Current Electricity

Chapters 1 and 2 dealt with charges at rest. This chapter is about charges in motion, that is, electric current, and it stays within steady currents: currents that do not change with time. Almost everything you use, from a torch to a lift motor, runs on current, so the chapter sits close to daily life, and the board treats it as a chapter of definitions, one central derivation and circuit problems.

The story runs in a fixed order. First we define current and see what carries it in a metal. Then Ohm's law is stated as an experimental fact and explained by the drift of free electrons, which is the derivation examiners ask for most. From that model we get resistivity, its dependence on temperature, and the materials used for heaters and standard resistances. Then energy and power. Then the source that keeps a current going, the cell, with its emf and internal resistance, and combinations of cells. Finally two general tools, Kirchhoff's rules and the Wheatstone bridge, that let you solve circuits that are not simple series or parallel networks.

Two things were removed in the rationalised syllabus and are not examined now: the meter bridge and the potentiometer, and the colour code of carbon resistors. Series and parallel combinations of resistors are assumed from Class 10 and still get used inside problems.

Electric current through an area is the rate of flow of charge across that area. If a net charge q flows across the area in time t, the average current is I = q/t; if the flow varies, the instantaneous current is I = dq/dt, the limit of ΔQ/Δt as Δt → 0. The SI unit is the ampere (A): 1 A = 1 C s⁻¹. Currents in a household wire are of the order of an ampere, in a nerve a microampere, and in lightning tens of thousands of amperes.

By convention the direction of current is the direction in which positive charge would move, so in a metal wire the current points opposite to the motion of the electrons that actually carry it. Note two facts the examiner likes. Current is a scalar, even though it has a direction in a wire: currents add by ordinary algebra at a junction, not by the parallelogram law. And the same current flows at every cross-section of a series circuit, because charge does not pile up anywhere in a steady state.

Along with current, this chapter uses current density j, the current per unit area through a surface held normal to the flow: j = I/A, unit A m⁻². Unlike current, j is a vector, pointing along the direction of flow of positive charge. If the normal to the area makes an angle θ with the direction of flow, I = jA cos θ. Current density is what links the microscopic picture (electrons drifting) to the macroscopic one (current in a wire), and it appears in the vector form of Ohm's law in section 3.4.

A charge placed in an electric field moves, so a current needs both mobile charges and a field. In a solid metal the positive ions are locked into a lattice and only the outermost electrons, the free or conduction electrons, can move. In an electrolyte both positive and negative ions move. In an insulator there are almost no free carriers, so even a strong field gives a negligible current.

With no field applied, the free electrons in a metal move randomly with thermal speeds of about 10⁵ m s⁻¹, colliding with the ions. The velocities point in all directions, so at any instant as many electrons cross a plane one way as the other and the net current is zero. Now imagine a cylinder of the metal with a thin layer of positive charge on one end face and negative charge on the other: there is a field inside, electrons are pushed towards the positive face, and a current flows. But as they arrive they neutralise the charge, the field dies away and the current stops in a very short time. To keep a steady current we need something that keeps replenishing the charges on the two ends, and that is what a cell or battery does: it maintains a steady potential difference between the ends of the conductor.

So the picture for the rest of the chapter is: a source maintains a constant field inside the conductor; the free electrons acquire a small steady average velocity opposite to the field on top of their random motion; and that average velocity, the drift velocity, is what constitutes the current. In a steady current the electron gas as a whole does not accelerate indefinitely because collisions with the ions keep resetting its velocity, which is exactly the origin of resistance treated in section 3.5.

Ohm's law (1827) is an experimental statement: at constant temperature and other physical conditions, the current through a conductor is directly proportional to the potential difference across its ends, V ∝ I, or V = IR. The constant R is the resistance of the conductor, unit ohm (Ω): 1 Ω = 1 V A⁻¹. Resistance depends on the material, the dimensions and the temperature of the conductor, not on V or I.

How R depends on dimensions can be argued directly. Put two identical slabs of length l in series: the same current flows through a slab of length 2l with twice the potential difference, so R ∝ l. Put them side by side: the same V drives twice the current through an area 2A, so R ∝ 1/A. Hence R = ρl/A, where ρ is the resistivity, a property of the material alone, unit Ω m. Its reciprocal σ = 1/ρ is the conductivity, unit S m⁻¹ (Ω⁻¹ m⁻¹). The reciprocal of resistance, G = 1/R, is the conductance, unit siemens (S).

The board also expects the vector form of Ohm's law, which links the field inside the conductor to the current density. Take a conductor of length l and area A carrying current I with a uniform field E along it. Then V = El and I = jA. Substituting in V = IR = I ρl/A gives El = jρl, so E = ρj, or equivalently j = σE. This form has no reference to the shape of the conductor and is the statement we will reproduce from the electron model in the next section.

A conductor obeying Ohm's law is called ohmic, and its V–I graph is a straight line through the origin whose slope (V against I) is R. Ohm's law is not a fundamental law of nature; it is a property of many materials over a range of conditions, and section 3.6 lists where it fails.

This is the derivation the board asks for most from this chapter. Write it in these steps.

Step 1, drift velocity. Consider a metal with n free electrons per unit volume, each of mass m and charge −e, in a field E. Each electron feels a force −eE and an acceleration a = −eE/m. Between two collisions with the ions an electron accelerates; at each collision its velocity is reset to a random thermal value. If the average time between successive collisions is τ, the relaxation time, then at any instant the velocities of the electrons since their last collision have been increased on average by aτ. The random thermal velocities average to zero, so the average velocity of the electrons, the drift velocity, is v_d = −eEτ/m. It is opposite to E and, unlike a free acceleration, it is constant.

Step 2, current in terms of drift velocity. In a wire of cross-section A, the electrons that cross a given section in time Δt are those within a length v_dΔt of it: number = n A v_d Δt, charge = neAv_dΔt. Hence I = neAv_d, and the current density j = I/A = nev_d.

Step 3, Ohm's law and resistivity. Substituting v_d = eEτ/m in magnitude: j = (ne²τ/m)E. Comparing with j = σE gives σ = ne²τ/m, so ρ = m/(ne²τ). Since n and τ do not depend on E, ρ is constant and V = IR follows, which is Ohm's law.

Put in numbers for copper (Example 3.1): a 1.5 A current in a wire of area 10⁻⁷ m² gives v_d ≈ 1 mm s⁻¹, far smaller than the thermal speed (~10⁵ m s⁻¹) and utterly unlike the speed at which the field itself is set up along the wire (close to the speed of light). That is why a bulb lights the instant a switch is closed: electrons everywhere in the circuit start drifting at once.

3.5.1 Mobility. The mobility μ of a charge carrier is the magnitude of its drift velocity per unit electric field, μ = |v_d|/E = eτ/m. Its SI unit is m² V⁻¹ s⁻¹. In terms of mobility, σ = neμ; for a material with more than one kind of carrier (a semiconductor, an electrolyte) the conductivities of the carriers add.

Ohm's law holds for metals and many alloys over a wide range of conditions, but it is not universal. NCERT lists three ways in which a material or device can deviate, and each has a characteristic V–I graph that you should be able to sketch.

(a) V stops being proportional to I. For a metallic wire that heats up as the current increases, the resistance rises, and the V–I line bends upwards at large currents (more V for the same increase in I). The graph is a straight line at small currents and curves away from it later.

(b) The relation depends on the sign of V. For a p–n junction diode, reversing the polarity of V does not reverse the current with the same magnitude: a small forward voltage gives a large current, while the same reverse voltage gives almost none. The V–I graph is not symmetric about the origin.

(c) The relation is not unique. For some materials, such as gallium arsenide (GaAs), there is a region where increasing V actually decreases I (negative resistance), so more than one value of V gives the same current. The graph rises, reaches a peak and then falls before rising again.

Devices that do not obey Ohm's law are called non-ohmic, and they are the basis of electronics (Chapter 14). Remember that even for a non-ohmic device the ratio V/I at any point can still be called its resistance at that operating point; what fails is the constancy of that ratio. When the examiner asks 'state the conditions under which Ohm's law is valid', the expected words are: constant temperature and unchanged physical conditions (mechanical strain, pressure) of the conductor.

Resistivity spans about 24 orders of magnitude, and materials are grouped by it. Conductors (metals) have ρ between about 10⁻⁸ and 10⁻⁶ Ω m: silver is the best at 1.6 × 10⁻⁸ Ω m, then copper 1.7 × 10⁻⁸, aluminium 2.7 × 10⁻⁸. Insulators have ρ of the order of 10¹⁰ to 10¹⁶ Ω m (glass, hard rubber, fused quartz at the top of that range). Semiconductors such as germanium and silicon lie in between, and their resistivity is very sensitive to temperature and to tiny amounts of impurities, which is what makes them useful in electronics.

Two families of alloys matter for exam questions. Nichrome (ρ ≈ 1.0 × 10⁻⁶ Ω m, about 60 times copper) has high resistivity, a high melting point and does not oxidise easily at red heat, so it is used for the heating elements of heaters, toasters and irons. Manganin and constantan have high resistivity and, more importantly, an almost zero temperature coefficient of resistivity, so their resistance stays fixed as they warm up in use; they are used for standard resistances, resistance boxes and wire-wound resistors.

Commercial resistors are of two kinds: wire-wound resistors made from alloys like manganin for values up to about 100 kΩ, and carbon resistors, which are cheap, compact and used in electronic circuits for higher values. The colour-code convention for reading a carbon resistor's value was part of the older syllabus and is no longer examined.

A short-answer question here usually asks why a particular alloy is chosen: link the property to the use. Heating element needs high ρ (large heat per unit length for a manageable length) and stability at high temperature. Standard resistance needs a value that does not drift, so low α, and a resistivity high enough that the coil is not impractically long.

Over a limited range of temperatures not too far from room temperature, the resistivity of a metal rises linearly with temperature: ρ_T = ρ₀[1 + α(T − T₀)], where ρ₀ is the resistivity at a reference temperature T₀ and α is the temperature coefficient of resistivity, unit °C⁻¹ or K⁻¹. Since R ∝ ρ for a given conductor, the same relation holds for resistance: R_T = R₀[1 + α(T − T₀)]. This is the formula behind every 'find the steady temperature of the heating element' numerical: find R at the two temperatures from V/I, then solve for T.

The model of section 3.5 explains the trend. ρ = m/(ne²τ), so ρ depends on the carrier density n and the relaxation time τ. In a metal, n is fixed by the material and does not change with temperature, but as T rises the ions vibrate more, collisions become more frequent, τ falls and ρ rises: α is positive (copper α ≈ 3.9 × 10⁻³ °C⁻¹). The ρ–T graph for copper is nearly a straight line rising with T, curving slightly at low temperatures. For alloys such as nichrome and manganin, ρ is high and nearly independent of temperature, so the graph is almost flat (manganin α ≈ 0.002 × 10⁻³ °C⁻¹). For semiconductors and insulators, heating frees many more carriers, so n increases with T much faster than τ falls, and ρ decreases with temperature: α is negative. The ρ–T graph of a semiconductor falls steeply as T rises.

A worked pattern (Example 3.4): a platinum resistance thermometer has R = 5 Ω at the ice point, 5.23 Ω at the steam point and 5.795 Ω in a bath, so the bath temperature is t = (R_t − R₀)/(R₁₀₀ − R₀) × 100 = (0.795/0.23) × 100 = 345.65 °C. The board also asks the reverse: given α, R₀ at 27 °C and the steady current from a 230 V supply, find the working temperature. Set up R_T = V/I, then T − T₀ = (R_T − R₀)/(R₀α).

Consider a conductor with a potential difference V between its ends A and B (V = V_A − V_B > 0) carrying current I from A to B. In time Δt a charge ΔQ = IΔt moves from A to B, and its potential energy falls by ΔU_pot = ΔQ(V_B − V_A) = −IVΔt. By energy conservation the charges should gain kinetic energy equal to this, but in a conductor they collide with the ions and hand the energy over to the lattice as heat. So the energy dissipated in the conductor in time Δt is ΔW = IVΔt, and the power dissipated is P = IV. With Ohm's law this becomes P = I²R = V²/R. This is the Joule heating that runs heaters, bulbs and fuses. The source (a cell, a generator) supplies this power by doing work on the charges.

Which form to use depends on what is held fixed. In a series connection the current is common, so P = I²R and the larger resistance dissipates more. In a parallel connection the voltage is common, so P = V²/R and the smaller resistance dissipates more. That is why a 60 W bulb glows brighter than a 100 W bulb when the two (both rated 220 V) are put in series: the 60 W bulb has the larger resistance (R = V²/P = 807 Ω against 484 Ω) and the same current passes through both.

The other standard question is power transmission. A power station delivers power P at voltage V to a load through cables of total resistance R_c. The current in the cables is I = P/V, so the power wasted as heat in the cables is P_c = I²R_c = P²R_c/V². For a fixed power to be delivered, the loss falls as 1/V², so transmission is done at very high voltages (hundreds of kilovolts) and stepped down by transformers near the consumer (Chapter 7). Thick low-resistance cables help too, but the voltage is the decisive factor.

A cell has two electrodes, positive (P) and negative (N), dipped in an electrolyte. Chemical action at the electrodes maintains a potential difference between them even when nothing is connected. The emf ε of the cell is the potential difference between its terminals when no current is drawn from it (open circuit); equivalently, it is the work done by the cell per unit charge in carrying positive charge from the negative to the positive terminal inside the cell. Despite the name, emf is a potential difference, not a force, and its unit is the volt.

The electrolyte itself opposes the flow of charge; this is the internal resistance r of the cell. It depends on the nature and concentration of the electrolyte, the area of the electrodes, their separation and the temperature; it is small for a car battery and large for a dry cell.

Connect an external resistance R across the cell. The current is I = ε/(R + r), and the potential difference across the terminals, the terminal voltage, is V = IR = ε − Ir. So V is less than ε whenever the cell delivers current; the two are equal only when I = 0 (open circuit, or r = 0). The maximum current the cell can give is ε/r, when R = 0 (short circuit). If the cell is being charged, current is driven into it through the positive terminal and the terminal voltage exceeds the emf: V = ε + Ir.

Two graphs are examined. Terminal voltage against current I is a straight line with intercept ε on the V-axis and slope −r, so a V–I plot of a cell gives both ε and r. Terminal voltage against external resistance R rises from 0 at R = 0 and approaches ε as R becomes large. From two readings of I with known R₁ and R₂ you solve ε = I₁(R₁ + r) = I₂(R₂ + r) for r and ε. Another useful rearrangement: r = (ε/V − 1)R.

Series. Let cell 1 (ε₁, r₁) and cell 2 (ε₂, r₂) be joined with the negative terminal of the first to the positive of the second, carrying the same current I. Across the first, V_AB = ε₁ − Ir₁; across the second, V_BC = ε₂ − Ir₂. Adding, V_AC = (ε₁ + ε₂) − I(r₁ + r₂). Comparing with V = ε_eq − Ir_eq: ε_eq = ε₁ + ε₂ and r_eq = r₁ + r₂. If the second cell is connected the other way round (negative to negative), ε_eq = ε₁ − ε₂ with the same r_eq. For n identical cells in series, ε_eq = nε, r_eq = nr.

Parallel (the derivation the board asks for). Now connect the two positive terminals at B₁ and the two negatives at B₂, and let the cells supply currents I₁ and I₂ to the external circuit, so I = I₁ + I₂. Both cells have the same terminal voltage V across them: V = ε₁ − I₁r₁ and V = ε₂ − I₂r₂. Hence I₁ = (ε₁ − V)/r₁, I₂ = (ε₂ − V)/r₂, and

I = (ε₁/r₁ + ε₂/r₂) − V(1/r₁ + 1/r₂).

Solving for V: V = (ε₁r₂ + ε₂r₁)/(r₁ + r₂) − I r₁r₂/(r₁ + r₂). Comparing with V = ε_eq − Ir_eq:

ε_eq = (ε₁r₂ + ε₂r₁)/(r₁ + r₂), r_eq = r₁r₂/(r₁ + r₂), i.e. 1/r_eq = 1/r₁ + 1/r₂.

A neat form to remember: ε_eq/r_eq = ε₁/r₁ + ε₂/r₂. If one cell is reversed, replace ε₂ by −ε₂. For n identical cells in parallel, ε_eq = ε and r_eq = r/n: parallel cells do not increase the voltage, they lower the internal resistance and can supply a larger current. A 1.5 V cell (r = 0.2 Ω) and a 2.0 V cell (r = 0.3 Ω) in parallel give ε_eq = (1.5 × 0.3 + 2.0 × 0.2)/0.5 = 0.85/0.5 = 1.7 V and r_eq = 0.06/0.5 = 0.12 Ω.

Series and parallel formulas cannot handle every circuit, for instance one with two cells in different branches. Kirchhoff's two rules (1845) can. First label every branch with a current and an assumed direction; if the answer comes out negative, the current actually flows the other way.

Junction rule (first rule). At any junction the sum of the currents entering equals the sum of the currents leaving. It follows from conservation of charge: in a steady current, charge cannot accumulate at a junction, so what flows in must flow out.

Loop rule (second rule). The algebraic sum of the changes in potential around any closed loop of the circuit is zero. It follows from the fact that electric potential is single-valued: going round a loop and returning to the starting point must bring you back to the same potential. It is equivalent to conservation of energy in the circuit.

Sign conventions (state them in the answer). Going round the loop in the chosen direction: crossing a resistor in the direction of the current is a drop, −IR; against the current it is a rise, +IR. Crossing a cell from its negative to its positive terminal is a rise, +ε; from positive to negative, −ε. The internal resistance of a cell is treated as an ordinary resistor in series with it.

Method for a numerical. (1) Mark currents so that the junction rule is used as early as possible (for two branches joining a third, call the third I₁ + I₂). (2) Write the loop rule for as many independent loops as there are unknowns. (3) Solve. For example, take a 12 V cell (2 Ω) and a 6 V cell (1 Ω) in two parallel branches feeding a 4 Ω resistor. With I₁, I₂ from the two cells and I₁ + I₂ through 4 Ω: 12 − 2I₁ − 4(I₁ + I₂) = 0 and 6 − I₂ − 4(I₁ + I₂) = 0, giving I₁ = 18/7 A, I₂ = −6/7 A: the 6 V cell is being charged by the 12 V cell. NCERT Example 3.5 uses the same rules with symmetry to find the resistance of a cube of twelve 1 Ω edges across a body diagonal: 5/6 Ω.

The Wheatstone bridge is the standard application of Kirchhoff's rules and the usual 3-mark derivation. Four resistors R₁, R₂, R₃, R₄ form a quadrilateral ABCD, with R₁ in arm AB, R₂ in arm BC, R₃ in arm AD and R₄ in arm DC. A cell is connected between A and C, and a galvanometer of resistance G between B and D. The bridge is balanced when the galvanometer shows no deflection, I_g = 0, which happens when B and D are at the same potential.

Derivation. Let the cell send current I into A, splitting into I₁ through R₁ and I₂ through R₃. At balance I_g = 0, so by the junction rule at B the current I₁ continues through R₂, and at D the current I₂ continues through R₄.

Loop rule for loop ADBA: −I₂R₃ + 0·G + I₁R₁ = 0, so I₁R₁ = I₂R₃.

Loop rule for loop CBDC: I₂R₄ + 0·G − I₁R₂ = 0, so I₁R₂ = I₂R₄.

Dividing the second by the first: R₂/R₁ = R₄/R₃. This is the balance condition. Written as P/Q = R/S it says the ratio of the resistances in the two arms meeting at B equals the ratio in the two arms meeting at D; the galvanometer resistance and the cell's emf and internal resistance drop out of it, which is the whole point.

Use. Take R₄ as the unknown, R₃ a known standard, and adjust R₂ (with R₁ fixed) until the galvanometer reads zero; then R₄ = R₂R₃/R₁. The measurement is a null method, so it does not depend on the calibration of the galvanometer, and it is most sensitive when the four resistances are of the same order. Two facts often asked: interchanging the cell and the galvanometer does not disturb the balance (the condition is symmetric), and at balance the galvanometer and cell branch can be removed or short-circuited without changing the currents in the arms. When the bridge is not balanced, you must go back to Kirchhoff's rules for all three loops, as in NCERT Example 3.7.