Current Electricity
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Direct answer
Electrons drift through copper at under a millimetre per second, yet the lamp lights the instant you flip the switch, because the electric field — not the electrons — propagates at near light speed; current is I = nAev_d. Ohm's law V = IR with R = ρl/A then governs the steady circuit, Kirchhoff's two rules extend it to networks, and the exam concentrates its marks on emf versus terminal voltage, the Wheatstone bridge and the potentiometer.
What you must remember
- Drift velocity v_d = I/(nAe); for a copper wire of 1 mm^2 carrying 1 A it is of order 10^-4 m/s — the switch responds to the field, not to the electron commute.
- Resistance R = ρl/A: longer and thinner wires resist more; resistivity ρ is the material's property and rises with temperature in metals, while semiconductors fall.
- Ohm's law is a material behaviour, not a law of nature; it fails for diodes, electrolytes and at extreme conditions — those are non-ohmic.
- A real cell: terminal voltage V = ε - Ir while discharging, V = ε when the circuit is open, and V = ε + Ir while being charged; internal resistance wastes power I^2r.
- Kirchhoff's junction rule (ΣI = 0) expresses charge conservation and the loop rule (ΣΔV = 0) energy conservation; every network problem is solved by writing both honestly.
- Wheatstone bridge balances when P/Q = R/S; at balance the galvanometer current is zero and the bridge is insensitive to the galvanometer's own resistance — the metre bridge measures unknown resistance from a balance length.
- The potentiometer compares emfs as ε1/ε2 = l1/l2 at null — no current drawn at balance, which is why it beats a voltmeter, which always draws some.
- Electrical power P = VI = I^2R = V^2/R; the commercial energy unit is the kilowatt-hour, 1 kWh = 3.6 × 10^6 J.
Inside a cell under load
Connect a 6 V cell of internal resistance 2 Ω across a 4 Ω resistor. The circuit current is I = ε/(R + r) = 6/6 = 1 A. The terminal voltage is what the external resistor actually receives: V = IR = 4 V, confirmed by the bookkeeping V = ε - Ir = 6 - 2 = 4 V. Of the 6 W the cell generates, 4 W reaches the load and 2 W warms the cell itself — an efficiency of R/(R + r) = 2/3. Two conclusions generalise. First, terminal voltage equals emf only when I = 0, which is why a battery tester reads full voltage until loaded. Second, the load receives maximum power exactly when R = r (here 2 Ω, giving 4.5 W); beyond that matching point efficiency keeps improving while delivered power declines — the distinction between maximum power transfer and maximum efficiency that question setters enjoy probing.
Where students slip
The ammeter is a low-resistance device inserted in series (a parallel shunt protects its galvanometer), while the voltmeter is a high-resistance device placed in parallel; swapping them — or asking what an ammeter across a battery would read — is a standard trap, since an ammeter across a battery is effectively a short circuit. Second, series and parallel outcomes: students mix up "resistances add" with "conductances add"; series resistances always exceed the largest, parallel always dips below the smallest. Third, temperature: a metal's resistance rises when heated, a semiconductor's falls because carrier numbers explode — the filament lamp's non-ohmic curve in NCERT is exactly this. Fourth, in potentiometer problems the driver cell must exceed the tested emf or no null exists; and at the null the tested cell delivers nothing, which is the entire point of the null method. Finally, drift-velocity questions punish the assumption that electrons race around the circuit carrying energy like couriers; the energy travels in the field.
Frequently asked questions
Why does a lamp glow immediately although drift velocity is tiny?
The electric field establishing itself around the circuit travels at nearly the speed of light; electrons everywhere in the filament start drifting almost at once.
When does terminal voltage equal the emf of a cell?
Only on open circuit, when I = 0. Under load V = ε - Ir, so the terminal voltage always falls below the emf while the cell delivers current.
What is the balance condition of a Wheatstone bridge?
P/Q = R/S; the galvanometer then carries no current, making the measurement independent of the galvanometer's resistance and of the cell's emf.
Why does a potentiometer measure emf more accurately than a voltmeter?
At the null point the potentiometer draws no current from the cell under test, so it reads the true emf; a voltmeter always draws some current and shows only the reduced terminal voltage.
How do metals and semiconductors respond to heating?
Metal resistivity rises (lattice vibrations scatter electrons harder); semiconductor resistivity falls, because thermal energy frees far more charge carriers than the extra scattering removes.