Current and circuits

Every result in the first five lessons was obtained by waiting until the charges had stopped moving, and nothing in them explains why, in a wire connected to a battery, they never do.

That is the gap this lesson fills. It needs the field, the potential and Gauss's law from earlier in the course, and it will end with a differential equation of exactly the kind Differential Equations treats first, solved by separating the variables. It is also where the subject stops being about isolated charged spheres and starts being about the objects that electricity is actually used for.

Current

Current is the rate at which charge crosses a surface:

I = \frac{d Q}{d t}

measured in coulombs per second, called amperes. One ampere is a large current in laboratory terms and an ordinary one in domestic terms; a 100 W lamp on a 230 V supply draws about 0.43 A.

Two things about the definition are worth pausing on. First, current is not a vector, despite having a direction along the wire: it is a flux, a scalar counting how much crosses a surface, in the same way that the flow rate of a river in litres per second is a scalar while the velocity of the water is not. Second, the direction assigned to a current is the direction positive charge would move, which as the first lesson explained is opposite to the actual motion of the electrons in a metal. Nothing goes wrong as long as the convention is kept, because a positive charge moving right and a negative charge moving left transport charge the same way.

Spread the current over the cross section and the local quantity is the current density $J = I / A$ , in amperes per square metre, which unlike $I$ does have a direction and is the thing that appears in material laws.

How fast do the charges actually move

Suppose a wire of cross section $A$ contains $n$ mobile charge carriers per cubic metre, each of charge $q$ , drifting at an average speed $v_{d}$ . In a time $d t$ every carrier within a distance $v_{d} d t$ of a chosen cross section will pass it, so the charge crossing is $n q A v_{d} d t$ , and

I = n q v_{d} A

For copper the carrier count is not a free parameter. Copper has a density of 8960 kg/m³ and a molar mass of 63.546 g/mol, and contributes very nearly one conduction electron per atom, so

n = \frac{8960}{0.063546} \times 6.022 \times 10^{23} = 8.49 \times 10^{28} m^{- 3}

Now put 1 A through a wire of 1 mm² cross section, which is ordinary domestic wiring:

v_{d} = \frac{I}{n q A} = \frac{1}{(8.49 \times 10^{28}) (1.602 \times 10^{- 19}) (10^{- 6})} = 7.35 \times 10^{- 5} m/s

That is 0.074 millimetres per second. An electron entering one end of a one metre wire arrives at the other end nearly four hours later.

Three speeds are being confused whenever this result surprises someone, and they differ by ten orders of magnitude each. The drift speed is the tenth of a millimetre per second just computed. The actual speed of an individual conduction electron in copper, set by quantum mechanics rather than by temperature, is about $1.6 \times 10^{6}$ m/s, some twenty billion times faster, but it is directed randomly and averages to almost nothing: the drift is a barely perceptible bias on a violent random motion. And the speed at which a lamp comes on when the switch is thrown is the speed at which the electric field propagates down the wire, a large fraction of the speed of light, because the field does not have to wait for any particular electron to arrive. The wire is already full of charge everywhere along it, and all that has to travel is the instruction to move.

Example. A copper wire of cross section 2.0 mm² carries 5.0 A. What is the drift speed?

$v_{d} = I / (n q A) = 5.0 / [(8.49 \times 10^{28}) (1.602 \times 10^{- 19}) (2.0 \times 10^{- 6})] = 1.84 \times 10^{- 4}$ m/s, about a fifth of a millimetre per second.

Now you. A copper wire of cross section 1.5 mm² carries 3.0 A. What is the drift speed?

Answer

$v_{d} = 3.0 / [(8.49 \times 10^{28}) (1.602 \times 10^{- 19}) (1.5 \times 10^{- 6})] = 1.47 \times 10^{- 4}$ m/s.

Resistance, and why Ohm's law is not a law

Something must keep the drift going. Left alone, a moving charge in a conductor would be accelerated by the field without limit; instead it scatters off lattice vibrations and impurities every $10^{- 14}$ seconds or so, losing its acquired velocity and starting again. The result is a steady drift proportional to the field rather than a steady acceleration, and for a wide class of materials the proportionality is remarkably good:

J = σ E

with $σ$ the conductivity of the material, or equivalently $E = ρ J$ with $ρ = 1 / σ$ the resistivity. This is Ohm's law in its useful form, discovered by Georg Ohm in 1827 and published to general indifference.

For a uniform wire of length $L$ and cross section $A$ , substitute $E = V / L$ and $J = I / A$ :

V = (\frac{ρ L}{A}) I = I R, R = \frac{ρ L}{A}

which is the familiar form. The resistance $R$ , in ohms, is a property of a particular object; the resistivity $ρ$ , in ohm metres, is a property of a material, and it is the one to reason with.

The range is the widest of any common physical property. Copper is $1.68 \times 10^{- 8}$ Ω m at 20 degrees Celsius, nichrome about $1.10 \times 10^{- 6}$ , silicon around $10^{3}$ pure, and good glass $10^{12}$ or more: twenty orders of magnitude between the wire and its insulation, which is why electrical engineering is possible at all.

Now the honesty. $V = I R$ is not a law of nature, it is a description of the materials that happen to obey it. A diode does not: its current rises exponentially with voltage in one direction and hardly at all in the other. A filament lamp does not, because it heats up and copper's resistivity rises by about 0.39 per cent per kelvin, so the resistance of a lamp at 2500 K is roughly ten times its cold value. An ionised gas does not, and can have a resistance that falls as current rises, which is why a fluorescent tube needs a ballast to stop it destroying itself. A superconductor has $ρ$ exactly zero below its transition temperature, which no amount of extrapolating Ohm's law would predict. Ohmic behaviour is common, useful and contingent.

A last observation that catches people out. The field inside a current-carrying copper wire is $E = ρ J$ , and for 1 A in 1 mm² that is $(1.68 \times 10^{- 8}) (10^{6}) = 0.017$ V/m. The electrostatics of the earlier lessons dealt in fields of $10^{4}$ V/m and up. Conduction is what happens when a tiny field acts on an enormous number of very mobile charges.

Example. A kettle element is made from nichrome wire of cross section 0.50 mm² and length 10 m, resistivity $1.10 \times 10^{- 6}$ Ω m. What is its resistance, and what power does it draw at 230 V?

$R = ρ L / A = (1.10 \times 10^{- 6}) (10) / (0.50 \times 10^{- 6}) = 22$ Ω. The power is $P = V^{2} / R = (230)^{2} / 22 = 2.40 \times 10^{3}$ W, which is a normal kettle. The current is $V / R = 10.5$ A, and the whole 2.4 kW appears as heat in the wire, because in a purely resistive element every joule delivered is dissipated.

Now you. A heating element of nichrome, cross section 0.30 mm² and length 8.0 m, is run at 230 V. What is its resistance and its power?

Answer

$R = (1.10 \times 10^{- 6}) (8.0) / (0.30 \times 10^{- 6}) = 29.3$ Ω, and $P = (230)^{2} / 29.3 = 1.80 \times 10^{3}$ W.

EMF, and what a battery does

A resistor connected to nothing carries no current, so a circuit needs a device that pushes charge from low potential to high, against the electric field, in the way a pump raises water. That device supplies an electromotive force, symbol $E$ , which is a poorly chosen name since it is not a force at all but an energy per unit charge, measured in volts.

In a chemical cell the pushing is done by a reaction that is energetically favourable, so a zinc atom gives up electrons at one electrode and a reduction consumes them at the other, and the energy released per electron transferred fixes the EMF. That is why cell voltages are set by chemistry and cluster around one to two volts: they are, in effect, reaction energies in electronvolts.

A real source has resistance of its own, the internal resistance $r$ , so the voltage appearing at its terminals is less than its EMF whenever current flows:

V_{terminal} = E - I r

A 12 V car battery with $r = 0.050$ Ω driving a 2.0 Ω load carries $I = 12 / 2.05 = 5.85$ A, so its terminals sit at $12 - (5.85) (0.050) = 11.7$ V and the load receives $I^{2} R = 68.5$ W while 1.7 W is wasted inside the battery. Starting the engine draws several hundred amperes, and the same 0.050 Ω then drops several volts, which is why the headlights dim.

Power in any two-terminal element is $P = V I$ , since $V$ is energy per charge and $I$ is charge per time. For a resistor, substituting $V = I R$ gives the two other forms, $P = I^{2} R = V^{2} / R$ . The first is the one to use when the current is known, the second when the voltage is, and mixing them up is the commonest arithmetic error in the subject.

Two rules that solve any network

Gustav Kirchhoff published in 1845, at twenty-one, two statements that reduce any network of sources and resistors to simultaneous linear equations.

The junction rule says the currents entering any junction sum to the currents leaving. This is charge conservation together with the observation that charge does not accumulate anywhere in a steady circuit: if it did, the resulting field would immediately stop it.

The loop rule says that the sum of the potential changes around any closed loop is zero. This is the statement that the electrostatic potential is a function of position: go round a loop and return to the same point and you return to the same potential. It is exactly the path independence of the fourth lesson, and it is worth registering now that this is the one result in the whole lesson with an expiry date. The ninth lesson produces an electric field for which the loop sum is not zero, and Kirchhoff's second rule survives there only by being rewritten.

Applying them needs one discipline: choose a direction for each unknown current before starting, and let the algebra return a negative number if the choice was wrong. Guessing correctly is not required and not worth trying.

The RC circuit

Connect a battery of EMF $E$ , a resistor $R$ and an uncharged capacitor $C$ in series and something genuinely time-dependent finally happens. Let $q$ be the charge on the capacitor. The loop rule gives

E - I R - \frac{q}{C} = 0

and since $I = d q / d t$ , this is a first order differential equation:

R \frac{d q}{d t} = E - \frac{q}{C}

Separate the variables, integrate, and impose $q = 0$ at $t = 0$ :

q (t) = C E (1 - e^{- t / R C})

The current is the derivative, $I = (E / R) e^{- t / R C}$ : full at the instant of connection, when the empty capacitor offers no back voltage, and decaying to nothing as it fills. For discharge through the same resistor with no battery, the same equation without $E$ gives $q = Q_{0} e^{- t / R C}$ .

The product $R C$ has units of seconds and is the time constant $τ$ . After one time constant the discharge has fallen to $1 / e$ , about 37 per cent; after three, to 5 per cent; after five, to 0.7 per cent, which is the usual working definition of finished. Note that $τ$ says nothing about how much charge is involved, only how fast.

Example. A 100 μF capacitor charged to 12 V is discharged through 47 kΩ. What is the time constant, the voltage after 10 s, and the time to fall to 1.0 V?

$τ = R C = (47 \times 10^{3}) (100 \times 10^{- 6}) = 4.7$ s. Voltage follows the same exponential as charge, so $V = 12 e^{- 10 / 4.7} = 12 e^{- 2.128} = 12 \times 0.1191 = 1.43$ V. For the time to reach 1.0 V, invert: $t = τ \ln (12 / 1.0) = 4.7 \times 2.485 = 11.7$ s.

Now you. A 220 μF capacitor charged to 9.0 V is discharged through 22 kΩ. Find the time constant and the voltage after 8.0 s.

Answer

$τ = (22 \times 10^{3}) (220 \times 10^{- 6}) = 4.84$ s. Then $V = 9.0 e^{- 8.0 / 4.84} = 9.0 e^{- 1.653} = 9.0 \times 0.1915 = 1.72$ V.

Where the current picture is about to fail

Steady currents look like a closed subject: charge conserved, energy accounted for, networks solvable, and one equation with a time in it. They are not, and the reason is an experiment that nothing in this lesson would predict.

Run two long parallel wires side by side and pass a current through both. They push on each other, attracting when the currents run the same way and repelling when they oppose. The force is easily measurable and it is not electrostatic: both wires are electrically neutral, containing exactly as many protons as electrons, so Coulomb's law says the force between them should be zero.

Something about charge in motion produces a force that charge at rest does not. It cannot be described by a field that acts along the line between the objects, because the force turns out to act sideways, and it cannot be described by anything depending on position alone, because it depends on how fast the charges are moving. It needs a second field, with its own force law and its own rules, and that field occupies the next four lessons.