Current Electricity

Picture a metal as a lattice of heavy, fixed positive ions with a sea of free electrons, each of charge $-e$ and mass $m$, with $n$ of them per unit volume. Even with no field applied, these electrons are far from still: they race around at high thermal speeds, repeatedly colliding with the ions, and each collision sends an electron off in a completely random new direction. Because the directions are random, the velocities of all $N$ electrons average to zero, and there is no net current:

Now switch on a field $\vec{E}$. Every electron feels a force $-e\vec{E}$, so between collisions it accelerates at

Follow one electron, number $i$. Suppose it left its last collision with velocity $\vec{v}_i$, and a time $t_i$ has passed since then. Accelerating uniformly for that time, its velocity now is

The drift velocity $\vec{v}_d$ is the average of $\vec{V}_i$ over all the electrons at one instant. Average each term separately. The first term, $\vec{v}_i$, averages to zero for exactly the reason above: right after a collision the direction is random. In the second term, $e\vec{E}/m$ is the same for every electron, so only $t_i$ needs averaging. Collisions happen at random times, so some electrons were hit a long while ago and some only just now, but the average time since the last collision is a fixed property of the metal, the relaxation time $\tau$:

The minus sign says the electrons drift against the field, as negative charges should. The surprise is that the result is a constant velocity, even though every electron is being accelerated the whole time. Each one picks up a little extra velocity along $-\vec{E}$, loses it to a randomising collision, picks it up again, and so on. Averaged over the whole crowd, all that is left is a steady drift. A quick unit check: $eE\tau/m$ has units of (N)(s)/(kg) = m/s, a velocity, as it should.

Take a flat area $A$ inside the conductor, at right angles to the field. In a short time $\Delta t$, every free electron within a distance $v_d\,\Delta t$ of that area, on the upstream side, drifts across it. Those electrons fill a cylinder of base $A$ and length $v_d\,\Delta t$, so their number is $n\,(A\,v_d\,\Delta t)$, and each carries charge of magnitude $e$:

Dividing by $\Delta t$ gives the current:

which is exactly the $I = nqv_dA$ stated in §3.1, with $q=e$ for electrons.

Now put the two results together. Substituting $v_d = eE\tau/m$ into $I = neAv_d$:

Dividing both sides by $A$ gives the current per unit area, the current density $j = I/A$, which points along $\vec{E}$ (conventional current runs opposite to the electron drift):

where $\sigma$ is the conductivity. To connect this with the resistance of a real wire, take a wire of length $\ell$ and cross-section $A$ with a potential difference $V$ across it, so that the uniform field inside is $E = V/\ell$:

Rearranging for $V$:

Two things fall out at once. First, $V$ is proportional to $I$, which is Ohm’s law, as long as $n$ and $\tau$ do not themselves depend on the field. Second, comparing with $V = IR = I\rho\ell/A$ identifies the resistivity directly:

As a unit check, $m/(ne^2\tau)$ comes out in kg/(m⁻³·C²·s) = kg m³ A⁻² s⁻³, which is exactly the unit of Ω m. This formula also explains the temperature behaviour described in §3.3. In a metal, $n$ barely changes with temperature, but hotter ions vibrate more and the electrons move faster, so collisions come more often and $\tau$ shrinks, which makes $\rho$ rise. In a semiconductor, heating frees many more carriers, so $n$ climbs steeply, and that growth outweighs the fall in $\tau$, which makes $\rho$ fall.

Same current flows through every resistor; voltages add:

always larger than the largest individual resistor.

Same voltage across every resistor; currents add, so conductance (1/R) adds:

always smaller than the smallest individual resistor, since every extra branch opens another path for charge to flow, not another obstacle.

At any junction, total current in equals total current out, charge can’t pile up or vanish at a point.

Around any closed loop, the sum of EMFs equals the sum of IR drops, energy is conserved going around and coming back to the same point.

Applying the loop rule is really just walking around a loop and keeping a running total of how the potential changes, which must come back to zero where you started, since potential has only one value at each point. Four sign conventions do all the work. Crossing a resistor $R$ in the same direction as the current $I$ you assumed for it, the potential drops, so write $-IR$; crossing it against that current, write $+IR$. Crossing a cell from its negative terminal to its positive terminal, the potential rises, so write $+\varepsilon$; crossing from positive to negative, write $-\varepsilon$. A cell’s internal resistance is treated as just another resistor.

The procedure is then always the same. Draw an arrow for the current in each branch, guessing the direction if you have to, and use the junction rule straight away so that you introduce as few unknown currents as possible. Then apply the loop rule to as many independent loops as there are unknowns left, and solve the simultaneous equations. A guessed direction never needs to be right: if a current comes out negative, it simply flows the opposite way to its arrow, with the same magnitude.

Join $n$ cells, with EMFs $\varepsilon_1, \ldots, \varepsilon_n$ and internal resistances $r_1, \ldots, r_n$, the positive terminal of each to the negative terminal of the next, and connect the two free ends across an external resistor $R$. There is only one path, so the same current $I$ passes through every cell and through $R$. Walk round the loop in the direction of the current and apply the loop rule of §3.8: each cell is crossed from $-$ to $+$, gaining its EMF but losing $Ir_k$ across its own internal resistance, and the external resistor costs $IR$:

The right-hand side, $IR$, is the potential difference across the external resistor, which is the terminal voltage $V$ of the whole combination. A single equivalent cell would give $V = \varepsilon_{eq} - Ir_{eq}$. For the two expressions to agree for every value of $I$, the part that does not depend on $I$ and the part proportional to $I$ must match separately:

For $n$ identical cells this becomes $\varepsilon_{eq} = n\varepsilon$ and $r_{eq} = nr$, so the current through $R$ is $I = n\varepsilon/(R + nr)$. If one cell is connected the wrong way round, the walk crosses it from $+$ to $-$, so its EMF enters the sum with a minus sign, while its internal resistance still adds, because the current still has to pass through it.

Now join two cells ($\varepsilon_1, r_1$ and $\varepsilon_2, r_2$) positive to positive at one point and negative to negative at another, and draw a total current $I$ from the pair. Let $I_1$ and $I_2$ be the currents leaving the two positive terminals. The junction rule at the common positive point gives

Both cells are connected across the same two points, so they must have the same terminal voltage $V$. Writing $V = \varepsilon - Ir$ for each cell and solving for its current:

Add the two currents:

and rearrange for the terminal voltage:

This has exactly the form $V = \varepsilon_{eq} - Ir_{eq}$. Matching the term proportional to $I$ gives the internal resistance, and the constant term then gives the EMF (multiplying top and bottom by $r_1r_2$ to tidy it):

The same steps with $n$ cells just add more terms to each sum, since $I = \sum_k(\varepsilon_k - V)/r_k$, giving the general rule

For $n$ identical cells, the first gives $1/r_{eq} = n/r$, so $r_{eq} = r/n$, and the second gives $\varepsilon_{eq} = (r/n)(n\varepsilon/r)$:

So the two arrangements do different jobs. Series stacking raises the EMF but piles up internal resistance, which pays off when the external resistance is large compared with $nr$. Parallel grouping leaves the EMF unchanged but divides the internal resistance, which pays off when the external resistance is small, and it also shares the current between the cells so that each one is worked less hard.

Label the four corners of the diamond A, B, C and D. Resistor $P$ runs from A to B and $Q$ from B to C along one side; $R$ runs from A to D and $S$ from D to C along the other. A cell is connected across A and C (the battery arm), and a galvanometer of resistance $G$ across B and D (the galvanometer arm). Let a current $I_1$ flow from A into $P$, a current $I_2$ from A into $R$, and a current $I_g$ from B to D through the galvanometer. The junction rule then fixes the other two currents: at B, $Q$ carries $I_1 - I_g$, and at D, $S$ carries $I_2 + I_g$.

Apply the loop rule to loop ABDA (through $P$, then the galvanometer, then back through $R$ against its current):

and to loop BCDB (through $Q$, back through $S$ against its current, then back through the galvanometer against $I_g$):

These two equations hold for any resistor values. Now impose the balance condition, $I_g = 0$. The galvanometer terms vanish, and the equations reduce to

Dividing the first equation by the second cancels both currents:

Look at what has dropped out: the cell’s EMF, its internal resistance, and the galvanometer’s resistance $G$ are all gone. That is the real reason the method is so precise. Physically, $I_1P = I_2R$ says the potential falls by the same amount from A to B as from A to D, so B and D sit at the same potential, and with no potential difference across it the galvanometer has no reason to carry any current. In use, $P$ and $Q$ are known ratio arms, $R$ is a variable known resistance adjusted until the galvanometer reads zero, and the unknown is then $S = (Q/P)\,R$. (NCERT writes the same condition as $R_2/R_1 = R_4/R_3$ with its own labels; in both cases the ratio of the two resistors on one side of the galvanometer equals the ratio of the two on the other side.)

The heart of the instrument is a long wire AB, of length $L$ and total resistance $R_w$, made of one material with a perfectly uniform cross-section. A driver cell (EMF $\varepsilon_D$, internal resistance $r_D$), in series with a rheostat set to $R_h$, pushes a steady current through the wire from A to B:

Because the wire is uniform, its resistance is shared out evenly along its length: a piece of length $\ell$ has resistance $(R_w/L)\,\ell$. With the same current $I$ in every part of the wire, the potential drop from A to a point J a distance $\ell$ along it is

The quantity $k$, in volts per metre, is the potential gradient. As long as the current stays steady, $k$ is the same everywhere along the wire, so the wire turns length into a direct, linear measure of potential difference. Its units check out: (A)(Ω)/(m) = V/m.

To measure a cell of EMF $\varepsilon$ and internal resistance $r$, connect its positive terminal to A (the same end as the driver’s positive terminal) and its negative terminal, through a galvanometer of resistance $G$, to a sliding contact (the jockey) touching the wire at J. Let $I_g$ be the current flowing from J through the galvanometer and the test cell back to A. The loop rule, going from A along the wire to J, then through the galvanometer, then through the test cell from $-$ to $+$ back to A, gives

So the galvanometer deflects one way when the jockey is too close to A ($V_{AJ} < \varepsilon$) and the other way when it is too far along ($V_{AJ} > \varepsilon$). Slide the jockey until the deflection is exactly zero. At this balance point, a distance $\ell$ from A, $I_g = 0$: the test circuit takes no current from the wire and adds none to it, so the wire current is still $I$, the drop to J is still exactly $k\ell$, and no current means no $Ir$ loss inside the test cell. Therefore

the full EMF, not the terminal voltage. A balance point can only exist on the wire if the drop across its whole length is larger than the EMF being measured, $kL > \varepsilon$, so the driver cell must be the stronger one. And the smaller $k$ is made (a longer wire, or less current), the longer the balance length for a given EMF, so the same reading error in the length becomes a smaller fraction of it: a lower potential gradient makes a more sensitive potentiometer.

Connect two cells, of EMFs $\varepsilon_1$ and $\varepsilon_2$, so that a two-way key can switch either one into the galvanometer circuit, and find the balance length for each, $\ell_1$ and $\ell_2$, without touching the driver circuit, so that $k$ stays the same. Each balance gives

Divide one by the other, and $k$ cancels:

The ratio of EMFs comes straight from a ratio of two lengths, with no need to know the current, the wire’s resistance, or the driver cell’s EMF.

First find the balance length $\ell_1$ for the cell on its own, with nothing else connected across it. No current flows through the cell, so

Now connect a known resistance $R$ directly across the cell’s terminals (through a key, closed for this step). The cell now drives a current $I'$ through $R$; at balance the galvanometer arm carries nothing, so all of the cell’s current goes through $R$, and

The potentiometer now balances against the cell’s terminal voltage, $V = I'R$, which is smaller than $\varepsilon$, so the new balance length $\ell_2$ is shorter:

Divide the two balance conditions, so $k$ cancels, and write both $\varepsilon$ and $V$ in terms of $I'$:

Solving for $r$:

Again only lengths and one known resistance appear, and the result is a resistance multiplied by a pure number, as it must be.