Current Electricity

Electric Current & Drift Velocity 🔉✕⇢

When charges are allowed to move, they constitute an electric current, and current electricity is the study of that motion in its steady form. In a torch, a wall clock, or the filament of a bulb, charge crosses every cross-section of the wire at a constant rate, held there by a cell. Before we can predict how circuits behave we need a sharp definition of 'how much charge is flowing', and a physical picture of what the carriers are actually doing inside the metal. Both turn out to hinge on a single, surprisingly slow quantity: the drift velocity of the conduction electrons. 🔉✕⇢

Full derivation, worked example and interactive 3D on the Electric Current & Drift Velocity tab →

Mobility and Current Density 🔉✕⇢

Current is a bulk, whole-conductor quantity. It counts the net charge crossing a chosen cross-section per second, and a single number $I$ describes an entire wire. That is enough for circuit bookkeeping, but it hides what is happening at a point. A thick busbar and a thin fuse wire may both carry 1 A, yet the electrons inside the fuse wire are being driven far harder and the wire may melt while the busbar stays cold. To describe conduction locally we need a quantity defined at every interior point of the material rather than for the object as a whole, and that quantity is the current density. 🔉✕⇢

NCERT defines it as the current per unit area, the area being taken normal to the current: $j=I/A$, with SI unit ampere per square metre. The phrase taken normal is not decoration. If the surface element is tilted relative to the flow, only the projection of that element perpendicular to the flow actually intercepts the moving charge, so the area in the denominator must be the perpendicular one. Once that is understood, the natural next step is to promote $j$ to a vector by handing it the direction of flow of positive charge. Inside an isotropic conductor that direction is the direction of the local electric field, so one writes $\mathbf{j}=j\,\mathbf{E}/E$. 🔉✕⇢

The microscopic content of $\mathbf{j}$ comes from the drift picture. Let $n$ be the number of free carriers per unit volume and $v_d$ their common drift speed. In a time $\Delta t$ every carrier lying within a slant cylinder of base area $A$ and length $v_d\Delta t$ behind the surface will cross it, so the number crossing is $nAv_d\Delta t$ and the charge crossing is $neAv_d\Delta t$. Dividing by $\Delta t$ gives $I=neAv_d$, and dividing again by $A$ gives the local statement $j=nev_d$, or in vector form $\mathbf{j}=ne\mathbf{v}_d$ for positive carriers. 🔉✕⇢

There is a sign subtlety worth slowing down for, because it is where most students first stumble. In a metal the carriers are electrons, whose charge is $-e$, and they drift antiparallel to $\mathbf{E}$ because the force on them is $-e\mathbf{E}$. The current density is the product of a negative charge and a velocity pointing against the field, and the two minus signs cancel, so $\mathbf{j}$ ends up pointing along $\mathbf{E}$ after all. This is exactly why conventional current can be drawn along the field without ever mentioning that the actual carriers are running the other way. 🔉✕⇢

Combining $j=nev_d$ with the drift result $v_d=eE\tau/m$ gives $j=(ne^2\tau/m)E$, which is linear in $E$ with a coefficient built only from material constants. Naming that coefficient the conductivity $\sigma$ produces the local or microscopic form of Ohm law, $\mathbf{j}=\sigma\mathbf{E}$, equivalently $\mathbf{E}=\rho\mathbf{j}$ with $\rho=1/\sigma$. This form is strictly more powerful than $V=IR$: it is a point relation between two field quantities, it holds inside a conductor of any shape, and it survives situations where the very notions of a single current and a single potential difference across a two-terminal object break down. 🔉✕⇢

Mobility is introduced to answer a different question: not how much current flows, but how responsive a given carrier is. NCERT defines it as the magnitude of the drift velocity per unit electric field, $\mu=|v_d|/E$. The SI unit follows directly from that ratio, metre per second divided by volt per metre, which simplifies to $\mathrm{m^2\,V^{-1}\,s^{-1}}$. Practical tables usually quote mobility in $\mathrm{cm^2\,V^{-1}\,s^{-1}}$, and since one square metre is $10^4$ square centimetres, the SI number is smaller than the practical number by a factor of $10^4$. 🔉✕⇢

Substituting the drift result gives the compact microscopic expression $\mu=e\tau/m$. Everything in it is a property of the carrier and its environment: the carrier charge, the carrier mass, and the mean free time between collisions. A carrier is mobile when it is light, strongly charged, and left alone for a long time between scattering events. Because $\tau$ shortens as the lattice is heated and its ions vibrate more violently, mobility in a metal falls with rising temperature, and that single fact is the seed of the whole temperature story told in a later card. 🔉✕⇢

Multiplying mobility by carrier charge and carrier density recovers the conductivity: $\sigma=ne\mu$, since $ne\mu=ne\cdot e\tau/m=ne^2\tau/m$. This factorisation is the most useful sentence in the card, because it cleanly separates the two independent ways a material can conduct well. It can have many carriers, a large $n$, or it can have nimble carriers, a large $\mu$. Metals win on $n$, with roughly one free electron per atom. Semiconductors have a far smaller $n$ but often a much larger $\mu$, and doping is precisely the business of raising $n$ by many orders of magnitude while leaving $\mu$ broadly intact. 🔉✕⇢

When more than one species of carrier is present the conductivities simply add, because the current densities add: $\sigma=\sum_i n_i q_i \mu_i$. NCERT is explicit that the mobile carriers differ by medium. In metals they are electrons alone, in an ionised gas they are electrons together with positive ions, and in an electrolyte both positive and negative ions move. In an electrolyte the positive ions drift along $\mathbf{E}$ and the negative ions drift against it, and because their charges also have opposite signs, both species contribute current density in the same direction and their contributions reinforce rather than cancel. 🔉✕⇢

It is worth putting numbers on all of this using the copper wire NCERT works out. With cross-sectional area $1.0\times10^{-7}\,\mathrm{m^2}$ carrying 1.5 A, the current density is $j=I/A=1.5\times10^{7}\,\mathrm{A\,m^{-2}}$, a large number by everyday standards. The free-electron density is $n=8.5\times10^{28}\,\mathrm{m^{-3}}$ and the drift speed works out to only $1.1\,\mathrm{mm\,s^{-1}}$. The consistency check is immediate: $nev_d=8.5\times10^{28}\times1.6\times10^{-19}\times1.1\times10^{-3}\approx1.5\times10^{7}\,\mathrm{A\,m^{-2}}$, matching $I/A$ exactly, which is the whole content of $j=nev_d$. 🔉✕⇢

The same numbers give a feel for mobility and relaxation time. Copper conductivity is about $5.9\times10^{7}\,\mathrm{S\,m^{-1}}$, so $\mu=\sigma/(ne)\approx5.9\times10^{7}/(8.5\times10^{28}\times1.6\times10^{-19})\approx4.3\times10^{-3}\,\mathrm{m^2\,V^{-1}\,s^{-1}}$. Inverting $\mu=e\tau/m$ gives $\tau=\mu m/e\approx4.3\times10^{-3}\times9.1\times10^{-31}/1.6\times10^{-19}\approx2.5\times10^{-14}\,\mathrm{s}$, a few tens of femtoseconds. An electron in copper is therefore scattered some $10^{14}$ times per second, which is precisely why it never builds up speed and instead settles to a steady crawl. 🔉✕⇢

In problems, these relations are usually tested by chaining them. A typical question hands you $I$, the wire diameter, and the density and atomic mass of the metal, expects $n$ from Avogadro number, then $v_d=I/(neA)$, then $\mu$ once $E$ or $\rho$ is supplied. The reliable habit is to write the chain $n \to j \to v_d \to \mu \to \sigma$ and fill it in, checking units at each link. Note also that $j$ is fixed by $I$ and $A$ alone, so if a wire tapers, $j$ rises in the narrow section even though $I$ is the same everywhere, and $v_d$ rises with it. 🔉✕⇢

Ohm Law and Its Limitations 🔉✕⇢

A basic law about the flow of current was found by G. S. Ohm in 1828, long before anyone knew what actually carried the charge. That historical order matters: Ohm had no electrons, no lattice, no relaxation time. He had wires, a source, and a way to measure, and he reported a pattern in the data. Everything in the drift picture that now explains the law was built afterwards, and the law would remain an experimental fact about a large class of materials even if the explanation were wrong. 🔉✕⇢

The statement is that for a conductor carrying current $I$ with potential difference $V$ across its ends, $V\propto I$, which is written $V=IR$ with the constant of proportionality $R$ called the resistance, measured in ohm. The essential word in that sentence is constant. The proportionality is a claim about the conductor: as you turn the source up and down, the ratio $V/I$ does not budge. 🔉✕⇢

This is the single most important conceptual point in the whole topic, and it is almost always taught backwards. The equation $V=IR$, read as a bare formula, asserts nothing whatsoever. Given any conductor, any device, a diode, a filament lamp, an electrolytic cell, you may always measure $V$ and $I$ at some operating point and simply name their ratio $R$. That is a definition, and definitions cannot be falsified by experiment. Written as $R\equiv V/I$, it is true of everything, always, and so it says nothing about nature. 🔉✕⇢

Ohm law is the extra, independent, and entirely falsifiable claim that this ratio comes out the same at every operating point. It is that claim which experiment can test and which many devices fail. So the honest phrasing is: $V=IR$ defines resistance, whereas Ohm law asserts that $R$ so defined is a constant for the given conductor at a given temperature. Graphically, the law says the $V$ versus $I$ characteristic is a straight line passing through the origin, and its slope is $R$. 🔉✕⇢

The origin condition carries its own content. A straight line that does not pass through the origin, $V=IR+V_0$, is not ohmic either, because then $V/I$ still depends on $I$ even though the graph is straight. A useful diagnostic is the distinction between the static or chord resistance $V/I$, the slope of the line joining the operating point to the origin, and the dynamic or slope resistance $dV/dI$, the tangent. A conductor is ohmic if and only if the two are equal at every point. 🔉✕⇢

The microscopic explanation arrived a century later. Combining $j=nev_d$ with $v_d=eE\tau/m$ gives $j=(ne^2\tau/m)E$, that is $\mathbf{j}=\sigma\mathbf{E}$ or equivalently $\mathbf{E}=\rho\mathbf{j}$. This local form is Ohm law expressed at a point, and it makes clear exactly what has to be true for the law to hold: $n$ and $\tau$ must not themselves depend on $E$. Linearity is not a theorem, it is a consequence of that assumption, and the assumption is the thing that fails in real devices. 🔉✕⇢

It also shows why heating breaks the law in practice even for an ordinary metal. Pushing more current dissipates more power, the wire warms, $\tau$ shortens, $\rho$ rises and so does $R$. A tungsten filament lamp is the standard laboratory example: its $V$ versus $I$ curve bends over noticeably, not because tungsten is exotic but because the filament runs at a couple of thousand kelvin at full power and its resistance there is roughly ten times its cold resistance. This is why Ohm law is always quoted with the proviso that physical conditions, chiefly temperature, are unchanged. 🔉✕⇢

NCERT sorts the genuine deviations into three types, and they are worth keeping distinct because each fails differently. The first type is simple non-linearity: $V$ ceases to be proportional to $I$. The characteristic is still a single-valued, monotonic curve, and reversing the source still reverses the current symmetrically, but the curve bends, so the chord resistance drifts as you move along it. A good conductor at high current, where heating matters, behaves this way. 🔉✕⇢

The second type is asymmetry: the relation between $V$ and $I$ depends on the sign of $V$. If a certain $V$ drives a current $I$, then reversing the direction of $V$ while keeping its magnitude fixed does not produce a current of the same magnitude in the opposite direction. The diode is the canonical example, and NCERT points forward to the semiconductor chapter for it. In forward bias a diode passes milliamperes for a volt or less, while in reverse bias the same magnitude of voltage passes only microamperes or less, so the characteristic has to be drawn with different scales on the two sides of the origin. 🔉✕⇢

The third type is the most violent: the relation is not even unique, meaning there is more than one value of $V$ for the same current $I$. Gallium arsenide behaves this way, and the physical content is that over part of its range the material has a negative differential resistance, where increasing the voltage actually decreases the current. Nothing about $V=IR$ can survive this, since a function returning several values of $V$ for one $I$ cannot be summarised by any single number $R$. 🔉✕⇢

None of this makes such materials marginal. NCERT is careful to say that materials and devices which do not obey Ohm law are actually widely used in electronic circuits, and the point is easy to underrate. Every rectifier, every transistor, every logic gate, and every solar cell depends on non-ohmic behaviour. A world of purely ohmic components could carry power and dissipate heat but could not switch, amplify, or compute. Ohm law describes the plumbing; the interesting devices are the ones that break it deliberately. 🔉✕⇢

For examinations, the reliable moves are these. If asked to state Ohm law, state the proportionality and the constancy of $R$, and name the condition of constant physical state. If handed a $V$ versus $I$ graph, check straightness and check that it passes through the origin before calling anything ohmic. If asked for resistance at a point on a curved characteristic, ask whether the chord value $V/I$ or the slope value $dV/dI$ is wanted, since they differ. And never present $V=IR$ alone as a statement of the law. 🔉✕⇢

Resistance and Resistivity 🔉✕⇢

Once $R$ is defined as $V/I$, the obvious next question is what it depends on. Experiment answers immediately that it depends on two quite different things at once. Change the metal and $R$ changes; keep the metal and change the dimensions and $R$ changes too. Because these are logically separate influences, physics separates them: geometry is factored out explicitly, and what remains is a number characterising the substance. That number is the resistivity. 🔉✕⇢

NCERT establishes the geometric dependence with a neat argument that needs no new physics, only the definition of resistance and the fact that identical objects behave identically. Take a slab of length $l$ and cross-sectional area $A$ carrying current $I$ with potential difference $V$. Place a second, identical slab end to end with it, so the combination has length $2l$. The same current $I$ flows through both, since whatever enters the first leaves it and enters the second. 🔉✕⇢

Because the second slab is identical to the first and carries the same current, the potential difference across it is also $V$. The potential difference across the combination is therefore $2V$, while the current is still $I$, so the combination has resistance $R_C=2V/I=2R$. Thus doubling the length of a conductor doubles the resistance, and since the argument can be repeated for any number of identical slabs, $R\propto l$ in general. 🔉✕⇢

The area dependence is obtained by cutting rather than stacking. Slice the same slab lengthwise into two identical half-slabs, each still of length $l$ but each of cross-sectional area $A/2$. Apply the same $V$ across the original slab. By symmetry the total current $I$ divides equally, so each half carries $I/2$, while the potential difference across each half is the full $V$ because both ends are still connected to the same two terminals. 🔉✕⇢

The resistance of each half is therefore $R_1=V/(I/2)=2V/I=2R$. Halving the cross-sectional area doubles the resistance, so $R\propto1/A$. Combining the two proportionalities gives $R\propto l/A$, and introducing the constant of proportionality for a given conductor yields $R=\rho l/A$. The constant $\rho$ depends on the material of the conductor but not on its dimensions, and it is called the resistivity. 🔉✕⇢

Rearranging as $\rho=RA/l$ fixes the SI unit as ohm metre. It is worth pausing on why the unit looks strange: it is ohm times metre squared divided by metre, which is ohm metre, not ohm per metre. Writing $\Omega\,\mathrm{m^{-1}}$ is a common and costly slip. The reciprocal $\sigma=1/\rho$ is the conductivity, with unit siemens per metre or equivalently $\Omega^{-1}\mathrm{m^{-1}}$, and it is the more natural quantity when the local law is written as $\mathbf{j}=\sigma\mathbf{E}$. 🔉✕⇢

Resistivity is what sorts all materials into three classes, and the spread is enormous. Metals sit at the low end, in the range $10^{-8}$ to $10^{-6}$ ohm metre. Insulators such as ceramic, rubber and plastics sit at the other extreme, with resistivities $10^{18}$ times greater than metals or more. Semiconductors lie in between, and they carry an additional signature that distinguishes them from both: their resistivity characteristically decreases with a rise in temperature, and it can be lowered deliberately by adding small amounts of suitable impurities, which is the feature exploited in every electronic device. 🔉✕⇢

The microscopic expression for resistivity follows from the drift derivation. Inverting $\sigma=ne^2\tau/m$ gives $\rho=m/(ne^2\tau)$. This is a genuinely useful formula rather than a decorative one, because it names the only three things a material can vary: the carrier density $n$, the carrier mass $m$, and the mean free time $\tau$. Nothing about length or area appears in it, which is the formal reason resistivity is a material property. It also predicts the entire temperature story, since $\tau$ and $n$ are the quantities that respond to heating. 🔉✕⇢

The distinction between object and material becomes vivid when a wire is stretched. Drawing a wire through a die keeps the volume of metal constant, $V_{ol}=Al$, so if the length becomes $kl$ the area must become $A/k$. Then $R'=\rho(kl)/(A/k)=k^2\rho l/A=k^2R$. Stretching to twice the length quadruples the resistance, and at constant volume $R\propto l^2$. Equivalently, writing $A=V_{ol}/l$ gives $R=\rho l^2/V_{ol}$ directly. The resistivity has not changed by a hair; only the shape has. 🔉✕⇢

The same idea can be run the other way to catch a classic trap. If a wire is cut into $n$ equal pieces and those pieces are bundled in parallel, the length of each is $l/n$ and the combined area is $nA$, so the bundle has resistance $\rho(l/n)/(nA)=R/n^2$. Melting and recasting a wire into a different length behaves like stretching, since the volume is again conserved. In every one of these problems the single reliable move is to track $l$ and $A$ separately, remember what is being held fixed, and leave $\rho$ alone. 🔉✕⇢

It is also worth noting when $R=\rho l/A$ may be used at all. The derivation assumed a uniform cross-section and a uniform current density across it, so the formula applies to a straight wire or a prism but not directly to a cone or a spherical shell. For a tapering conductor the correct procedure is to integrate, treating the body as a series stack of thin slabs, $R=\int \rho\,dl/A(l)$. The same caution applies to the shape of the terminals: the formula assumes the current enters over the whole face, not at a point. 🔉✕⇢

A final habit that prevents most errors: state clearly which of the two quantities a question is about before computing anything. Resistivity belongs to copper; resistance belongs to this particular piece of copper wire. Two wires of the same material always have the same $\rho$ and almost never the same $R$. A question that says the wire is replaced by a thicker one of the same material is telling you that $\rho$ is fixed and $A$ has changed, and that alone usually solves it. 🔉✕⇢

Temperature Dependence of Resistivity 🔉✕⇢

The resistivity of a material is found to be dependent on the temperature, and different materials do not exhibit the same dependence. This is not a small correction bolted onto an otherwise clean law; it is the reason Ohm law must always be quoted with the proviso of constant physical conditions, and it is the mechanism behind everything from a toaster element to a platinum thermometer to the difference between a metal and a semiconductor. 🔉✕⇢

Over a range of temperature that is not too large, the resistivity of a metallic conductor is given approximately by $\rho_T=\rho_0[1+\alpha(T-T_0)]$, where $\rho_T$ is the resistivity at temperature $T$ and $\rho_0$ is its value at a chosen reference temperature $T_0$. The quantity $\alpha$ is the temperature coefficient of resistivity. Since $\alpha(T-T_0)$ must be dimensionless, the dimension of $\alpha$ is inverse temperature, and it is quoted per degree Celsius or per kelvin, the two being numerically identical because only a temperature difference appears. 🔉✕⇢

For metals $\alpha$ is positive, so resistivity rises with temperature. The relation implies that a graph of $\rho_T$ against $T$ is a straight line, and for copper this holds well over ordinary laboratory ranges. At temperatures much lower than zero degrees Celsius, however, the graph deviates considerably from a straight line, so the formula is a local linearisation valid near the chosen $T_0$ rather than a global truth. Choosing a reference point near the temperatures of interest is therefore part of using it correctly. 🔉✕⇢

Because the geometry factor $l/A$ is common to both sides and changes only slightly with thermal expansion, the same linear form is used for resistance: $R_T=R_0[1+\alpha(T-T_0)]$. Nearly every numerical problem is set up this way, since resistance is what an ohmmeter reads. The mild inconsistency of ignoring expansion is deliberate and harmless, because for a metal the fractional change in $\rho$ with temperature outruns the fractional change in $l/A$ by two or three orders of magnitude. 🔉✕⇢

The physical cause is read straight off the microscopic formula $\rho=m/(ne^2\tau)$. Resistivity depends inversely on both the number $n$ of free electrons per unit volume and the average time $\tau$ between collisions. As temperature rises, the average speed of the electrons increases and the lattice ions vibrate more vigorously, so collisions become more frequent and $\tau$ decreases. In a metal $n$ does not depend on temperature to any appreciable extent, since the conduction electrons are already all liberated at any ordinary temperature. With $n$ fixed and $\tau$ falling, $\rho$ must rise, which is exactly what is observed. 🔉✕⇢

For insulators and semiconductors the competition comes out the other way. Their carriers must be thermally excited across an energy gap, so $n$ increases with temperature, and it increases very rapidly, roughly exponentially. That increase more than compensates any decrease in $\tau$, so for such materials $\rho$ decreases with temperature and $\alpha$ is effectively negative. This is the defining signature of a semiconductor: resistivity characteristically decreasing with a rise in temperature, in flat contradiction to the metallic behaviour. 🔉✕⇢

Between these extremes sit materials engineered to do almost nothing. Nichrome, an alloy of nickel, iron and chromium, exhibits a very weak dependence of resistivity on temperature, and manganin and constantan have similar properties. Their $\alpha$ values are one to two orders of magnitude smaller than those of pure metals, because in a disordered alloy the dominant scattering is off the random arrangement of the atoms themselves, a temperature-independent process, which swamps the temperature-dependent scattering off lattice vibrations. That is why these alloys are widely used in wire-bound standard resistors: their resistance values change very little with temperature. 🔉✕⇢

The NCERT toaster problem shows the whole scheme in action. A toaster uses nichrome for its heating element. When a negligibly small current passes, so that heating effects can be ignored and the element sits at room temperature $T_1=27.0$ degrees Celsius, its resistance is measured to be $R_1=75.3\,\Omega$. The element is then connected to a 230 V supply, and after a few seconds the current settles to a steady value of 2.68 A. The temperature coefficient of resistance of nichrome, averaged over the range involved, is $\alpha=1.70\times10^{-4}\,{}^\circ\mathrm{C}^{-1}$. The question asks for the steady temperature of the element. 🔉✕⇢

The reasoning behind the word settles deserves attention, because it is the physics of the problem. When the toaster is first switched on its current is slightly higher than the steady value, but the heating effect raises the temperature, which raises the resistance, which lowers the current. This is a self-limiting feedback loop. Within a few seconds a steady state is reached in which the temperature rises no further, because the heat generated electrically equals the heat lost to the surroundings, and both the resistance and the current have settled. 🔉✕⇢

Now the arithmetic. The resistance at the steady temperature $T_2$ is $R_2=230\,\mathrm{V}/2.68\,\mathrm{A}=85.8\,\Omega$. Using $R_2=R_1[1+\alpha(T_2-T_1)]$ and solving for the temperature difference gives $T_2-T_1=(R_2-R_1)/(R_1\alpha)=(85.8-75.3)/(75.3\times1.70\times10^{-4})$. The numerator is $10.5\,\Omega$ and the denominator is $1.28\times10^{-2}\,\Omega\,{}^\circ\mathrm{C}^{-1}$, so $T_2-T_1=820\,{}^\circ\mathrm{C}$. Hence $T_2=820+27.0=847\,{}^\circ\mathrm{C}$, which is the steady temperature of the heating element. 🔉✕⇢

Two features of that answer are worth noticing. First, the resistance changed by only about 14 percent even though the temperature changed by more than 800 degrees, which is precisely the weak dependence that makes nichrome a good heating alloy: the element does not run away as it heats. Second, had the same calculation been done with copper, whose $\alpha$ is about $4\times10^{-3}$ per degree, the same 14 percent change in resistance would have corresponded to a rise of only about 35 degrees, so the small $\alpha$ of nichrome is exactly what allows it to reach incandescent temperatures under control. 🔉✕⇢

The same linear law run backwards turns a resistor into a thermometer, which is the platinum resistance thermometer. Given $R_0$ at the ice point and $R_{100}$ at the steam point, an unknown temperature follows from the interpolation $t=(R_t-R_0)\times100/(R_{100}-R_0)$. With NCERT numbers $R_0=5\,\Omega$, $R_{100}=5.23\,\Omega$ and $R_t=5.795\,\Omega$, this gives $t=0.795\times100/0.23=345.65$ degrees Celsius. Platinum is chosen because its response is close to linear over a wide range and it is chemically stable, not because its $\alpha$ is especially large. 🔉✕⇢

Series & Parallel Resistors 🔉✕⇢

Almost every circuit problem in this chapter begins by collapsing a tangle of resistors into a single equivalent resistance, and the two rules that make this possible — the series rule and the parallel rule — are the workhorses of the subject. They follow directly from the two things that are conserved in a steady circuit: charge (so current is continuous) and energy (so potential differences add up around a path). 🔉✕⇢

Resistors are in series when they are joined end to end so that the same current must pass through each in turn, with no junction in between to let current escape. Because charge is not created or destroyed at the joins, the current $I$ is identical in every series resistor. The potential differences, however, add: the drop across the chain is the sum of the drops across the parts, $V=V_1+V_2+\cdots=IR_1+IR_2+\cdots$. Dividing by the common current gives the series rule $R_{eq}=R_1+R_2+\cdots=\sum R_i$. A series combination always has a larger resistance than any single member, because you are simply making the current fight through more material in a row. 🔉✕⇢

The NCERT slab argument makes this feel inevitable. Imagine a conducting slab of length $l$ and area $A$ with some resistance $R$. Place a second identical slab end to end with the first, so the combined length is $2l$. The same current flows through both, and the potential difference across the pair is the sum of the two equal drops, hence twice as large; so the resistance of the combination is $2R$. Stacking conductors in a line doubles the length and doubles the resistance — which is the geometric root of both $R\propto l$ and the series rule. 🔉✕⇢

Resistors are in parallel when they are connected between the same two nodes, so that each one experiences the identical potential difference $V$ across its ends. Now it is the current that splits: the total current entering the node divides among the branches, $I=I_1+I_2+\cdots$, by the junction rule. Since each branch obeys $I_i=V/R_i$, we have $I=V/R_1+V/R_2+\cdots$, and dividing by the common voltage gives the parallel rule $\dfrac{1}{R_{eq}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+\cdots=\sum\dfrac{1}{R_i}$. 🔉✕⇢

The companion NCERT slab argument covers parallel. Take the original slab and split it lengthwise into two half-slabs, each of the same length $l$ but half the cross-sectional area, $A/2$. Each half now carries half the current at the same voltage, so each has resistance $2R$; but the two half-slabs side by side are just the original slab, whose resistance is $R$. Two resistances of $2R$ in parallel give $R$ — consistent with the parallel rule, and the geometric origin of $R\propto 1/A$. 🔉✕⇢

A parallel combination always has a resistance smaller than the smallest member, because adding another path can only give the current more room to flow. This leads to a useful mental check that trips up the unwary: in parallel it is the smallest resistor that dominates the equivalent resistance, not the largest. For two resistors the handy closed form is $R_{eq}=\dfrac{R_1R_2}{R_1+R_2}$ — the product over the sum — which is worth memorising because two-resistor parallels appear constantly. 🔉✕⇢

Current and power sharing follow from the same ideas and are frequently examined. In series, the current is common, so the power dissipated in each resistor, $P_i=I^2R_i$, is proportional to its resistance — the biggest resistor gets hottest. In parallel, the voltage is common, so the power in each branch, $P_i=V^2/R_i$, is inversely proportional to its resistance — now the smallest resistor dissipates the most and carries the largest share of current, since the branch current $I_i=V/R_i$ is largest where $R_i$ is smallest. 🔉✕⇢

For networks that are neither purely series nor purely parallel, the strategy is to reduce step by step from the inside out. Identify a pair of resistors that are unambiguously in series or in parallel, replace them with their equivalent, redraw the simpler circuit, and repeat until a single resistor remains. Ladder networks, cube-of-resistors problems and bridge networks are all handled either by this reduction or, when symmetry or a balance condition applies, by exploiting equal-potential nodes to merge or delete branches. 🔉✕⇢

A final practical caution: only genuinely series or genuinely parallel groupings may be combined by these rules. Two resistors that share just one node are not in parallel, and two resistors with a junction between them that leaks current elsewhere are not in series. When a network resists reduction — most famously an unbalanced Wheatstone bridge — no amount of series–parallel manipulation will simplify it, and one must fall back on Kirchhoff's rules to write and solve the loop and junction equations directly. 🔉✕⇢

EMF, Internal Resistance and Terminal Voltage 🔉✕⇢

A steady current needs something to maintain a steady field in the conductor, and the simplest such device is the electrolytic cell. A cell has two electrodes, a positive P and a negative N, immersed in an electrolytic solution. Dipped in the solution, each electrode exchanges charge with the electrolyte until an equilibrium is reached. The positive electrode ends up at a potential $V_+$ above the electrolyte immediately adjacent to it, and the negative electrode ends up at a potential $V_-$ below the electrolyte adjacent to it, with both $V_+$ and $V_-$ non-negative. 🔉✕⇢

When no current flows, the electrolyte has the same potential throughout, so the potential difference between P and N is simply the sum of the two electrode drops, $V_+-(-V_-)=V_++V_-$. This difference is called the electromotive force of the cell and is denoted $\varepsilon$, so $\varepsilon=V_++V_-$, a positive quantity. The energy source behind it is chemical: the cell converts chemical energy into the work needed to carry charge from the low-potential terminal to the high-potential one inside itself, against the electrostatic force. 🔉✕⇢

NCERT is unusually blunt about the name, and the point deserves repeating. The emf is actually a potential difference and not a force. The word force survives only for historical reasons, from a time when the phenomenon was not understood properly. Its unit is the volt, not the newton, and it is measured per unit charge. Any answer that treats emf as a force, or assigns it newtons, or tries to add it to a mechanical force, has misread the name for the thing. 🔉✕⇢

The operational definition follows from the no-current case. Consider first the situation when the external resistance $R$ is infinite, so that $I=0$. Walking from P to N through the cell, the potential difference is the drop from P into the electrolyte, plus the change across the electrolyte, plus the change from the electrolyte to N. With no current the middle term vanishes, and the total is exactly $\varepsilon$. Hence emf is the potential difference between the positive and negative electrodes in an open circuit, when no current is flowing through the cell. 🔉✕⇢

The electrolyte through which the current flows has a finite resistance $r$, called the internal resistance of the cell. It is not a manufacturing defect; it is the unavoidable resistance of the conducting path inside the cell, and it depends on the separation and area of the electrodes, the concentration of the electrolyte, and the temperature. NCERT notes that internal resistances vary widely from cell to cell, and that the internal resistance of dry cells is much higher than that of common electrolytic cells. 🔉✕⇢

Now close the circuit. If $R$ is finite, $I$ is not zero, and the current flows through the electrolyte from N to P, which is from B to A in the cell interior. The potential difference between the terminals becomes $V=V_++V_--Ir=\varepsilon-Ir$. The minus sign is the whole content of the equation: some of the energy the cell supplies per unit charge is spent driving that charge through the cell own resistance, and only the remainder is delivered to the outside world. The quantity $V$ is called the terminal voltage. 🔉✕⇢

Since $V$ is also the potential difference across the external resistor, Ohm law gives $V=IR$. Setting the two expressions equal, $IR=\varepsilon-Ir$, and collecting terms gives $I(R+r)=\varepsilon$, so $I=\varepsilon/(R+r)$. This is the master formula of single-loop circuits, and it reads exactly as it should: the emf drives the current through the total resistance of the loop, internal and external in series. Substituting back gives the terminal voltage $V=IR=\varepsilon R/(R+r)$. 🔉✕⇢

The two limiting cases anchor everything else. On open circuit, $R$ tends to infinity, $I$ tends to zero, and $V$ tends to $\varepsilon$. This is why a voltmeter of very high resistance placed across an isolated cell reads essentially the emf, and why emf and terminal voltage are so easily confused: in the commonest measurement they nearly coincide. On short circuit, $R=0$, and the current reaches its maximum possible value $I_{max}=\varepsilon/r$, with the terminal voltage collapsing to zero. In most cells the maximum allowed current is kept far below this to prevent permanent damage. 🔉✕⇢

The expression $V=\varepsilon R/(R+r)=\varepsilon/(1+r/R)$ shows that the terminal voltage always falls short of the emf while current is drawn, and that the shortfall is governed entirely by the ratio $r/R$. In practical calculations the internal resistance may be neglected when the current is small enough that $\varepsilon$ greatly exceeds $Ir$, which is why a fresh cell driving a high-resistance load can be treated as ideal. It is also why a car battery, with $r$ of a few milliohms, holds its terminal voltage under a headlamp load but visibly dips when the starter motor demands hundreds of amperes. 🔉✕⇢

Measuring $r$ follows directly from the same equation. The simplest method uses two known loads: with $R_1$ the terminal voltage is $V_1$ and with $R_2$ it is $V_2$, and eliminating $\varepsilon$ from $\varepsilon=V_1+V_1r/R_1=V_2+V_2r/R_2$ gives $r$. A single-load version is even quicker if the emf is known independently: from $V=\varepsilon-Ir$ and $I=V/R$ one obtains $r=R(\varepsilon-V)/V=R(\varepsilon/V-1)$. The classical laboratory route is a potentiometer, which compares the open-circuit and loaded balancing lengths and so measures $r$ without drawing current from the cell during the emf reading. 🔉✕⇢

A useful consistency check is the energy balance. Multiplying $\varepsilon=I(R+r)$ by $I$ gives $\varepsilon I=I^2R+I^2r$. The left side is the total power the chemical reaction supplies, the first term on the right is the power delivered to the external circuit, and the second is the power wasted heating the cell itself. Nothing is unaccounted for. This also explains why a heavily loaded cell gets warm and why its efficiency, the ratio $R/(R+r)$, drops as the load resistance is lowered. 🔉✕⇢

For problem solving, the habits that matter are these. Always draw the internal resistance explicitly in series with the cell, so that it cannot be forgotten when summing loop resistances. Read the words carefully: emf, open-circuit voltage, and the reading of an ideal voltmeter across an isolated cell all mean $\varepsilon$, whereas terminal voltage, potential difference across the cell, and voltage across the external resistor all mean $V$. And if a question says a cell of emf 2 V has 1.8 V across its terminals, it has already told you that $Ir=0.2$ V. 🔉✕⇢

Cells in Series and in Parallel 🔉✕⇢

Like resistors, cells can be combined in a circuit, and like resistors a combination of cells can be replaced by a single equivalent cell for the purpose of calculating currents and voltages elsewhere. The equivalence is defined by behaviour at the two external terminals: the replacement cell must produce the same terminal voltage as the original combination for every current drawn. That definition is what makes the derivation mechanical, because both the combination and its replacement obey a relation of the form $V=\varepsilon-Ir$, and matching coefficients does the rest. 🔉✕⇢

Take two cells in series first, with emfs $\varepsilon_1$ and $\varepsilon_2$ and internal resistances $r_1$ and $r_2$. The negative terminal of the first is joined to the positive terminal of the second, leaving the outer terminals A and C free, with the junction at B. The same current $I$ flows through both cells, since there is nowhere else for it to go. Applying the single-cell result to each in turn, the potential differences are $V_{AB}=\varepsilon_1-Ir_1$ and $V_{BC}=\varepsilon_2-Ir_2$. 🔉✕⇢

Potential differences along a path simply add, so $V_{AC}=V_{AB}+V_{BC}=(\varepsilon_1+\varepsilon_2)-I(r_1+r_2)$. Comparing this with the defining form $V_{AC}=\varepsilon_{eq}-Ir_{eq}$ and matching the constant and the coefficient of $I$ gives $\varepsilon_{eq}=\varepsilon_1+\varepsilon_2$ and $r_{eq}=r_1+r_2$. In words, the equivalent emf of a series combination of $n$ cells is just the sum of their individual emfs, and the equivalent internal resistance is just the sum of their internal resistances. 🔉✕⇢

That rule carries a crucial condition: it holds when the current leaves each cell from its positive electrode, meaning the cells are aligned to push in the same direction. If instead the two negative terminals are joined, the second cell is reversed, and its terminal relation becomes $V_{BC}=-\varepsilon_2-Ir_2$. The algebra then delivers $\varepsilon_{eq}=\varepsilon_1-\varepsilon_2$. Stated generally: if in the combination the current leaves any cell from its negative electrode, the emf of that cell enters the expression for $\varepsilon_{eq}$ with a negative sign. The internal resistances, being resistances, always add regardless of orientation. 🔉✕⇢

This sign rule has a consequence worth stating explicitly, because it is where marks are lost. Reversing a cell subtracts twice its emf from the total, not once. Two identical 1.5 V cells aiding give 3.0 V; reverse one and you get 0 V, not 1.5 V. And the reversed cell is being charged rather than discharged, absorbing energy at the rate $\varepsilon_2 I$ from the rest of the circuit, which is exactly what happens when a battery is put into a charger. 🔉✕⇢

Now consider the parallel combination, in which both positive terminals are joined at $B_1$ and both negative terminals at $B_2$, with the external circuit connected across $B_1$ and $B_2$. Here the two cells share the same terminal voltage but split the current. Let $I_1$ and $I_2$ be the currents leaving the positive electrodes of the two cells. Since as much charge flows into the junction as flows out, $I=I_1+I_2$, which is just the junction rule. 🔉✕⇢

Because both cells span the same pair of points, each obeys $V=\varepsilon_1-I_1r_1$ and $V=\varepsilon_2-I_2r_2$ with the same $V$. Solving each for its current gives $I_1=(\varepsilon_1-V)/r_1$ and $I_2=(\varepsilon_2-V)/r_2$. Adding them, $I=(\varepsilon_1/r_1+\varepsilon_2/r_2)-V(1/r_1+1/r_2)$. Rearranging to isolate $V$ yields $V=(\varepsilon_1r_2+\varepsilon_2r_1)/(r_1+r_2)-I\,r_1r_2/(r_1+r_2)$. 🔉✕⇢

Matching this against $V=\varepsilon_{eq}-Ir_{eq}$ identifies $\varepsilon_{eq}=(\varepsilon_1r_2+\varepsilon_2r_1)/(r_1+r_2)$ and $r_{eq}=r_1r_2/(r_1+r_2)$. These are more memorably written as $1/r_{eq}=1/r_1+1/r_2$ and $\varepsilon_{eq}/r_{eq}=\varepsilon_1/r_1+\varepsilon_2/r_2$. The second form is the one to remember, because it generalises immediately to any number of cells and because it makes plain that what adds in parallel is not emf but emf divided by internal resistance, which is a current. If the second cell is connected with its negative terminal to the positive of the first, the same equations hold with $\varepsilon_2$ replaced by $-\varepsilon_2$. 🔉✕⇢

Extending to $n$ cells in parallel gives $1/r_{eq}=1/r_1+\ldots+1/r_n$ and $\varepsilon_{eq}/r_{eq}=\varepsilon_1/r_1+\ldots+\varepsilon_n/r_n$. Notice that $\varepsilon_{eq}$ is a weighted average of the individual emfs, with weights $1/r_i$, so it always lies between the largest and smallest emf present. It can never exceed the largest. That single observation kills an entire family of wrong answers. 🔉✕⇢

The identical-cell cases are the ones that appear most often. For $n$ identical cells of emf $\varepsilon$ and internal resistance $r$ in series, $\varepsilon_{eq}=n\varepsilon$ and $r_{eq}=nr$, so the current through an external $R$ is $I=n\varepsilon/(R+nr)$. For $m$ identical cells in parallel, the weighted average collapses to $\varepsilon_{eq}=\varepsilon$ and $r_{eq}=r/m$, giving $I=\varepsilon/(R+r/m)=m\varepsilon/(mR+r)$. The parallel bank offers no extra voltage at all; what it offers is a lower internal resistance. 🔉✕⇢

That difference decides which arrangement to use, and the criterion is a comparison between $R$ and $r$. Series is the right choice when the external resistance is large compared with the internal resistance, since then $R+nr$ is dominated by $R$ and the current scales almost as $n\varepsilon/R$: stacking cells buys voltage, which is what a high-resistance load needs. Parallel is the right choice when $R$ is small compared with $r$, since then dividing the internal resistance by $m$ substantially raises the current: a bank of cells in parallel buys current-delivering capacity and shares the load, so each cell heats less and lasts longer. 🔉✕⇢

A practical warning attaches to parallel connection. Because $\varepsilon_{eq}$ is a weighted average, cells of unequal emf connected in parallel do not merely average out politely; a circulating current flows between them even with no external load. Setting $I=0$ in the junction equation gives $I_1=-I_2=(\varepsilon_1-\varepsilon_2)/(r_1+r_2)$, so the stronger cell drives the weaker one backwards, wasting energy and heating both. This is the physical reason manufacturers insist that only cells of matched type and charge state be paralleled. 🔉✕⇢

Kirchhoff's Rules 🔉✕⇢

Electric circuits generally consist of a number of resistors and cells interconnected sometimes in a complicated way. The formulae derived earlier for series and parallel combinations of resistors are not always sufficient to determine all the currents and potential differences in the circuit. As soon as a network has more than one independent loop — two cells that drive current through a shared resistor, say — there is no single pair of points across which we can reduce the entire network by the series and parallel rules. Two rules, called Kirchhoff's rules, are very useful for analysis of such electric circuits. They are not new physics: the first is conservation of charge for steady currents, and the second is conservation of energy written for a closed loop. 🔉✕⇢

Full derivation, worked example and interactive 3D on the Kirchhoff's Rules tab →

Wheatstone Bridge 🔉✕⇢

As an application of Kirchhoff's rules consider the circuit called the Wheatstone bridge. The bridge has four resistors $R_1$, $R_2$, $R_3$ and $R_4$. Across one pair of diagonally opposite points, $A$ and $C$, a source is connected; this is called the battery arm. Between the other two vertices, $B$ and $D$, a galvanometer $G$, which is a device to detect currents, is connected; this line is called the galvanometer arm. The value of the arrangement is that it compares one resistance against others with great accuracy, without our having to know either the current supplied by the cell or the potential difference across its terminals. 🔉✕⇢

Full derivation, worked example and interactive 3D on the Wheatstone Bridge tab →

Metre Bridge 🔉✕⇢

A practical device using this principle is called the meter bridge, and it turns the balance condition of the Wheatstone bridge into a measured length. The device consists of a wire of uniform cross-section, exactly one metre long, placed along a scale graduated in millimetres. Because the resistance of a uniform wire is proportional to its length, the two portions of the wire on either side of a moving point of connection are automatically two of the four arms of the bridge, and the ratio of those two resistances is varied smoothly just by moving that point of connection along the wire. 🔉✕⇢

The arrangement maps directly onto the four arms of the Wheatstone bridge. Two gaps at the top hold a known resistance $R$, usually a resistance box, in one gap and the unknown resistance $S$ in the other. The one metre wire forms the other two arms. A cell drives current from one end of the wire to the other, and a galvanometer is connected between the junction of $R$ and $S$ at the top and a sharp moving contact, called the jockey, which can be pressed down on the wire at any point along its length. The jockey plays the part of the variable resistance $R_3$ of the standard bridge, except that here the variation is continuous and is read as a length. 🔉✕⇢

To determine the unknown resistance the jockey is pressed down at various points and the galvanometer is observed. At most points the galvanometer deflects one way or the other; there is one special point, called the balance point or the null point, where the deflection is exactly zero. Let this point divide the wire into a length $l$ on the left and a length $100-l$ on the right, both measured in cm. The resistances of these two portions are in the ratio $l$ to $100-l$, because resistance is proportional to length for a wire of uniform cross-section. 🔉✕⇢

Applying the balance condition of the Wheatstone bridge, with the arms at the top being $R$ and $S$ and the two portions of the wire in the ratio $l$ to $100-l$, gives the working relation of the device, $R/S=l/(100-l)$. Solving for the unknown, $S=R(100-l)/l$. Everything on the right is known or measured: $R$ is read from the resistance box and $l$ from the scale, so $S$ follows at once. Note again that the emf of the cell and the resistance of the galvanometer never appear in the result. The metre bridge takes over from the Wheatstone bridge this useful property of a null method: the value obtained depends only on ratios and on a measured length. 🔉✕⇢

The choice of the known resistance $R$ decides how accurately the null point can be located. The galvanometer responds most strongly to a small movement of the jockey when the balance point falls near the middle of the wire, close to the 50 cm mark. If $R$ is chosen so badly that balance occurs near one end, say at 5 cm or at 95 cm, then one of the two portions of the wire is very short, the fractional change in its resistance per millimetre of travel is small, and the null point is hard to locate. Good practice is to select a value of $R$ comparable to the expected value of $S$, so that $l$ lies roughly between 30 cm and 70 cm. 🔉✕⇢

Real metre bridges have deviations from the ideal relation, and these are examined occasionally. The most important is the correction at the two ends of the wire: the joints and the thick copper strips at the two ends contribute small resistances which are not part of the measured length, so the zero of the scale is slightly shifted. Taking a second reading with $R$ and $S$ interchanged and averaging removes these end effects to a first approximation. A second source of deviation is a cross-section that is not uniform along the wire, which breaks the assumed proportionality between resistance and length; this is why the wire is drawn with care and protected from being scratched. 🔉✕⇢

Heating is a third deviation and is easily overlooked. If too large a current flows for too long, the wire becomes warm, its resistivity increases, since the temperature coefficient of resistivity of a metal is positive, and the balance point drifts while the reading is being taken. The remedy is to keep the current small, to include a resistance in series with the cell, and to press the jockey down only briefly rather than dragging it along the wire, which also protects the wire from wear that would change its cross-section over time. 🔉✕⇢

The metre bridge shows the same symmetry as the bridge it is derived from. Interchanging $R$ and $S$ between the two gaps moves the balance point from $l$ to $100-l$, and comparing the two readings is exactly how the end corrections are obtained in careful work. There is a further consequence worth noting: since $S=R(100-l)/l$, a small uncertainty in the measured $l$ produces the smallest fractional uncertainty in $S$ when $l$ is near 50 cm, which is the same conclusion reached earlier from the deflection of the galvanometer. Two different arguments, one about the response of the galvanometer and one about the arithmetic, point to the same working rule. 🔉✕⇢

In summary, the metre bridge is the Wheatstone bridge with two of its arms replaced by the two portions of a uniform wire, so that a ratio of resistances is read as a ratio of lengths. Its single working relation $S=R(100-l)/l$ carries all the physics; its accuracy is governed by keeping the null point near the middle of the wire and by correcting for the ends; and its virtue, shared with every null method, is that the value obtained does not depend on the emf of the cell, on the resistance of the galvanometer, or on the current supplied. 🔉✕⇢

Electrical Energy, Power and Heating 🔉✕⇢

Consider a conductor with end points A and B carrying a current $I$ from A to B. Since current flows from A to B, the potential at A exceeds that at B, so the potential difference $V=V(A)-V(B)$ is positive. In a time interval $\Delta t$, an amount of charge $\Delta Q=I\Delta t$ travels from A to B. Its potential energy at A was $\Delta Q\,V(A)$ and at B it is $\Delta Q\,V(B)$, so the change in potential energy is $\Delta U=\Delta Q[V(B)-V(A)]=-\Delta Q\,V=-IV\Delta t$, a negative quantity. The charge loses potential energy in transit. 🔉✕⇢

What happens to that energy depends entirely on whether the carriers collide. If charges moved freely through the conductor without collisions, conservation of energy would give $\Delta K=-\Delta U=IV\Delta t$, so their kinetic energy would grow steadily as they moved: the conductor would act as an accelerator, and the carriers would arrive at B moving fast. This is what happens in a vacuum tube or an accelerator beamline, and it is precisely what does not happen in a wire. 🔉✕⇢

In an actual conductor the carriers do not accelerate on average; they move with a steady drift velocity, because of collisions with the ions and atoms during transit. During each collision the energy gained by the charge since its last collision is handed over to the lattice. The atoms vibrate more vigorously, which is to say the conductor heats up. So the energy lost from the electrostatic potential energy of the charge does not accumulate as kinetic energy of the carriers at all; it flows into the thermal energy of the material. In the time $\Delta t$, the energy dissipated as heat is $\Delta W=IV\Delta t$. 🔉✕⇢

The energy dissipated per unit time is the power dissipated, $P=\Delta W/\Delta t$, and therefore $P=VI$. This is the most general of the power formulae and the only one that is always valid, because it was derived from the definition of potential difference alone and never used Ohm law. It applies to a resistor, a motor, a cell being charged, a diode, an electrolytic bath, anything at all: the rate at which electrical energy is delivered to a two-terminal element is the product of the voltage across it and the current through it. 🔉✕⇢

For an ohmic resistor, and only then, Ohm law lets $V=IR$ be substituted to give the two derived forms $P=I^2R$ and $P=V^2/R$. These are the ohmic loss, sometimes called Joule heating, and it is this power which heats the coil of an electric bulb to incandescence so that it radiates out heat and light. All three expressions give the same number in any situation where all three of $V$, $I$ and $R$ are defined and consistent; they are not alternative physical claims but algebraic rearrangements of one. 🔉✕⇢

Choosing among them is a matter of convenience, and the rule is to pick the form whose two symbols are the ones actually held fixed or known in the problem. If a resistor is in series with others, the current through it is the shared quantity, so $P=I^2R$ is natural. If it is in parallel across a supply, the voltage is shared, so $P=V^2/R$ is natural. Reaching for the wrong one is not an error of algebra but an invitation to the misconception discussed below. 🔉✕⇢

Where does this power come from? A steady current requires an external source to maintain the field, and it is that source which must supply the power. In a simple circuit consisting of a cell and a resistor, it is the chemical energy of the electrolyte that is drawn down, and it continues for as long as the reactants last. The full accounting is $\varepsilon I=I^2R+I^2r$: the cell converts chemical energy at the rate $\varepsilon I$, delivers $I^2R$ to the external resistor, and wastes $I^2r$ heating itself. 🔉✕⇢

Energy, as distinct from power, is the time integral: for a steady current, $W=Pt=VIt=I^2Rt$. The SI unit is the joule, but the commercial unit is the kilowatt hour, the energy consumed by a one kilowatt device running for one hour, which is $1\,\mathrm{kWh}=1000\,\mathrm{W}\times3600\,\mathrm{s}=3.6\times10^{6}\,\mathrm{J}$. A domestic meter reads in these units. A 2 kW heater run for three hours consumes 6 kWh, which is $2.16\times10^{7}$ joules, and a 60 W bulb left on for a full day consumes 1.44 kWh. 🔉✕⇢

Appliance ratings encode the same relations and are often misread. A bulb marked 100 W, 220 V is telling you the power it draws at its rated voltage, from which its operating resistance follows as $R=V^2/P=220^2/100=484\,\Omega$ and its rated current as $I=P/V=0.45$ A. The rating is not an intrinsic property: run that bulb at 110 V and, ignoring the temperature dependence of the filament, it draws $V^2/R=110^2/484=25$ W, a quarter of its rated power. Halving the voltage quarters the power because power goes as the square of the voltage at fixed resistance. 🔉✕⇢

Equation $P=I^2R$ has an important application to power transmission. Electrical power is carried from generating stations to homes and factories that may be hundreds of miles away, over cables whose resistance $R_c$ is considerable, and the loss in those cables is pure waste. Suppose a power $P$ is to be delivered to a device at voltage $V$, so that the current in the line is $I=P/V$. The power dissipated in the connecting wires is then $P_c=I^2R_c=P^2R_c/V^2$. 🔉✕⇢

The conclusion is the reason the grid looks the way it does: to deliver a fixed power $P$, the power wasted in the connecting wires is inversely proportional to the square of the transmission voltage. Raise $V$ by a factor of ten and the loss falls by a factor of a hundred, with no change to the cable at all. This is why transmission lines carry current at enormous values of $V$, hundreds of kilovolts, and why high voltage danger signs are a common sight as one moves away from populated areas. Using electricity at such voltages is not safe, so at the receiving end a transformer lowers it to a value suitable for use. 🔉✕⇢

The alternative strategy, thickening the cable to cut $R_c=\rho l/A$, works in principle but scales badly: halving the loss means doubling the mass of copper or aluminium over hundreds of miles, at enormous cost and with towers strong enough to carry it. Raising the voltage costs only insulation and transformers, which are one-time and local. That asymmetry, not any deep physics, is why the entire world settled on high-voltage alternating-current transmission. 🔉✕⇢

Drift Speed versus Signal Speed 🔉✕⇢

This is the single largest conceptual trap in current electricity, and it is worth confronting head on. Nearly everyone arrives with a picture of electrons racing from the switch to the bulb, and the lamp lighting when the first of them arrives. That picture is wrong by about eleven orders of magnitude, and it also fails to explain how alternating current works at all. Replacing it correctly requires two separate numbers and a clear statement of which one does which job. 🔉✕⇢

The first number is the drift speed. From $I=neAv_d$ the drift speed is $v_d=I/(neA)$, and NCERT evaluates it for a copper wire of cross-sectional area $1.0\times10^{-7}\,\mathrm{m^2}$ carrying 1.5 A. Assuming each copper atom contributes roughly one conduction electron, the carrier density is obtained from the density of copper, $9.0\times10^{3}\,\mathrm{kg\,m^{-3}}$, and its atomic mass of 63.5 u, giving $n=(6.0\times10^{23}/63.5)\times9.0\times10^{6}=8.5\times10^{28}\,\mathrm{m^{-3}}$. 🔉✕⇢

Substituting, $v_d=1.5/(8.5\times10^{28}\times1.6\times10^{-19}\times1.0\times10^{-7})=1.1\times10^{-3}\,\mathrm{m\,s^{-1}}$, that is 1.1 millimetres per second. This is a slower crawl than almost any motion one meets in daily life. It is slower than the minute hand of a large clock, slower than a snail, comparable to the growth of a stalactite over a human lifetime only in the sense that both are unimpressive. And yet it is the drift that constitutes the entire 1.5 A of current. 🔉✕⇢

Two comparisons put that figure in context, both drawn from NCERT. The thermal speed of a copper atom at 300 K, from the equipartition estimate $\tfrac12Mv^2=\tfrac32k_BT$, is about $2\times10^{2}\,\mathrm{m\,s^{-1}}$, so the drift speed is about $10^{-5}$ times the typical thermal speed at ordinary temperatures. The electrons themselves move even faster and far more randomly. The drift is therefore a tiny systematic bias superposed on a violent random motion, not a stately procession. 🔉✕⇢

The second comparison is the decisive one. An electric field travelling along the conductor propagates at the speed of an electromagnetic wave, namely $3.0\times10^{8}\,\mathrm{m\,s^{-1}}$. The drift speed is smaller than this by a factor of about $10^{-11}$. These two speeds differ so enormously that they cannot possibly be describing the same process, and recognising that they describe two genuinely different processes is the whole resolution. 🔉✕⇢

NCERT states the resolution in one sentence, in answer to the question of how current is established almost the instant a circuit is closed. The electric field is established throughout the circuit almost instantly, with the speed of light, causing at every point a local electron drift. Establishment of a current does not have to wait for electrons from one end of the conductor travelling to the other end. It does, however, take a little while for the current to reach its steady value. 🔉✕⇢

The key phrase is at every point. When the switch closes, surface charges redistribute over the conductors and set up the field throughout the circuit within nanoseconds. Every electron in the wire, including the ones already sitting inside the bulb filament, feels that field essentially simultaneously and begins to drift at once. The filament does not wait for a specific electron from the battery; it already contains an enormous supply of its own, and they start moving as soon as they are pushed. 🔉✕⇢

Numbers make the contrast concrete. Over a 1 metre length of the copper wire above, the time for an individual electron to traverse end to end is $t=L/v_d=1/(1.1\times10^{-3})\approx9.1\times10^{2}\,\mathrm{s}$, roughly fifteen minutes. The time for the field to traverse the same metre is $t=L/c=1/(3.0\times10^{8})\approx3.3\times10^{-9}\,\mathrm{s}$, about three nanoseconds. If lighting a lamp required an electron to make the journey, switching on a room light would be a fifteen-minute affair, and a transcontinental telephone call would be impossible. 🔉✕⇢

The complementary puzzle is how such a feeble drift can deliver amperes at all, and the answer is the sheer number of carriers. NCERT puts it simply: the electron number density is enormous, of order $10^{29}$ per cubic metre. Each electron carries only $1.6\times10^{-19}\,\mathrm{C}$ and moves only a millimetre per second, but the product $neAv_d$ contains that huge $n$, and the tiny factors are overwhelmed. Current is a case of a vast crowd shuffling, not a few sprinters. 🔉✕⇢

It should also be said that the electrons do not move in lockstep. Asked whether all the free electrons of a metal move in the same direction when they drift from lower to higher potential, NCERT answers by no means: the drift velocity is superposed over the large random velocities of the electrons. At any instant an individual electron may well be heading backwards. Only the average over the enormous population is biased, and only that bias is the current. Relatedly, the paths between collisions are straight lines when no field is present and curved when a field is applied, since the field supplies a constant transverse acceleration. 🔉✕⇢

Alternating current settles the matter beyond argument. At 50 Hz the field reverses a hundred times a second, so the drift velocity oscillates about zero, and the amplitude of an electron actual excursion is roughly $v_d/\omega=1.1\times10^{-3}/(2\pi\times50)\approx3.5\times10^{-6}\,\mathrm{m}$, a few micrometres. An electron in your mains wiring jiggles back and forth across a distance smaller than a red blood cell and never gets anywhere at all, yet the lamp is fully lit. Whatever is being delivered from the power station, it is plainly not the electrons themselves. 🔉✕⇢

The right mental model, then, is a long pipe already full of water, or a bicycle chain already threaded round both sprockets. Pushing at one end makes the far end move immediately, not because any particular water molecule or chain link has travelled the length of the system, but because the whole medium is already in place and the disturbance propagates through it. The energy in fact travels in the electromagnetic field in the space around the wires, at nearly $c$; the wires guide it. The electrons only shuffle, and what they do is transfer energy locally to the lattice. 🔉✕⇢