Current Electricity

Twelve identical resistors of resistance $R$ are soldered along the twelve edges of a cube. Find the resistance between two opposite corners.

Look for two resistors in series. There are none. Look for two in parallel. There are none either. Every reduction rule taught at Main level is simply unavailable.

The answer is $\frac{5R}{6}$, and it comes from symmetry, not from reduction. Feed current in at corner $A$ and take it out at the diagonally opposite corner $G$. The three edges leaving $A$ are indistinguishable, so each carries $I/3$ and the three nodes they reach are at the same potential. The same argument applies to the three nodes adjacent to $G$.

Nodes at equal potential can be merged without changing anything, because no current flows between them. The cube collapses to three resistors in series:

$R_{AG}=\frac{R}{3}+\frac{R}{6}+\frac{R}{3}=\frac{5R}{6}$

That is the shape of the whole chapter. Main gives circuits that reduce. Advanced gives circuits that do not, conductors whose cross-section varies, and circuits caught mid-transient. The tools are symmetry, self-similarity, integration and the initial-and-final-state trick.

1. Symmetry: merging and splitting nodes

Three moves solve most irreducible networks.

Merge equal-potential nodes. If two nodes are at the same potential, joining them with a wire changes nothing, and the network usually becomes reducible. Symmetry of the network about the line joining the terminals is what guarantees the equality.

Cut a node in two. The converse move: a node that symmetry says carries no net current across a plane can be split, again without changing anything.

Delete a bridge. In a balanced Wheatstone bridge, $\frac{P}{Q}=\frac{R}{S}$ makes the galvanometer branch carry no current, so it can be removed or short-circuited, whichever makes the reduction easier.

Illustration 1

Find the resistance of a cube of twelve $R$ resistors across a face diagonal and across an edge.

Face diagonal: by symmetry two nodes are equipotential and drop out; the network reduces to $\frac{3R}{4}$.

Edge: the symmetry plane contains both terminals, and the reduction gives $\frac{7R}{12}$.

The ordering is a useful sanity check. Nearer terminals mean more parallel paths of shorter length, so $\frac{7R}{12}<\frac{3R}{4}<\frac{5R}{6}$ — exactly what the three answers show.

Illustration 2

Five resistors form a Wheatstone bridge with $P=10\Omega$, $Q=20\Omega$, $R=15\Omega$, $S=30\Omega$ and a $50\Omega$ galvanometer across the bridge. Find the equivalent resistance.

$\frac{P}{Q}=\frac{10}{20}=0.5$ and $\frac{R}{S}=\frac{15}{30}=0.5$, so the bridge is balanced.

Remove the galvanometer: $\left(10+20\right)$ in parallel with $\left(15+30\right)$:

$R_{eq}=\frac{30\times 45}{75}=18\Omega$

The galvanometer's own resistance never entered. That is the entire point of balancing: whatever sits across the bridge is irrelevant when no current flows through it.

2. Self-similar networks and the star-delta transformation

An infinite ladder is solved by exploiting the fact that it looks the same after one section is removed. If the input resistance is $X$, then

$X=R+\frac{RX}{R+X}\Rightarrow X^2-RX-R^2=0\Rightarrow X=\frac{R\left(1+\sqrt{5}\right)}{2}$

the golden ratio times $R$. The negative root is discarded because resistance cannot be negative.

The star-delta transformation handles networks with no symmetry at all. A delta of $R_{AB}$, $R_{BC}$, $R_{CA}$ becomes an equivalent star with

$R_A=\frac{R_{AB}{R}_{CA}}{R_{AB}+R_{BC}+R_{CA}}$

and cyclic permutations for $R_B$ and $R_C$. Converting one delta to a star almost always exposes series and parallel combinations that were hidden.

Illustration 3

An infinite ladder has series elements $2R$ and shunt elements $R$. Find its input resistance.

$X=2R+\frac{RX}{R+X}$

$X\left(R+X\right)=2R\left(R+X\right)+RX\Rightarrow X^2-2RX-2R^2=0$

$X=R\left(1+\sqrt{3}\right)=2.73R$

The self-similarity trick fails for a finite ladder, where the last section has nothing beyond it. Finite ladders need either recursion from the far end or a difference equation.

3. Conductors whose cross-section changes

Resistance is $\rho L/A$ only when $A$ is constant. Otherwise, add resistances of thin slices in series:

$R=\int \frac{\rho dx}{A(x)}$

A truncated cone of end radii $r_1$ and $r_2$ and length $L$ gives the neat result

$R=\frac{\rho L}{\pi r_1{r}_2}$

which is the geometric mean of the two end areas, not the arithmetic mean.

When current flows radially, the slices are shells and the area grows with radius:

$R_{coaxial}=\frac{\rho \ln \left(b/a\right)}{2\pi L},R_{spherical}=\frac{\rho }{4\pi }\left(\frac{1}{a}-\frac{1}{b}\right)$

Illustration 4

A material of resistivity $\rho$ fills the space between two coaxial cylinders of radii $a$ and $b$ and length $L$. Find the resistance for radial current flow.

A shell at radius $r$ of thickness $dr$ has area $2\pi rL$:

$dR=\frac{\rho dr}{2\pi rL}\Rightarrow R=\frac{\rho }{2\pi L}{\int }_a^b\frac{dr}{r}=\frac{\rho \ln \left(b/a\right)}{2\pi L}$

The logarithm means the inner region dominates. Doubling $b$ adds only $\rho \ln 2/2\pi L$, however large $b$ already is, which is why cable insulation testing is so sensitive to the inner conductor's surface.

Illustration 5

A conductor tapers linearly from radius $2$ mm to $4$ mm over a length of $0.1$ m, with $\rho =1.7\times 10^{-8}\Omega$ m. Find its resistance, and compare with using the mean radius.

$R=\frac{\rho L}{\pi r_1{r}_2}=\frac{1.7\times 10^{-8}\times 0.1}{\pi \left(2\times 10^{-3}\right)\left(4\times 10^{-3}\right)}=6.8\times 10^{-5}\Omega$

Using the mean radius $3$ mm would give $\frac{\rho L}{\pi {\left(3\times 10^{-3}\right)}^2}=6.0\times 10^{-5}\Omega$.

The mean-radius estimate is low by $12\%$, because resistance weights the narrow end much more heavily than the wide one.

4. Reducing a network: superposition and maximum power

Any network of sources and resistors, seen from two terminals, behaves as a single emf in series with a single resistance. The emf is the open-circuit voltage; the resistance is what you measure with all sources removed — emfs shorted, current sources opened.

That reduction makes the maximum-power question immediate. Delivering power to a load $R$ from a source of emf $\epsilon$ and internal resistance $r$:

$P=\frac{{\epsilon }^{2}R}{{\left(R+r\right)}^2},\frac{dP}{dR}=0\Rightarrow R=r,P_{max}=\frac{{\epsilon }^2}{4r}$

At that point the efficiency is only $50\%$ — half the energy is burnt inside the source. Maximum power and maximum efficiency are different goals, which is why power grids deliberately operate far from the matched condition.

Illustration 6

A battery of emf $12$ V and internal resistance $2\Omega$ drives a variable load. Find the load for maximum power, the maximum power, and the efficiency at $R=8\Omega$.

Maximum power at $R=r=2\Omega$:

$P_{max}=\frac{144}{8}=18$ W, at efficiency $50\%$

At $R=8\Omega$: $I=\frac{12}{10}=1.2$ A, so $P=1.2^2\times 8=11.5$ W at efficiency $\frac{8}{10}=80\%$.

Less power, delivered far more efficiently. Which one you want depends entirely on whether the source's energy is cheap.

5. Non-ohmic elements and the load line

Ohm's law is a property of certain materials at constant temperature, not a law of nature. For a filament lamp, a diode or a thermistor, $V/I$ is not constant and two different resistances have to be distinguished:

$R_{static}=\frac{V}{I},r_{dynamic}=\frac{dV}{dI}$

The static value is what a power calculation needs; the dynamic value is what a small-signal calculation needs, and for a diode in forward bias the two differ by orders of magnitude.

Such a circuit cannot be solved algebraically, because the element has no formula. It is solved graphically. The device supplies its own $I$-$V$ characteristic, and the rest of the circuit supplies the constraint

$V=\epsilon -IR$

a straight line called the load line. The operating point is where the two curves cross.

A filament lamp is the standard case, because its resistance climbs steeply with temperature:

$R=R_0\left(1+\alpha \Delta T\right)$

so its cold resistance is a small fraction of its working value.

Illustration 7

A lamp is rated $100$ W at $220$ V. Estimate its working resistance and the current drawn at the instant of switching on, taking $\alpha =5\times 10^{-3}$ K${}^{-1}$ and a temperature rise of $2000$ K.

Working resistance: $R=\frac{V^2}{P}=\frac{48400}{100}=484\Omega$, at a working current of $0.45$ A.

Cold resistance: $R_0=\frac{484}{1+\left(5\times 10^{-3}\right)(2000)}=\frac{484}{11}=44\Omega$

Switch-on current: $\frac{220}{44}=5.0$ A

More than ten times the running current, which is why filament lamps almost always fail at the moment they are switched on rather than during use.

Illustration 8

A device has the characteristic $I=k{V}^2$ with $k=0.02$ A V${}^{-2}$, and is driven by a $6$ V source through a $10\Omega$ resistor. Find the operating current.

Load line: $V=6-10I$. Substituting:

$I=0.02{\left(6-10I\right)}^2\Rightarrow 2I^2-3.4I+0.72=0$

$I=\frac{3.4\pm \sqrt{11.56-5.76}}{4}=\frac{3.4\pm 2.408}{4}$

The physical root is $I=0.248$ A, giving $V=3.52$ V; the other root would need $V<0$.

Always discard roots that put the operating point off the device's real characteristic. The algebra does not know that a diode conducts in only one direction.

6. Transients: the first instant and the last

For a capacitor charging through a resistor,

$q=CV\left(1-e^{-t/RC}\right),i=\frac{V}{R}{e}^{-t/RC}$

and for discharging, both decay as $e^{-t/RC}$ from their initial values. Two limits do most of the work in an exam.

At $t=0$ an uncharged capacitor holds no voltage, so it behaves as a short circuit. At $t\to \infty$ no current flows into it, so it behaves as an open circuit. Replacing every capacitor accordingly turns a transient problem into two ordinary resistive circuits.

The energy accounting is worth memorising. Charging a capacitor through any resistance from a battery of emf $V$ costs the battery $C{V}^2$, stores $\frac{1}{2}C{V}^2$ and dissipates $\frac{1}{2}C{V}^2$ — regardless of $R$. The resistance sets only how long it takes.

Illustration 9

A $10\mu$F capacitor is charged through a $2\text{k}\Omega$ resistor by a $50$ V battery. Find the time constant, the energy stored and the energy dissipated.

$\tau =RC=\left(2\times 10^3\right)\left(10\times 10^{-6}\right)=2\times 10^{-2}$ s

$U_{stored}=\frac{1}{2}C{V}^2=\frac{1}{2}\left(10^{-5}\right)\left(2500\right)=1.25\times 10^{-2}$ J

$U_{dissipated}$ is the same, $1.25\times 10^{-2}$ J.

Halving the resistor would halve the time and change nothing else. The $50\%$ efficiency of capacitor charging through a resistor is unavoidable, which is why switch-mode supplies use inductors instead.

Illustration 10

In a circuit a battery of emf $\epsilon$ drives two resistors $R_1$ in series with a parallel combination of $R_2$ and an uncharged capacitor $C$. Find the current drawn immediately after closing the switch and long afterwards.

At $t=0$: the capacitor is a short, so $R_2$ is shorted out.

$i_0=\frac{\epsilon }{R_1}$

At $t\to \infty$: the capacitor is an open circuit, so the current flows through both resistors.

$i_{\infty }=\frac{\epsilon }{R_1+R_2}$

The current always falls when a capacitor charges in a shunt branch, and always rises when it charges in a series branch. Identifying which case you have takes seconds and settles the qualitative answer.

7. Measurement: shunts, multipliers and the potentiometer

A galvanometer of resistance $G$ and full-scale current $I_g$ becomes an ammeter with a parallel shunt and a voltmeter with a series multiplier:

$S=\frac{I_{g}G}{I-I_g},R_{series}=\frac{V}{I_g}-G$

Both instruments disturb the circuit they measure, which is the whole reason the potentiometer exists. At balance it draws no current from the cell under test, so it reads the true emf rather than the terminal voltage. A voltmeter, drawing current through the internal resistance, always reads low.

Illustration 11

A galvanometer of $60\Omega$ gives full-scale deflection at $2$ mA. Convert it into an ammeter reading $5$ A and into a voltmeter reading $10$ V.

Shunt: $S=\frac{\left(2\times 10^{-3}\right)\left(60\right)}{5-0.002}=\frac{0.12}{4.998}=0.024\Omega$

Multiplier: $R=\frac{10}{2\times 10^{-3}}-60=5000-60=4940\Omega$

A good ammeter has almost no resistance and a good voltmeter almost infinite resistance, and the two numbers here — $0.024\Omega$ and $4940\Omega$ — show how far apart the same instrument can be pushed.

Illustration 12

A cell of emf $2.0$ V and internal resistance $1.0\Omega$ is measured by a voltmeter of resistance $99\Omega$, and separately by a potentiometer. What does each read?

Voltmeter: $I=\frac{2.0}{100}=0.02$ A, so it reads $0.02\times 99=1.98$ V.

Potentiometer: at balance no current is drawn, so it reads the true $2.00$ V.

The voltmeter's $1\%$ error is not a defect of the instrument. It is the terminal voltage, which is genuinely what appears across the cell when it is delivering current.

8. Drift, mobility and current density

Current is the flux of charge, and the microscopic statement is

$\vec{J}=ne{\vec{v}}_d=\sigma \vec{E},v_d=\frac{eE\tau }{m},\mu =\frac{v_d}{E}=\frac{e\tau }{m}$

so conductivity is $\sigma =n{e}^2\tau /m$. The drift speed in a household wire is under a millimetre per second, yet a lamp lights instantly — because the field propagates at nearly the speed of light and sets every electron in the circuit moving at once.

In a conductor of varying cross-section, the current $I$ is the same everywhere but $J$, $E$ and $v_d$ are all larger where the conductor is narrow.

Illustration 13

A copper wire of area $1$ mm${}^2$ carries $3$ A. Find the drift speed, taking $n=8.5\times 10^{28}$ m${}^{-3}$.

$v_d=\frac{I}{neA}=\frac{3}{\left(8.5\times 10^{28}\right)\left(1.6\times 10^{-19}\right)\left(10^{-6}\right)}$

$v_d=2.2\times 10^{-4}$ m s${}^{-1}$

About one metre every eighty minutes. The electron that lights your lamp was already sitting in the filament; the signal, not the electron, is what travels.

Illustration 14

A wire tapers so that its area at one end is half that at the other. Compare the current, current density and drift speed at the two ends.

Current is identical at both ends, since charge cannot accumulate.

$J=I/A$ is twice as large at the narrow end.

$v_d=J/ne$ is likewise twice as large there, and so is the field $E=J/\sigma$.

Only the current is conserved. Everything derived from it varies inversely with area, which is why thin sections of a conductor run hot.

Summary

• Symmetry beats reduction: merge equipotential nodes, split zero-current nodes, and delete or short a balanced bridge.
• Cube of twelve $R$: $\frac{7R}{12}$ across an edge, $\frac{3R}{4}$ across a face diagonal, $\frac{5R}{6}$ across the body diagonal.
• A balanced bridge makes the galvanometer branch irrelevant, whatever its resistance.
• Infinite ladders are self-similar: set the whole equal to one section plus the whole, and solve the quadratic.
• Star-delta: $R_A=\frac{R_{AB}{R}_{CA}}{R_{AB}+R_{BC}+R_{CA}}$, which exposes hidden series and parallel groups.
• Varying cross-section: $R=\int \frac{\rho dx}{A(x)}$. A truncated cone gives $\frac{\rho L}{\pi r_1{r}_2}$, the geometric mean of the end areas.
• Radial flow: $\frac{\rho \ln \left(b/a\right)}{2\pi L}$ coaxially, $\frac{\rho }{4\pi }\left(\frac{1}{a}-\frac{1}{b}\right)$ spherically.
• Any two-terminal network reduces to one emf plus one resistance; short the emfs to find the resistance.
• Maximum power at $R=r$ gives $\frac{{\epsilon }^2}{4r}$ at only $50\%$ efficiency — power and efficiency are different goals.
• Non-ohmic elements need a load line: the operating point is where $V=\epsilon -IR$ crosses the device's own characteristic.
• $R=R_0\left(1+\alpha \Delta T\right)$: a lamp's cold resistance can be a tenth of its hot value, giving a large switch-on surge.
• Transients: at $t=0$ an uncharged capacitor is a short; at $t\to \infty$ it is an open circuit.
• Charging through any $R$ costs $C{V}^2$, stores $\frac{1}{2}C{V}^2$ and wastes $\frac{1}{2}C{V}^2$ — independent of $R$.
• $S=\frac{I_{g}G}{I-I_g}$ for an ammeter, $\frac{V}{I_g}-G$ for a voltmeter; a potentiometer draws no current and so reads the true emf.
• $\vec{J}=ne{\vec{v}}_d=\sigma \vec{E}$ with $\sigma =n{e}^2\tau /m$; in a tapering wire only $I$ is conserved, while $J$, $E$ and $v_d$ all rise where it narrows.