-19} \times 2.0\times10^{-6}) \approx 2.7 \times 10^{-4}\,\mathrm{m/s}\).
Two wires made of the same metal carry the same current. Wire P has twice the diameter of wire Q. Compare their current densities and drift velocities.
Area scales as diameter squared, so \(A_P = 4A_Q\). Since \(J = I/A\): \(J_Q = 4J_P\). Since \(n\) is the same: \(v_{dQ} = 4\,v_{dP}\). The thinner wire has four times the current density and four times the drift velocity.
A student argues: “If drift velocity in copper is only \(\sim 10^{-4}\,\mathrm{m/s}\), then the electrons in a one-metre wire take hours to reach the other end, so the bulb should take hours to light up.” Identify the error.
The student confuses drift velocity with signal propagation speed. The electric field is established throughout the circuit at nearly the speed of light (\(\sim 2\times10^8\,\mathrm{m/s}\)), so all electrons begin drifting almost simultaneously. The bulb lights up within nanoseconds; no single electron needs to travel the full length of the wire.

Solved examples

Five fully-worked problems from this chapter, free — solution and answer shown in full. The complete set of worked examples is in the full book.

Example 1 · Charge Flow and Electron Count

A wire carries a steady current of \(3.2\,\mathrm{A}\).

How much charge passes through a cross-section of the wire in \(5\,\mathrm{s}\)?
How many electrons cross the cross-section in this time?
If the current increases linearly from zero to \(3.2\,\mathrm{A}\) over \(5\,\mathrm{s}\) (instead of being steady), how much charge passes through?

Solution. (a) For steady current: \[ q = It = (3.2)(5) = \boxed{16\,\mathrm{C}}. \] (b) Each electron carries charge \(e = 1.6 \times 10^{-19}\, \mathrm{C}\). The number of electrons: \[ n = \frac{q}{e} = \frac{16}{1.6 \times 10^{-19}} = \boxed{10^{20}}. \] This enormous number illustrates the collective nature of current: \(10^{20}\) electrons, each crawling at a fraction of a millimetre per second, together produce a modest household-scale current. (c) When current varies with time, we integrate: \[ q = \int_0^{5} I(t)\,dt. \] The current increases linearly: \(I(t) = (3.2/5)\,t = 0.64\,t\, \mathrm{A}\). \[ q = \int_0^{5} 0.64\,t\,dt = 0.64 \cdot \frac{t^2}{2}\bigg|_0^5 = 0.64 \times 12.5 = \boxed{8\,\mathrm{C}}. \] Interpretation: The linearly increasing current delivers exactly half the charge of the steady current, as expected from the area under a triangular \(I\)-\(t\) graph versus a rectangular one.

Example 2 · Drift Velocity in a Copper Wire

A copper wire of cross-sectional area \(A = 1.0 \times 10^{-6}\, \mathrm{m^2}\) (\(1\,\mathrm{mm^2}\)) carries a steady current of \(I = 1.5\,\mathrm{A}\). Copper has one free electron per atom, molar mass \(M = 63.5\,\mathrm{g/mol}\), and density \(\rho = 8900\,\mathrm{kg/m^3}\).

Find the free-electron number density \(n\).
Calculate the drift velocity \(v_d\).
How long would it take a single electron to drift \(1\,\mathrm{m}\) along the wire?

Solution. (a) Number density of free electrons: Each atom contributes one free electron, so \(n\) equals the number of atoms per unit volume. \[ n = \frac{\rho N_A}{M} = \frac{8900 \times 6.022 \times 10^{23}}{63.5 \times 10^{-3}} \approx \boxed{8.44 \times 10^{28}\,\mathrm{m^{-3}}}. \] (b) Drift velocity: From \(I = neAv_d\): \[ v_d = \frac{I}{neA} = \frac{1.5}{(8.44 \times 10^{28})(1.6 \times 10^{-19}) (1.0 \times 10^{-6})}. \] Computing the denominator: \[ neA = 8.44 \times 10^{28} \times 1.6 \times 10^{-19} \times 1.0 \times 10^{-6} = 1.35 \times 10^{4}\,\mathrm{A}. \] \[ v_d = \frac{1.5}{1.35 \times 10^4} \approx \boxed{1.1 \times 10^{-4}\,\mathrm{m/s}} \approx 0.11\,\mathrm{mm/s}. \] This is extraordinarily slow—a snail outpaces an electron in a wire by a comfortable margin. (c) Time to drift 1 m: \[ t = \frac{d}{v_d} = \frac{1}{1.1 \times 10^{-4}} \approx 9.1 \times 10^{3}\,\mathrm{s} \approx \boxed{2.5\,\text{hours}}. \] Interpretation: An individual electron takes over two hours to travel one metre. Yet the bulb at the end of a one-metre wire turns on within nanoseconds of closing the switch, because the electric field propagates at nearly the speed of light, setting all the electrons into drift almost simultaneously. The slow drift of each electron and the fast response of the circuit are entirely compatible.

Example 3 · Resistance, Ohm's Law, and Scale Reasoning

A resistor is connected to a \(6.0\,\mathrm{V}\) battery and carries a current of \(120\,\mathrm{mA}\).

Find the resistance.
A \(9.0\,\mathrm{V}\) source is substituted at the same temperature. What current flows?
The temperature is raised so that the resistance increases to \(75\,\Omega\). The \(9.0\,\mathrm{V}\) source remains. What current flows now, and by what percentage has it changed from (b)?

Solution. (a) \[ R = \frac{V}{I} = \frac{6.0}{0.120} = \boxed{50\,\Omega}. \] (b) The resistor is metallic and Ohmic at constant temperature, so \(R = 50\,\Omega\) still applies: \[ I = \frac{V}{R} = \frac{9.0}{50} = \boxed{0.18\,\mathrm{A} = 180\,\mathrm{mA}}. \] (c) With \(R = 75\,\Omega\): \[ I = \frac{9.0}{75} = 0.12\,\mathrm{A} = 120\,\mathrm{mA}. \] The current has decreased from \(180\,\mathrm{mA}\) to \(120\,\mathrm{mA}\)—a reduction of \(33\%\)—even though the voltage is higher than in part (a). This illustrates the central role of resistance: at fixed voltage, increasing \(R\) suppresses \(I\). Heating the resistor did not “help” the current; it actively hindered it.

Example 4 · Comparing Wires: Same Material, Different Geometry

Two aluminium wires are cut from the same spool (\(\rho = 2.82\times10^{-8}\,\Omega\,\mathrm{m}\)). Wire A: length \(3\,\mathrm{m}\), diameter \(1\,\mathrm{mm}\). Wire B: length \(0.5\,\mathrm{m}\), diameter \(3\,\mathrm{mm}\).

Find the resistance of each wire.
A \(12\,\mathrm{V}\) source is applied across each wire separately. Find the current through each.
Compare the current densities.

Solution. (a) Cross-sectional areas: \[ A_A = \pi(0.5\times10^{-3})^2 = 7.854\times10^{-7}\,\mathrm{m^2}, \quad A_B = \pi(1.5\times10^{-3})^2 = 7.069\times10^{-6}\,\mathrm{m^2}. \] Resistances: \[ R_A = \frac{2.82\times10^{-8}\times3}{7.854\times10^{-7}} \approx \boxed{0.108\,\Omega}, \] \[ R_B = \frac{2.82\times10^{-8}\times0.5}{7.069\times10^{-6}} \approx \boxed{2.0\times10^{-3}\,\Omega}. \] Wire B has a resistance about 54 times smaller than wire A, despite being the same material. It is 6 times shorter (factor 6) and has \(9\times\) the area (factor 9): total reduction \(6\times9 = 54\). (b) Currents: \[ I_A = \frac{12}{0.108} \approx 111\,\mathrm{A}, \qquad I_B = \frac{12}{2.0\times10^{-3}} = 6000\,\mathrm{A}. \] (c) Current densities: \[ J_A = \frac{I_A}{A_A} \approx \frac{111}{7.854\times10^{-7}} \approx 1.41\times10^{8}\,\mathrm{A/m^2}, \] \[ J_B = \frac{I_B}{A_B} \approx \frac{6000}{7.069\times10^{-6}} \approx 8.49\times10^{8}\,\mathrm{A/m^2}. \] \[ \frac{J_B}{J_A} \approx 6 = \frac{L_A}{L_B}. \] Wire B carries 54 times the current of A but through 9 times the area, so the current density ratio is \(54/9 = 6\). This equals the length ratio \(L_A/L_B\), which can be verified from \(J = E/\rho = (V/L)/\rho\): same voltage, same material, ratio of \(J\) values equals ratio of \(1/L\) values.

Example 5 · Rated Power and Operating Conditions

A bulb is rated \(60\,\mathrm{W}\), \(240\,\mathrm{V}\). It is connected in series with a \(480\,\Omega\) resistor across a \(240\,\mathrm{V}\) supply.

Find the resistance of the bulb at its operating temperature.
Find the current through the circuit.
Find the actual power dissipated in the bulb.
Compare this to the rated \(60\,\mathrm{W}\) and explain the discrepancy.

Solution. (a) At rated conditions, \(P = V^2/R_{\text{bulb}}\): \[ R_{\text{bulb}} = \frac{V^2}{P} = \frac{(240)^2}{60} = \frac{57600}{60} = \boxed{960\,\Omega}. \] (b) In series with \(480\,\Omega\): \[ I = \frac{240}{960+480} = \frac{240}{1440} = \frac{1}{6}\,\mathrm{A} \approx \boxed{0.167\,\mathrm{A}}. \] (c) Actual power in bulb: \[ P_{\text{bulb}} = I^2 R_{\text{bulb}} = \left(\frac{1}{6}\right)^2\times960 = \frac{1}{36}\times960 = \boxed{26.7\,\mathrm{W}}. \] (d) The bulb dissipates less than half its rated power. The rated \(60\,\mathrm{W}\) applies only when \(240\,\mathrm{V}\) appears directly across the bulb. Here, the series resistor drops part of the supply voltage (\(V_{480} = (1/6)\times480 = 80\,\mathrm{V}\)) leaving only \(V_{\text{bulb}} = 160\,\mathrm{V}\) across the bulb—not \(240\, \mathrm{V}\). The bulb glows noticeably dimmer than rated. Note also that \(R_{\text{bulb}}\) changes with temperature; at lower power the filament is cooler and \(R_{\text{bulb}}\) would be somewhat lower than \(960\,\Omega\), so the actual current would be slightly higher than calculated. A complete analysis requires the temperature coefficient, but this calculation gives a good first estimate.

Problem bank

Five questions from this chapter’s 50-question bank, free — attempt each one before you reveal the answer. The rest of the bank, and the timed test that draws on all of it, are in the full book.

✎ Free practice questions

Charge and Electron Count
A wire carries a steady current of \(2.4\,\mathrm{A}\). How many electrons cross a cross-section of the wire every second?
\(N = I/e = 2.4/(1.6\times10^{-19}) = 1.5\times10^{19}\) electrons per second.
Time-Varying Current
The current through a branch is \(I(t) = (4t^2 - 2t)\,\mathrm{A}\). Find the total charge that flows between \(t = 1\,\mathrm{s}\) and \(t = 3\,\mathrm{s}\).
\(q = \int_1^3(4t^2-2t)\,dt = \bigl[\tfrac{4t^3}{3}-t^2\bigr]_1^3 = (36-9) - (\tfrac{4}{3}-1) = 27 - \tfrac{1}{3} = \tfrac{80}{3} \approx 26.7\,\mathrm{C}\).
Current Density
A cylindrical conductor of radius \(1.5\,\mathrm{mm}\) carries \(3.0\,\mathrm{A}\). Find the current density, assuming uniform distribution.
\(A = \pi(1.5\times10^{-3})^2 \approx 7.07\times10^{-6}\,\mathrm{m^2}\). \(J = I/A = 3.0/7.07\times10^{-6} \approx 4.24\times10^5\,\mathrm{A/m^2}\).
Drift Velocity and Comparison
Two wires, one copper and one aluminium, have the same length and the same cross-sectional area. They carry the same current \(I\). Given \(n_{\mathrm{Cu}} = 8.5\times10^{28}\,\mathrm{m^{-3}}\) and \(n_{\mathrm{Al}} = 6.0\times10^{28}\,\mathrm{m^{-3}}\), find the ratio of drift velocities \(v_{d,\mathrm{Al}}/v_{d,\mathrm{Cu}}\). Which metal has the higher current density?
Since \(J = nev_d\) and \(J = I/A\) is the same for both: \(v_d \propto 1/n\). \(v_{d,\mathrm{Al}}/v_{d,\mathrm{Cu}} = n_{\mathrm{Cu}}/n_{\mathrm{Al}} = 8.5/6.0 \approx 1.42\). The current density is the same in both (same \(I\), same \(A\)).
Three-Cell Circuit
Three cells, each of EMF \(2\,\mathrm{V}\) and internal resistance \(1\,\Omega\), are connected so that two are in parallel and this combination is in series with the third. This assembly is connected to an external resistance \(R = 2\,\Omega\).
1. Find the equivalent EMF and internal resistance of the combination.
2. Find the current through \(R\) and through each cell.
3. Find the terminal voltage of the parallel pair and of the series cell.
(a) Two parallel cells: \(\mathcal{E}_{\text{par}} = 2\,\mathrm{V}\), \(r_{\text{par}} = 0.5\,\Omega\). In series with third: \(\mathcal{E}_{\text{total}} = 4\,\mathrm{V}\), \(r_{\text{total}} = 1.5\,\Omega\). (b) \(I = 4/(2+1.5) = 4/3.5 = 8/7\,\mathrm{A}\) through \(R\) and through the series cell. Each of the two parallel cells carries \(I/2 = 4/7\,\mathrm{A}\). (c) Terminal voltage of parallel pair: \(V_{\text{par}} = \mathcal{E}_{\text{par}} - (I)(r_{\text{par}}) = 2 - (8/7)(0.5) = 2 - 4/7 = 10/7 \approx 1.43\,\mathrm{V}\). Terminal voltage of series cell: \(V_{\text{series}} = 2 - (8/7)(1) = 2 - 8/7 = 6/7 \approx 0.86\,\mathrm{V}\).