Electric current and currents in conductors
In the electrostatics chapter, every charge was assumed at rest. The moment charges move, an electric current appears. Lightning is the violent extreme; a torch bulb is the gentle one. Current is the net rate at which charge crosses any chosen cross-section of a conductor, oriented normal to the flow:
Why do metals carry a current while a glass rod refuses? In bulk metals, the outermost electrons of every atom are practically free — about 10²⁹ electrons per cubic metre form an "electron gas" that drifts through the lattice of fixed positive ions. In insulators, the electrons are bound tightly to their nuclei and refuse to move under any ordinary field. Semiconductors sit between the two and, with a small impurity dose, can be tuned. NCERT confines its detailed treatment to solid conductors, where current is carried by negatively charged electrons against a backdrop of fixed positive ions.
With no field applied, electrons inside a conductor are far from still — they zip about in random thermal motion at speeds of ~10⁵ m/s, but the average velocity over all electrons is zero. No preferred direction, no current. Switch on an electric field, and a tiny anisotropy appears: every electron experiences a force −eE against the field direction. The field is established at the speed of light along the entire wire, so every electron starts drifting almost instantaneously — which is why a bulb lights up the moment the switch is flipped, even though the drift itself is glacially slow.
Ohm's law
In 1828 — decades before electrons were even discovered — Georg Simon Ohm announced that for a metallic conductor at fixed temperature, the current is directly proportional to the voltage across its ends:
V = I R
Ohm's law — V proportional to I, with R the constant of proportionality
The constant of proportionality R is the resistance, measured in ohms (Ω): 1 Ω = 1 V·A⁻¹. Resistance is not a property of the material alone — it also depends on the geometry of the conductor. NCERT derives this by a beautiful slab argument: doubling the length doubles the resistance (more lattice to collide with), and halving the cross-sectional area also doubles it (electrons squeezed through a narrower channel). Combining the two,
Cast in terms of fields and current densities (where j = I/A is the current density and E = V/l the field), Ohm's law also reads j = σE — a local, point-by-point statement that proves more useful in advanced settings.
Four ways to change a resistor
R = ρl/A packages every NEET trap on resistance into four levers. Three are geometric (length, area, melting/stretching), one is the material itself. Temperature acts through ρ, treated separately below.
Length
R ∝ l
linear
Double the length, double the resistance — same material, same area.
PYQ pattern: stretched wireCross-section
R ∝ 1/A
inverse
Halve the area, double the resistance. A thicker wire conducts more easily.
Material (ρ)
10⁻⁸ → 10¹⁶ Ω·m
spans 24 orders of magnitude
Copper, silver, aluminium have low ρ. Glass, rubber, ceramic at the high end.
Temperature
RT = R0(1+αΔT)
α positive for metals
Metals: R rises with T. Semiconductors: R falls with T. Treated in detail in §6.
NEET trap: sign of αDrift velocity and the origin of resistivity
Why does a current obey Ohm's law? The microscopic answer is the free-electron model. Inside a metal, electrons move with two distinct motions superimposed: a large random thermal motion (~10⁵ m/s, no net direction) and a tiny ordered drift along the field. Between any two successive collisions with the lattice ions, an electron accelerates under the field by a = −eE/m. Let τ (the relaxation time) be the average time between collisions. Averaging over all electrons in the conductor at any instant, the random part cancels, and the drift speed is
Each electron carries charge −e, and there are n electrons per unit volume. In a time Δt, the number of electrons that cross an area A is n·vd·A·Δt, contributing a charge |q| = neAvdΔt. The current is therefore I = neAvd, and the current density
j = nevd = (ne²τ/m) E
Microscopic Ohm's law — j proportional to E, with conductivity σ = ne²τ/m
Compare with the macroscopic j = σE and the two pictures align exactly. The conductivity emerges as σ = ne²τ/m, and therefore the resistivity has a clean microscopic form:
A related quantity, the mobility μ, is the drift speed per unit field: μ = vd/E = eτ/m. Its SI unit is m²·V⁻¹·s⁻¹. Mobility lets you write the conductivity as σ = neμ — a form that becomes essential for semiconductors, where both electrons and holes contribute.
Limitations of Ohm's law
Ohm's law is empirical, not fundamental. It holds for metallic conductors at constant temperature over a wide range of voltages, but it fails — and fails interestingly — in three distinct ways. NEET often tests these as identify-the-graph questions.
- Non-linear V–I: V is not strictly proportional to I, even though both rise together. The V–I curve bends. Resistance becomes voltage-dependent.
- Direction-dependent V–I: Reversing the sign of V does not simply reverse the sign of I. This happens in a diode — it conducts freely in forward bias but blocks current in reverse bias.
- Non-unique V–I: The same current can correspond to more than one voltage. A famous example is gallium arsenide (GaAs), whose V–I curve doubles back on itself — used in tunnel diodes and microwave devices.
Crucially, even non-ohmic devices obey conservation of charge and energy — they merely refuse the linear proportionality. The wider laws (Kirchhoff's rules, energy conservation) still apply.
Resistivity of various materials
Materials are sorted by their resistivity into three families. The range is staggering: from 10⁻⁸ Ω·m in silver to 10¹⁶ Ω·m or higher in insulators — a span of 24 orders of magnitude, wider than almost any other physical property in the syllabus.
Conductors (metals)
10⁻⁸ to 10⁻⁶ Ω·m
very low ρ
Silver, copper, aluminium, gold. Large n (~10²⁹/m³). ρ rises with temperature.
Semiconductors
~10⁻⁵ to 10³ Ω·m
intermediate ρ
Silicon, germanium. Carrier density n thermally generated across a small band gap. ρ falls with temperature. Doping multiplies n by orders of magnitude.
Insulators
10¹⁰ to 10¹⁶+ Ω·m
enormous ρ
Glass, rubber, dry wood, ceramics. Few free carriers at room temperature. ρ also falls with temperature, but starts so high they remain insulating in practice.
A practical wrinkle: certain alloys — nichrome (Ni-Fe-Cr), manganin, constantan — are engineered to have an almost negligible temperature coefficient. Their resistance barely shifts when the wire heats up, which makes them the materials of choice for standard resistors and the heating elements of toasters and electric heaters, where high resistance must hold steady even at red heat.
Temperature dependence of resistivity
Over a limited temperature range, the resistivity of a metal varies almost linearly with temperature:
The microscopic reason is encoded in ρ = m/(ne²τ). For a metal, n is set by the chemistry and barely changes with T. But τ — the average time between collisions — shrinks as ions vibrate harder, knocking electrons more often. Less τ, more ρ, more R. For a semiconductor, n itself rises sharply with T (more electrons jump the band gap), and this overrides any τ decrease, so ρ falls.
NEET tests this constantly — directly as in the platinum-wire problem (NEET 2023 Q.41), as a conceptual contrast (NEET 2022 Q.9), and as graph-identification (NEET 2020 Q.133). The numerical version usually requires plugging into RT = R0(1 + αΔT) and solving for one unknown.
Electrical energy and power
When a charge ΔQ moves through a potential difference V, the source loses energy ΔU = V·ΔQ to the conductor. If charges moved freely, they would accelerate and gain kinetic energy. But collisions inside the conductor transfer this energy to the lattice ions, which vibrate harder — the conductor warms up. In time Δt, the heat dissipated is
P = V I = I²R = V²/R
Power dissipated in a resistor — three equivalent forms
This is Joule heating, or ohmic loss. The same formula heats the filament of an incandescent bulb to incandescence and warms the coil of an electric iron. Which form of the formula to use depends on what you know: I²R when current is fixed (series), V²/R when voltage is fixed (parallel).
One application reshapes the entire electric grid: power transmission. To deliver power P to a distant city over cables of resistance Rc, the wasted power in the cables is Pc = I²Rc = (P/V)²·Rc, inversely proportional to V². That is why transmission is done at hundreds of kilovolts and stepped down by transformers near the consumer. A factor-of-10 voltage rise cuts cable losses by a factor of 100.
Cells, EMF, and internal resistance
A cell maintains a steady current by converting chemical energy into electrical energy. It has two electrodes — a positive terminal P and a negative terminal N — dipped in an electrolyte. The electromotive force (EMF) ε is the potential difference between P and N when no current is being drawn — that is, in open circuit. Despite the misleading name "force," EMF is just the work done per unit charge by the source in transporting charge from N to P inside the cell. Its unit is the volt.
Once a current I is drawn through an external resistor R, the cell's internal electrolyte (with resistance r) drops the voltage as the current passes through it. The voltage available at the terminals — the terminal voltage V — is therefore
Connecting Kirchhoff's loop rule around the simple loop (cell + R) gives IR = ε − Ir, hence I = ε/(R + r). Internal resistance is therefore not a side detail — it sets the maximum current a cell can deliver, the energy wasted inside the cell, and the gap between the rated EMF and the voltage you actually see at the terminals.
Cells in series and parallel
Just as resistors can be combined, so can cells. When you stack two cells in series, joining the positive of one to the negative of the next, the EMFs and internal resistances simply add (assuming both cells push current in the same direction):
Now turn to resistor combinations themselves — the bread-and-butter of NEET numericals.
The classic NEET 2023 Q.39 exploits this elegantly: ten equal resistors in series give Rs = 10R; the same ten in parallel give Rp = R/10. For a battery of EMF E with negligible internal resistance, the parallel current is (E/(R/10)) / (E/10R) = 100 times the series current. NEET 2021 Q.21 runs the same logic in reverse — four wires in parallel give R/4 = 0.25 Ω, so R = 1 Ω and the series total is 4 Ω.
Kirchhoff's rules
Once a circuit grows beyond a simple loop — multiple cells, multiple branches, bridges, ladders — series-parallel reduction is no longer enough. The two rules of Gustav Robert Kirchhoff (1845) handle any circuit:
The junction rule says no charge piles up at any junction in steady state — what enters must leave. The loop rule says that walking once around a closed loop must bring you back to the same potential — the electrostatic field is conservative. Together, these are not new physics — they are charge conservation and energy conservation in disguise. But operationally, they let you solve any DC circuit by writing one equation per unknown current.
A four-step recipe for Kirchhoff loop analysis
The NEET 2023 Q.7 question — two cells of 10 V and 5 V opposing in a loop with two 5 Ω resistors — yielded a clean 0.5 A from A to B, derived by one application of the loop rule: (10 − 5)/(5 + 5) = 0.5 A. The trickier 2018 Q.13 problem (n cells in series, short-circuited) collapses to I = nε/(nr) = ε/r — independent of n, because both the EMF and the internal resistance scale together.
The Wheatstone bridge
The cleanest application of Kirchhoff's rules — and one of the most surveyed-for in NEET — is the Wheatstone bridge. Four resistors R₁, R₂, R₃, R₄ form a diamond; a cell drives current across one diagonal (A to C); a galvanometer detects current across the other diagonal (B to D). The bridge is said to be balanced when no current flows through the galvanometer (Ig = 0).
Derivation is a textbook two-loop application of Kirchhoff's rules. With Ig = 0, the junction rule gives I₁ = I₃ and I₂ = I₄. Apply the loop rule to ADBA: −I₁R₁ + I₂R₂ = 0, so I₁R₁ = I₂R₂. Apply the loop rule to CBDC: I₂R₄ − I₁R₃ = 0, so I₂R₄ = I₁R₃. Divide one by the other: R₁/R₃ = R₂/R₄, equivalent to P/Q = R/S.
The practical version of this circuit is the metre bridge: a one-metre uniform wire of constant resistance per unit length forms two of the four arms. Sliding a contact along the wire varies P/Q until the galvanometer reads zero — the balance length gives the unknown resistance directly. NEET 2020 Q.120 was a metre-bridge calculation (3:2 ratio, 10 Ω in one gap → unknown = 15 Ω over 1.5 m, giving 0.1 m per ohm).
A close cousin of the metre bridge is the potentiometer — a uniform wire across which a steady current produces a uniform potential gradient. At balance, no current flows through the unknown cell, so its terminal voltage equals its EMF. The potentiometer is therefore the only practical instrument that measures EMF directly, not terminal voltage. NEET 2016, 2017, and 2021 each asked a potentiometer numerical (comparing two EMFs from balance lengths, or finding the balance length given a known EMF).
NEET PYQ Snapshot
Five real NEET previous-year questions — solve before moving on.
Ten resistors, each of resistance R, are connected in series to a battery of EMF E and negligible internal resistance. They are then reconnected in parallel to the same battery. The current is increased n times. Find n.
Answer: (3) 100Why: Series: Req = 10R, Is = E/10R. Parallel: Req = R/10, Ip = 10E/R = 100 × Is. So n = 100. Tests resistor combination + Ohm's law in one breath.
The resistance of a platinum wire at 0 °C is 2 Ω and at 80 °C is 6.8 Ω. The temperature coefficient of resistance of the wire is —
Answer: (4) 3 × 10⁻² °C⁻¹Why: R = R₀(1 + αT). 6.8 = 2(1 + α·80) ⇒ 6.8/2 − 1 = 80α ⇒ 2.4 = 80α ⇒ α = 0.03 = 3 × 10⁻² °C⁻¹. Note: α positive — platinum is a metal.
As temperature increases, the electrical resistance —
Answer: (2)Why: In a conductor, n is fixed and τ falls with T → ρ rises → R rises. In a semiconductor, n rises sharply with T (band-gap excitation), overriding τ → ρ falls → R falls. Direct conceptual contrast.
Match Column-I (physical quantity) with Column-II (mathematical relation): (A) Drift velocity (B) Resistivity (C) Relaxation period (D) Current density. Options: (P) m/(ne²ρ), (Q) nevd, (R) eEτ/m, (S) E/J.
Answer: (2) A–R, B–S, C–P, D–QWhy: vd = eEτ/m (R). ρ = E/J from j = σE (S). τ = m/(ne²ρ) by inverting ρ = m/(ne²τ) (P). J = I/A = nevd (Q). Direct match-the-formula on the master derivation.
Two resistors of 100 Ω and 200 Ω are connected in parallel. The ratio of thermal energy developed in 100 Ω to that in 200 Ω in a given time is —
Answer: (1) 2 : 1Why: In parallel, same V across both. P = V²/R, so P ∝ 1/R. P100/P200 = 200/100 = 2:1. The smaller R dissipates more heat. (Trap: in series, the ratio would invert to 1:2 because P = I²R.)
Expert FAQs
Questions NEET has asked from this chapter, answered straight.
What is the SI unit of electric current and how is it defined?
Why does the drift velocity of electrons remain steady despite a constant electric field accelerating them?
How is resistivity related to the microscopic properties of the conductor?
Why does resistance of metals increase with temperature while that of semiconductors decreases?
What is the difference between EMF and terminal voltage of a cell?
What does Kirchhoff's junction rule conserve, and what does the loop rule conserve?
What is the balance condition of a Wheatstone bridge?
Why is electric power transmitted at very high voltages over long distances?
Go Deeper
Drill into the subtopics that NEET asks most often.