Current Electricity

This article follows the chapter in NCERT's own order, each idea resting on the one before it. It begins with the basic definition of electric current and current density, then states Ohm's law and shows how resistance depends on a conductor's length and cross-section through resistivity. It then goes underneath that empirical law to the microscopic drift-velocity picture of conduction, deriving resistivity from electron collisions and defining mobility, and catalogues how real devices such as diodes and GaAs depart from Ohm's law. The chapter then turns to how resistivity changes with temperature, differently for metals and semiconductors, before covering electrical power and the transmission-line argument for high-voltage power lines. The second half is about circuits: emf, internal resistance and terminal voltage; cells in series and parallel; Kirchhoff's junction and loop rules for networks too complex for series-parallel reduction; and the Wheatstone bridge. Three worked examples rebuild NCERT's own flagship numericals — drift speed, terminal voltage, and a Kirchhoff's-rules network — and a cheat-sheet closes the article.

Electric current and current density

Picture a small area held perpendicular to the direction charges are flowing. In a time interval $t$, let $q$ be the net charge (forward minus backward, positive minus negative) that crosses this area. Current is simply the rate of this flow:

where: $I$ is the current (SI unit ampere, A), $\Delta Q$ is the net charge crossing the area in the short interval $\Delta t$, and the limit form handles currents that are not perfectly steady. A domestic appliance typically draws currents of order one ampere; nerve currents run in microamperes; a lightning bolt can carry tens of thousands of amperes. Current is drawn with an arrow but is a scalar, not a vector — currents at a junction add and subtract algebraically, never by a parallelogram rule. For current spread over a finite cross-section, it is often more useful to describe the flow per unit area, the current density $j$. If $n$ is the number density of charge carriers, each of charge $q$, drifting with velocity $v_d$, then $j = nqv_d$, and if $j$ is uniform over a cross-section $A$ normal to the flow, the total current is $I = jA$. Current density is a vector, pointing along the direction of positive-charge flow, with SI unit $\text{A/m}^2$.

Ohm's law and the resistivity of a conductor

Georg Simon Ohm's 1828 discovery, made before anyone understood the physical mechanism behind it, is simply this: for a conductor carrying current $I$ with potential difference $V$ across its ends,

where: the constant of proportionality $R$ is the resistance of the conductor, with SI unit ohm ($\Omega$), where $1\ \Omega = 1\ \text{V A}^{-1}$. Resistance depends on the conductor's material, shape and size, a dependence worked out in the first derivation below by a simple thought experiment: placing two identical slabs end to end doubles the length and doubles the resistance, while splitting a slab lengthwise into two halves of half the cross-sectional area also doubles each half's resistance. Together these give

where: $l$ is the conductor's length along the current, $A$ is its cross-sectional area, and $\rho$ — the resistivity — depends on the material (and its temperature) but not on its dimensions, which is what makes copper a good conductor and rubber a good insulator regardless of shape. Its reciprocal, $\sigma \equiv 1/\rho$, is the conductivity. Combining $R = \rho l/A$ with $V=IR$, and writing $j=I/A$, $E=V/l$, turns Ohm's law into a local, point-by-point statement:

where: the vector form $\vec{j} = \sigma \vec{E}$ is the most compact statement of Ohm's law: at every point inside an ohmic conductor, current density is proportional to the local field. Resistivities span an enormous range: metals sit around $10^{-8}$ to $10^{-6}\ \Omega\text{m}$, insulators can be $10^{18}$ times higher, and semiconductors lie in between, with the added feature that their resistivity falls sharply with small amounts of the right impurity.

The microscopic picture: drift velocity, relaxation time and mobility

Ohm's law as stated above is purely empirical — it says that $V \propto I$, not why. The answer lies in what conduction electrons do inside a metal. With no field applied, each electron moves rapidly due to thermal agitation, colliding again and again with the fixed positive ions of the lattice; after each collision its velocity direction is completely randomised, so averaged over all free electrons the net velocity is zero and there is no current. Switch on a field $\vec{E}$ and every electron (charge $-e$, mass $m$) feels a steady force $-e\vec{E}$ and accelerates between collisions — but each collision again resets its velocity randomly, so the field never builds up an unbounded speed. What survives the averaging, derived in full below, is a small, constant drift velocity

where: $\tau$ is the relaxation time, the average time between collisions (order $10^{-14}\ \text{s}$ in a metal), and the minus sign shows electrons drift opposite to $\vec{E}$. The average velocity is independent of time even though every electron is continuously accelerated — collisions keep wiping the slate clean. The drift speed is tiny, a fraction of a millimetre per second, swamped by random thermal speeds of order $10^5\ \text{m/s}$; only because there are enormously many free electrons (order $10^{28}$–$10^{29}$ per cubic metre) does this glacial drift add up to currents of amperes. The ratio of drift speed to field strength defines the mobility of the charge carriers:

where: mobility has SI unit $\text{m}^2\text{V}^{-1}\text{s}^{-1}$ and is always positive by convention. It is mobility (through $\tau$) that changes with temperature or impurity content, and it is what engineers optimise when designing a conductor or semiconductor. Substituting the drift velocity into $j = nev_d$ reproduces Ohm's law exactly, with conductivity identified microscopically as $\sigma = ne^2\tau/m$ — the bridge between measurable resistivity and the microscopic collision physics of the free-electron gas.

Limitations of Ohm's law

Ohm's law is not a fundamental law of nature — it is an empirical regularity obeyed by a large class of materials over a limited range of conditions, and several important devices violate it. The departures fall into three broad types. First, $V$ may cease to be proportional to $I$ at all, so a $V$–$I$ graph curves away from a straight line. Second, the relation can depend on the sign of $V$: reversing the polarity while keeping magnitude fixed does not reverse the current symmetrically — the signature behaviour of a diode, which conducts easily one way and barely at all the other. Third, for materials such as gallium arsenide (GaAs), the $V$–$I$ relation is not even single-valued: more than one $V$ can correspond to the same $I$. It is worth being precise about what "Ohm's law" asserts, since this is one of the most commonly blurred points in the chapter. The equation $V = IR$ is not itself Ohm's law — it is simply the definition of resistance, writable for any conducting device, ohmic or not, at any instant, by taking $R \equiv V/I$ at that operating point. Ohm's law is the stronger, additional claim that this ratio $R$ stays constant — independent of $V$ — so a plot of $I$ against $V$ is a straight line through the origin. A resistor obeys this; a diode does not, even though $V=IR$ still defines its instantaneous resistance.

Temperature dependence of resistivity

Resistivity is not a fixed number for a material — it changes with temperature, and different classes of material respond in opposite ways. Over a limited temperature range around a reference temperature $T_0$, the resistivity of a metallic conductor is well approximated by a linear relation:

where: $\rho_T$ is the resistivity at $T$, $\rho_0$ is the resistivity at reference temperature $T_0$, and $\alpha$ is the temperature coefficient of resistivity, dimension (temperature)$^{-1}$, positive for metals. The microscopic formula $\rho = m/(ne^2\tau)$ explains why: raising temperature makes ions vibrate more, so electrons collide more often, shortening $\tau$; since free-electron density $n$ in a metal barely changes, the fall in $\tau$ alone drives $\rho$ up. Alloys such as nichrome, manganin and constantan are engineered to have a tiny $\alpha$, so their resistance barely drifts with temperature — exactly why they suit standard resistors and heating elements. Semiconductors and insulators behave oppositely: resistivity falls as temperature rises. Here $\tau$ still decreases, but the free-carrier density $n$ rises so sharply (electrons thermally excited across the energy gap) that it overwhelms the fall in $\tau$, and $\rho$ falls overall. This single competition — "more collisions" ($\tau$ down, $\rho$ up) versus "more carriers" ($n$ up, $\rho$ down) — explains every temperature-resistivity graph in the chapter.

Electrical energy and power

Consider a conductor with a current $I$ flowing from a point A at higher potential to a point B at lower potential, $V = V(A)-V(B) > 0$. In a time $\Delta t$, a charge $\Delta Q = I\Delta t$ moves from A to B and its potential energy falls by $\Delta Q\,V$. Without collisions this lost energy would simply become kinetic energy — the charges would accelerate. But collisions with the lattice constantly transfer that gained energy to the ions, heating the conductor instead. The energy dissipated as heat in time $\Delta t$ is therefore

where: using Ohm's law this can equally be written $P = I^2R = V^2/R$ — the familiar ohmic-loss formula behind every heating element and every calculation of wasted power in a wire. The energy ultimately comes from whatever external source maintains the current, such as the chemical energy of a cell. Equation (8) has a striking consequence for power transmission. To deliver a fixed power $P$ through a cable of resistance $R_c$, the power wasted in the cable is $P_c = I^2R_c$; since $P=VI$, the current needed is $I=P/V$, so

where: for fixed $P$ and $R_c$, the wasted power falls as the square of the transmission voltage $V$ — exactly why power stations transmit at very high voltage over long lines, then step it back down with a transformer before it reaches homes.

EMF, internal resistance and terminal voltage

A steady current needs a continuously maintained field, which requires an external agency that keeps moving charge from lower to higher potential energy inside itself — a cell or battery. Every cell has two electrodes immersed in an electrolyte; with no current flowing, the potential difference between the terminals is the electromotive force, or emf, $\varepsilon$. Despite the name, emf is not a force — the name is historical — it is a potential difference, the work done per unit charge by the source moving charge from one terminal to the other internally. The electrolyte has its own finite internal resistance $r$. When a current $I$ flows (an external resistor $R$ is connected), the terminal voltage falls short of the emf:

where: $Ir$ is the potential drop inside the cell, unavoidable since current also flows through the cell's own resistance. Combining with $V=IR$ for the external resistor gives the current actually delivered:

where: the maximum current occurs at $R=0$ (short circuit), $I_{\max}=\varepsilon/r$, though real cells are never allowed near this limit since the heating $I^2r$ would damage them. Internal resistance can be neglected whenever $\varepsilon \gg Ir$, but never in problems that specifically ask for terminal voltage or near-short-circuit current.

Cells in series and parallel

Cells, like resistors, can be combined, and it is often convenient to replace a combination by a single equivalent cell with emf $\varepsilon_{eq}$ and internal resistance $r_{eq}$. For $n$ cells in series, so the same current flows through each and the terminal potential differences add, the equivalent emf and internal resistance are simply the sums (when every cell's current leaves its positive terminal in the same sense round the loop):

where: if a cell is connected with reversed polarity, its emf enters with a negative sign — for two cells with the second reversed, $\varepsilon_{eq} = \varepsilon_1 - \varepsilon_2$ — while internal resistances still simply add, since resistance has no sign to flip. For cells in parallel, so the same terminal voltage $V$ appears across every cell, the combination rule follows from applying $V = \varepsilon_i - I_ir_i$ to each cell and demanding a common $V$ (derived in the second derivation below):

where: internal resistances combine like parallel resistors, while the equivalent emf is a resistance-weighted average of the individual emfs. Parallel combination is chosen when a circuit needs more current than one cell can safely supply; series combination when it needs more voltage.

Kirchhoff's rules

Many real circuits — several cells and resistors wired in a mesh — cannot be reduced to simple series-parallel combinations at all. Kirchhoff's rules, two conservation laws in disguise, solve any such network without needing any special symmetry. The junction rule: at any junction, the sum of currents entering equals the sum leaving — conservation of charge, since a steady current means no charge piles up anywhere. The loop rule: the algebraic sum of potential changes around any closed loop is zero, because electric potential is a well-defined function of position — returning to the same point must give zero net change, just as for gravitational potential energy on a path that returns to its starting height. Applying it means walking a chosen loop, adding $-IR$ for every resistor traversed along the assumed current direction ($+IR$ against it), and $+\varepsilon$ or $-\varepsilon$ for every cell depending on whether the walk crosses it from negative to positive terminal or the reverse. Using both rules, a network is solved by first applying the junction rule to minimise unknowns, then writing one loop equation per remaining unknown, choosing loops that each add at least one new, unconstrained element.

The Wheatstone bridge

The Wheatstone bridge is the chapter's signature application of Kirchhoff's rules. Four resistors $R_1$, $R_2$, $R_3$, $R_4$ are arranged in a diamond; a cell is connected across one pair of opposite corners (the battery arm), and a galvanometer — a sensitive current-detecting instrument — across the other pair (the galvanometer arm). In general current $I_g$ flows through the galvanometer, but for a special, tuned set of resistances the bridge is balanced: $I_g = 0$. Applying Kirchhoff's rules to this condition (derived in full below) gives the elegant result

where: $R_1$, $R_2$ sit in one pair of adjacent arms and $R_3$, $R_4$ in the other. This is the working principle of the meter bridge, a lab device for measuring an unknown resistance: with three resistances known and the fourth unknown, the bridge is balanced by adjusting one known resistor until the galvanometer shows zero deflection, and the unknown is read off from equation (14). Because balance depends only on a ratio, not on the driving cell's emf or internal resistance, this method beats a direct ammeter-and-voltmeter measurement.

Board level - drift speed of conduction electrons in a copper wire

Estimate the average drift speed of conduction electrons in a copper wire of cross-sectional area $1.0\times10^{-7}\ \text{m}^2$ carrying a current of $1.5\ \text{A}$. Assume one conduction electron per copper atom. Density of copper is $9.0\times10^3\ \text{kg/m}^3$, atomic mass of copper is $63.5\ \text{u}$, and $e = 1.6\times10^{-19}\ \text{C}$. Compare the drift speed with (a) the thermal speed of copper atoms at room temperature ($\approx 2\times10^2\ \text{m/s}$), and (b) the speed of light, which is roughly the speed at which the field itself is established along the wire.

Step 1. Find the free-electron number density $n$. One conduction electron per atom means $n$ equals the number of copper atoms per cubic metre. A cubic metre of copper has mass $9.0\times10^3\ \text{kg} = 9.0\times10^6\ \text{g}$, and since $6.0\times10^{23}$ copper atoms (Avogadro's number) have a mass of $63.5\ \text{g}$, $n = \dfrac{6.0\times10^{23}}{63.5}\times 9.0\times10^{6} = 8.5\times10^{28}\ \text{m}^{-3}$.

Step 2. Apply $v_d = I/(neA)$, rearranged from $I = neAv_d$: $v_d = \dfrac{1.5}{(8.5\times10^{28})(1.6\times10^{-19})(1.0\times10^{-7})} = \dfrac{1.5}{1.36\times10^{3}} = 1.1\times10^{-3}\ \text{m/s} = 1.1\ \text{mm/s}$.

Step 3. Compare with thermal speed: $v_d/v_{\text{thermal}} \approx (1.1\times10^{-3})/(2\times10^{2}) \approx 5\times10^{-6}$ — the drift is a tiny, biased whisper riding on enormous random agitation.

Step 4. Compare with the field-propagation speed, essentially $c = 3.0\times10^{8}\ \text{m/s}$: $v_d/c \approx 3.7\times10^{-12}$. This is why a bulb lights up almost instantly on closing a switch even though electrons crawl at a millimetre per second: the field is established at near light speed, so drift begins everywhere in the wire almost simultaneously. Answer: $v_d \approx 1.1\ \text{mm/s}$, about $5\times10^{-6}$ times the thermal speed of copper atoms and about $4\times10^{-12}$ times the speed of light.

Board level - terminal voltage and power balance for a cell with internal resistance

A battery of emf $\varepsilon = 10\ \text{V}$ and internal resistance $r = 3\ \Omega$ is connected to an external resistor, and the current in the circuit is found to be $0.5\ \text{A}$. Find the resistance of the external resistor, the terminal voltage of the battery, and verify that the power supplied by the battery equals the power delivered to the resistor plus the power lost internally.

Step 1. Find the external resistance from $I = \varepsilon/(R+r)$, rearranged: $R + r = \varepsilon/I = 10/0.5 = 20\ \Omega$, so $R = 20 - 3 = 17\ \Omega$.

Step 2. Find the terminal voltage using $V = \varepsilon - Ir$: $V = 10 - (0.5)(3) = 8.5\ \text{V}$. Check: this also equals $IR = (0.5)(17) = 8.5\ \text{V}$, as it must, since $V$ is simultaneously the potential difference across the cell's terminals and across $R$.

Step 3. Find the power delivered externally and lost internally: $P_{\text{external}} = I^2R = (0.5)^2(17) = 4.25\ \text{W}$, $P_{\text{internal}} = I^2r = (0.5)^2(3) = 0.75\ \text{W}$.

Step 4. Find total power from the battery's chemistry, $P_{\text{total}} = \varepsilon I = (10)(0.5) = 5\ \text{W}$, and check against Step 3: $4.25+0.75=5\ \text{W}$ — the balance closes exactly. Answer: $R = 17\ \Omega$, terminal voltage $= 8.5\ \text{V}$, with $4.25\ \text{W}$ delivered externally and $0.75\ \text{W}$ dissipated internally, summing to the full $5\ \text{W}$ supplied.

Competitive level (JEE / NEET) - Kirchhoff's rules for a two-cell resistor network

A network contains two cells, of emf $10\ \text{V}$ and $5\ \text{V}$, and five resistors arranged so that three independent loop currents $I_1$, $I_2$, $I_3$ (in amperes) satisfy the simultaneous equations obtained by applying Kirchhoff's loop rule to three closed loops of the network: $7I_1 - 6I_2 - 2I_3 = 10$, $I_1 + 6I_2 + 2I_3 = 10$, and $2I_1 - 4I_2 - 4I_3 = -5$. Find $I_1$, $I_2$, $I_3$, and hence the current in every branch of the network, where the branch currents are the combinations $I_1$, $I_2$, $(I_1 - I_2)$, $(I_2+I_3)$ and $(I_2+I_3-I_1)$.

Step 1. Eliminate $I_2$ and $I_3$ at once by adding the first two equations directly, since their coefficients are exact negatives: $(7I_1-6I_2-2I_3)+(I_1+6I_2+2I_3)=10+10$, giving $8I_1=20$, so $I_1 = 2.5\ \text{A}$.

Step 2. Substitute $I_1=2.5$ into the second equation: $2.5+6I_2+2I_3=10 \Rightarrow 3I_2+I_3=3.75$. Substitute into the third: $5-4I_2-4I_3=-5 \Rightarrow I_2+I_3=2.5$.

Step 3. Solve the pair from Step 2: $I_3 = 2.5-I_2$, so $3I_2+(2.5-I_2)=3.75 \Rightarrow 2I_2=1.25$, giving $I_2 = 0.625\ \text{A}$ and $I_3 = 1.875\ \text{A}$.

Step 4. Compute the five branch currents: $I_1 = 2.5\ \text{A}$; $I_2 = 0.625\ \text{A}$; $(I_1-I_2) = 1.875\ \text{A}$; $(I_2+I_3) = 2.5\ \text{A}$; and, most strikingly, $(I_2+I_3-I_1) = 0\ \text{A}$ exactly. Answer: $I_1=2.5\ \text{A}$, $I_2=0.625\ \text{A}$, $I_3=1.875\ \text{A}$, and the branch currents are $2.5\ \text{A}$, $0.625\ \text{A}$, $1.875\ \text{A}$, $2.5\ \text{A}$ and $0\ \text{A}$ — one branch carries no current at all despite sitting between two cells. Timing note: about three to four minutes. Look first for a pair of loop equations whose coefficients on two unknowns are already equal and opposite, as in Step 1 — that single addition collapses a three-variable system to two variables immediately. As a sanity check, apply the loop rule to any closed loop not used to generate the original three equations; NCERT notes this always comes out automatically satisfied, a fast way to catch an arithmetic slip.

• Treating current as a vector. Current is a scalar; it is drawn with an arrow only to fix a reference direction, and currents at a junction add and subtract algebraically, never by a parallelogram rule.
• Confusing emf with terminal voltage. The emf $\varepsilon$ is the potential difference in the open-circuit limit; the terminal voltage $V=\varepsilon-Ir$ is always smaller whenever current flows, and only the terminal voltage is what a loaded voltmeter reads.
• Sign errors in the loop rule. Going against the assumed current direction through a resistor flips the sign of its $IR$ term, and crossing a cell negative-to-positive versus positive-to-negative flips the sign of its emf term.
• Assuming Ohm's law universally applies. $V=IR$ always defines resistance at an operating point; Ohm's law is the extra claim that this ratio is constant, which diodes and GaAs do not satisfy.
• Mixing up the resistivity temperature-coefficient sign. $\alpha>0$ for metals but semiconductors and insulators trend the opposite way because carrier density dominates; quoting one sign for "everything" is a common slip.
• Forgetting internal resistance changes with load. A cell's rated voltage is its emf, not a fixed output; delivered voltage and current depend on the external resistance through $I=\varepsilon/(R+r)$.
• Mis-stating the Wheatstone bridge ratio. Balance is a ratio of adjacent arms, $R_1/R_2=R_3/R_4$; scrambling adjacent versus opposite arms is the most common meter-bridge error.
• Losing track of mobility units. Mobility appears in both $\text{m}^2\text{V}^{-1}\text{s}^{-1}$ (SI) and $\text{cm}^2\text{V}^{-1}\text{s}^{-1}$ (practical), a factor of $10^4$ apart; mixing the two mid-calculation silently introduces a large error.

• Define electric current and current density, and explain precisely why current is a scalar despite being drawn with a directional arrow.
• State Ohm's law, distinguish it clearly from the mere definition $R \equiv V/I$, and compute resistance from resistivity, length and cross-sectional area.
• Derive the drift-velocity picture of conduction from Newton's second law and the randomising effect of collisions, and use it to derive $\sigma = ne^2\tau/m$.
• Define mobility and relaxation time, and explain how each connects the microscopic collision picture to the macroscopic resistivity.
• Identify and describe the three characteristic ways a real device (diode, GaAs) can fail to obey Ohm's law.
• Use $\rho_T = \rho_0[1+\alpha(T-T_0)]$ to relate resistance change to temperature change, for both metals and semiconductors, with the correct sign of $\alpha$ in each case.
• Compute electrical power dissipated in a resistor using $P=IV=I^2R=V^2/R$, and explain why high-voltage transmission minimises line losses.
• Distinguish emf from terminal voltage, and compute current, terminal voltage and power balance for a cell with internal resistance connected to an external resistor.
• Combine cells in series and in parallel to find equivalent emf and internal resistance, including cases with reversed polarity.
• Apply Kirchhoff's junction and loop rules to solve a multi-loop resistor network, and derive and apply the Wheatstone bridge balance condition.

Ask the tutor: Want to work through a Kirchhoff's-rules network or a Wheatstone-bridge numerical from your own textbook together, step by step?